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40. Prove: If $A D, B E$ and $C F$ are the angle bisectors of $\triangle A B C$, then the area of $\triangle D E F$ does not exceed one-fourth of the area of $\triangle A B C$. (1981 GDR Mathematical Olympiad) | 40. Let $a=BC, b=CA, c=AB. S=S_{\triangle ABC}, S_{0}=S_{\triangle DEF}$, then by the property of angle bisectors we have
$$\frac{AF}{b}=\frac{BF}{a}=\frac{AF+BF}{b+a}=\frac{c}{a+b}$$
Thus, $AF=\frac{bc}{a+b}$, similarly, $AE=\frac{bc}{a+c}$, therefore,
$$\begin{array}{l}
S_{\triangle AEF}=\frac{1}{2} AE \cdot AF \sin... | S_{0} \leqslant \frac{1}{4} S | Geometry | proof | Yes | Yes | inequalities | false | 733,630 |
41. Through a point inside $\triangle ABC$, draw lines parallel to the three sides (as shown in the figure), $DE \parallel BC, FG \parallel CA, HI \parallel AB$, points $D, E, F, G, H, I$ are all on the sides of $\triangle ABC$, $S_{1}$ represents the area of hexagon DGHEFI, $S_{2}$ represents the area of $\triangle AB... | 41. To prove $S_{1} \geqslant \frac{2}{3} S_{2}$, it suffices to prove that $S_{\triangle A G H}+S_{\triangle A C H}+ S_{\triangle A G H} \leqslant \frac{1}{3} S_{2}$. Noting the parallelograms $A G O H, B I O D$, and $C E O F$, the solution to the proposition lies in proving
$$S_{\triangle O I F}+S_{\triangle O E H}+S... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,631 |
40. Given that $a, b, c$ are positive numbers, and $a+b+c=1$, prove: $\frac{1}{b c+a+\frac{1}{a}}+\frac{1}{c a+b+\frac{1}{b}}+$ $\frac{1}{a b+c+\frac{1}{c}} \leqslant \frac{27}{31} \cdot$ (2008 Serbian Mathematical Olympiad Problem) | 40. The original inequality is equivalent to $\frac{a}{a^{2}+p}+\frac{b}{b^{2}+p}+\frac{c}{c^{2}+p} \leqslant \frac{27}{31}(p=a b c+1)$.
Consider the function
$$f(x)=\frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}+3 x}-\frac{a}{a^{2}+x}-\frac{b}{b^{2}+x}-\frac{c}{c^{2}+x}$$
First, prove that for all $x \geqslant a b+b c+c a$, we h... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,633 |
42. As shown in the figure, line $l$ intersects side $AB$ of $\triangle ABC$ at $B_{1}$, and side $AC$ at $C_{1}$. The centroid $G$ of $\triangle ABC$ and point $A$ are on the same side of $l$ (including $G$ on $l$). Prove that $S_{B B_{1} G C_{1}} + S_{C C_{1} C B_{1}} \geqslant \frac{4}{9} S_{\triangle A B C}$. When ... | 42. Pay attention to the special position of point $G$ as the centroid of $\triangle ABC$. Take the midpoint $D$ of $BC$, and connect $DB_1$, $DC_1$, then
$$\begin{aligned}
S_{B B_{1} G C_{1}}+S_{C C_{1} G B_{1}}= & 2 S_{\triangle G B_{1} C_{1}}+S_{\triangle B C_{1} B_{1}}+S_{\triangle C B_{1} C_{1}}= \\
& 2 S_{\triang... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,634 |
43. Given that the four vertices of quadrilateral $P_{1} P_{2} P_{3} P_{4}$ lie on the sides of $\triangle A B C$. Prove that among the four triangles $\triangle P_{1} P_{2} P_{3}$, $\triangle P_{1} P_{2} P_{4}$, $\triangle P_{1} P_{3} P_{4}$, and $\triangle P_{2} P_{3} P_{4}$, at least one has an area no greater than ... | 43. Proof There are two cases: (1) four vertices on two sides; (2) four vertices on three sides.
(1) Without loss of generality, let $P_{1}, P_{4}$ be on $A B$, and $P_{2}, P_{3}$ be on $A C$, with $P_{1}, P_{2}$ on $A P_{4}, A P_{3}$ respectively. By moving $B$ to $P_{4}$ and $C$ to $P_{3}$, the area of triangle $A B ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,635 |
44. $M$ is a point inside $\triangle ABC$ satisfying $\angle AMC=90^{\circ}$, $\angle AMB=120^{\circ}$, $\angle BMC=150^{\circ}$, and let $P, Q, R$ be the circumcenters of $\triangle AMC$, $\triangle AMB$, $\triangle BMC$ respectively. Prove that $S_{\triangle PQR} > S_{\triangle ABC}$. | 44. As shown in the figure, let the midpoints of $A M, B M, C M$ be $U$, $V, W$ respectively. It is easy to see that $P Q \perp A M$ at $U, Q R \perp B M$ at $V, R P \perp C M$ at $W$. Let $M U=a, M V=b, M W=c$, then
$$\begin{array}{l}
S_{\triangle A B C}=4 S_{\triangle U V W}=4 \times\left(\frac{1}{2} a b \times \frac... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,636 |
45. Let $E$ be the intersection of the diagonals of convex quadrilateral $ABCD$, and let $F_{1}$, $F_{2}$, and $F_{3}$ be the areas of $\triangle ABE$, $\triangle CDE$, and quadrilateral $ABCD$, respectively. Prove that $\sqrt{F_{1}} + \sqrt{F_{2}} \leq \sqrt{F}$, and determine when equality holds. (2004 | 45. Let $a, b, c, d$ be the lengths of $E A, E B, E C, E D$ respectively, then
$$F_{1}=S_{\triangle A B E}=\frac{a}{a+c} S_{\triangle A B C}=\frac{a}{a+c} \cdot \frac{b}{b+d} \cdot S_{A B C D}$$
Similarly,
$$F_{2}=S_{\triangle C D E}=\frac{c}{a+c} \cdot \frac{d}{b+d} \cdot S_{A B C D}$$
Therefore,
$$\begin{array}{l}
... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,637 |
46. Let the circle $C_{a}$ be the circle that is internally tangent to the circumcircle of $\triangle A B C$ and tangent to the sides $A B, A C$. Denote $r_{a}$ as the radius of circle $C_{a}$, and similarly define $r_{b}, r_{c} ; r$ is the radius of the incircle of $\triangle A B C$. Prove: $r_{a}+r_{b}+r_{c} \geqslan... | 46. Let $C_{a}$ be tangent to $AB, AC, \triangle ABC$'s circumcircle at $D, E, F$, respectively. Let $M, N$ be the midpoints of $\overparen{AB}, \overparen{AC}$, and $I$ be the incenter of $\triangle ABC$. At this time, $F$ is the homothetic center of $C_{a}$ and $\triangle ABC$'s circumcircle, and the tangent through ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,638 |
48. Let the lengths of the three sides of a triangle be $a, b, c$, the lengths of the medians to these sides be $m_{a}$, $m_{b}$, $m_{c}$, and the diameter of the circumcircle be $D$. Prove: $\frac{a^{2}+b^{2}}{m_{c}}+\frac{b^{2}+c^{2}}{m_{a}}+\frac{c^{2}+a^{2}}{m_{b}} \leqslant 6 D$. (1994, No. 20 | 48. Extend the medians, intersecting the circumcircle of $\triangle ABC$ at $A_{1}, B_{1}$, and $C_{1}$. Clearly, $A A_{1} \leqslant D, B B_{1} \leqslant D, C C_{1} \leqslant D$,
i.e.,
$$m_{a}+A_{1} A_{2} \leqslant D, m_{b}+B_{1} B_{2} \leqslant D, m_{c}+C_{1} C_{2} \leqslant D$$
By the intersecting chords theorem, $... | \frac{a^{2}+b^{2}}{m_{c}}+\frac{b^{2}+c^{2}}{m_{a}}+\frac{c^{2}+a^{2}}{m_{b}} \leqslant 6 D | Inequalities | proof | Yes | Yes | inequalities | false | 733,640 |
49. On the sides $AB, BC, CA$ of $\triangle ABC$, take points $M, K, L$ (not coinciding with the vertices of $\triangle ABC$). Prove that at least one of the triangles $\triangle MAL, \triangle KBM, \triangle LCK$ has an area not greater than one-fourth of the area of $\triangle ABC$. (8th IMO Problem) | 49. Since the ratio of the areas of two triangles with one equal angle is equal to the ratio of the products of the sides enclosing this angle, we have
$$\begin{array}{l}
\frac{S_{\triangle K B M}}{S_{\triangle A B C}}=\frac{B K \cdot B M}{A B \cdot B C} \\
\frac{S_{\triangle M A L}}{S_{\triangle A B C}}=\frac{A M \cdo... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,641 |
50. Given that $I$ is the incenter of $\triangle ABC$ and $R$ is the circumradius of $\triangle ABC$, prove: $AI + BI + CI \leq 3R$. (2007 Moldova Training Team Problem) | 50. It is easy to get $r=A I \sin \frac{A}{2}=B I \sin \frac{B}{2}=C I \sin \frac{C}{2}$, and $r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$, so
$$A I+B I+C I=4 R\left(\sin \frac{B}{2} \sin \frac{C}{2}+\sin \frac{C}{2} \sin \frac{A}{2}+\sin \frac{A}{2} \sin \frac{B}{2}\right)$$
Thus, it is sufficient to pr... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,642 |
51. Prove that if the cyclic hexagon $A B C D E F$ satisfies $A B=B C, C D=D E, E F=F A$, then the area of $\triangle A C E$ does not exceed the area of $\triangle B D F$. (1974 Czechoslovak Mathematical Olympiad) | 51. Let $O$ be the center of the circumcircle of hexagon $A B C D E F$, with radius $R$, and let $\alpha=\angle C A E$, $\beta=\angle A E C$, $\gamma=\angle A C E$. From the given conditions about the sides, we have
$$\begin{array}{l}
\angle A O B=\angle B O C=\beta \\
\angle C O D=\angle D O E=\alpha \\
\angle E O F=\... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,643 |
53. A convex quadrilateral with area $S$ is inscribed in a circle, with the circle's center inside the quadrilateral. Prove: the area of the quadrilateral formed by the projections of the diagonals onto the sides does not exceed $\frac{S}{2}$. (1999 Bulgaria | 53. As shown in the figure, $O$ is the intersection of the diagonals of the cyclic quadrilateral $ABCD$, and its perpendicular feet to the four sides are $P, Q, R, S$, respectively. Therefore, $\angle 2=\angle 1=\angle 4=\angle 3$, so $\theta P$ bisects $\angle SPQ$.
Similarly, it can be proven that: $OQ, OR=OS$ bisec... | S_{PQRS} \leqslant \frac{S}{2} | Geometry | proof | Yes | Yes | inequalities | false | 733,646 |
54. Let $a, b, c, t_{a}, t_{b}, t_{c}$ be the side lengths and the lengths of the angle bisectors of $\triangle ABC$. Prove that $\frac{1}{t_{a} t_{b}}+\frac{1}{t_{b} t_{c}}+$ $\frac{1}{t_{c} t_{a}} \geqslant \frac{4}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}$. (2002 German National Team Problem: The perim... | 54. From the formula for the length of the angle bisector of a triangle $t_{a}=\frac{2 b c}{b+c} \cos \frac{A}{2}=\frac{2 b c}{b+c} \sqrt{\frac{1+\cos A}{2}}=\frac{2 b c}{b+c} \sqrt{\frac{(b+c)^{2}-a^{2}}{4 b c}}=\frac{2 \sqrt{b c p(p-a)}}{b+c}$, where $2 p=a+b+c$, similarly $t_{b}=\frac{2 \sqrt{c a p(p-b)}}{c+a}$, $t_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,647 |
55. Quadrilateral $ABCD$ is inscribed in a circle with radius $R$, and the lengths of the sides of the quadrilateral are $a$, $b$, $c$, and $d$ respectively. Prove that $16 R^{2} S^{2}=(a b+b c)(a c+b d)(a d+b c)$, and use this to prove the inequality: $\sqrt{2} R S \geqslant \sqrt[4]{(a b c d)^{3}} \cdot(2000$ Irish M... | 55. As shown in the figure, let $A B=a, B C=b, C D=c, D A=d$, $\angle B A C=\alpha, \angle A B C=\beta$, then $\angle B C D=\pi-\alpha$, $\angle C D A=\pi-\beta, S=\frac{1}{2} A B \cdot A D \sin \angle B A C+\frac{1}{2} B C$ $C D \sin \angle B C D=\frac{1}{2}(a d+b c) \sin \alpha$, similarly, $S=$ $\frac{1}{2}(a b+c d)... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,648 |
56. Given that $a, b, c$ are the three sides of a $\triangle ABC$, let $x=a+\frac{b}{2}, y=b+\frac{c}{2}, z=c+\frac{a}{2}$. Prove that a $\triangle XYZ$ can be constructed with $x, y, z$ as its sides, and that the area of $\triangle XYZ$ is not less than $\frac{9}{4}$ times the area of $\triangle ABC$ (2003 Indian Math... | 56. It is known that $x+y>z, y+z>x, x+z>y$. And
$$\begin{array}{c}
y+z-x=\left(b+\frac{c}{2}\right)+\left(c+\frac{a}{2}\right)-\left(a+\frac{b}{2}\right)= \\
\frac{1}{2}((b+c-a)+(b+c-a)+(c+a-b)) \geqslant \\
\frac{3}{2} \sqrt[3]{(b+c-a)^{2}(c+a-b)} \\
x+z-y \geqslant \frac{3}{2} \sqrt[3]{(a+b-c)(c+a-b)^{2}} \\
x+y-z \g... | S^{\prime} \geqslant \frac{9}{4} S | Geometry | proof | Yes | Yes | inequalities | false | 733,649 |
57. Let $D, E, F$ be points on the sides $BC, CA, AB$ of $\triangle ABC$, respectively. Let $\alpha, \beta, \gamma, \delta$ be the areas of $\triangle AEF, \triangle BFD, \triangle CDE$, and $\triangle DEF$, respectively. Prove that: $\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha} \geqslant \frac... | 57. Let's assume the area of $\triangle ABC$ is 1, and $x=\frac{AF}{AB}, y=\frac{BD}{BC}, z=\frac{CE}{CA}$, then
$$\alpha=\frac{S_{\triangle AEF}}{S_{\triangle ABC}}=\frac{\frac{1}{2} AE \cdot AF \sin \angle EAF}{\frac{1}{2} AB \cdot AC \sin \angle BAC}=\frac{AE \cdot AF}{AB \cdot AC}=x(1-z)$$
Similarly, we have $\bet... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,650 |
58. On the sides of triangle $ABC$, construct squares $ABHI$, $BCDE$, and $CAFG$ outwardly. Let $XYZ$ be the triangle formed by the segments $EF$, $DI$, and $GH$. Prove that $S_{\triangle XYZ} \leqslant (4-2\sqrt{3}) S_{\triangle ABC}$. (2003 | 58. As shown in the figure, draw auxiliary lines, we have $A H=$
$$\begin{array}{c}
B I=\sqrt{2} c, A G=C F=\sqrt{2} b, B D=C E= \\
\sqrt{2} a, (A B=c, B C=a, C A=b), \text { and } \angle H A G \\
=90^{\circ}+A, \text { hence } \\
H G=\sqrt{2} \sqrt{b^{2}+c^{2}+2 b c \sin A} \\
\sin \alpha=\frac{b \cos A}{\sqrt{b^{2}+c... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,651 |
61. In an acute triangle $\triangle ABC$, $H$ is its orthocenter, $L, M, N$ are the midpoints of sides $AB, BC, CA$, respectively. Prove that $H L^{2}+H M^{2}+H N^{2}<A L^{2}+B M^{2}+C N^{2}$. (2006 Italian National Training Team Selection Test) | 61. It is known that $A H=2 R \cos A, B H=2 R \cos B, C H=2 R \cos C$.
From the median length formula in a triangle, we get
$$\begin{aligned}
4 H M^{2}= & 2\left(H B^{2}+H C^{2}\right)-B C^{2}= \\
& 8 R^{2}\left(\cos ^{2} B+\cos ^{2} C\right)-B C^{2}= \\
& 8 R^{2}\left(\cos ^{2} B+\cos ^{2} C\right)-a^{2}= \\
& 8 R^{2... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,654 |
63. $\triangle A B C$ is an isosceles triangle, $A B=A C$, a line $l$ is drawn through $A$ parallel to $B C$, points $P, Q$ are on the perpendicular bisectors of $A B, A C$ respectively, and satisfy $P Q \perp B C, M, N$ are points on line $l$ such that $A P \perp P M, A Q \perp Q N$. Prove: $\frac{1}{A M}+\frac{1}{A N... | 63. As shown in the figure,
$$\begin{array}{c}
A N=\frac{A Q}{\cos \angle Q A N} \\
A M=\frac{A P}{\cos \angle P A M} \\
A Q=\frac{A C}{2 \cos \angle Q A C}=\frac{A B}{2 \cos \angle Q A C} \\
A P=\frac{A B}{2 \cos \angle P A B}
\end{array}$$
Therefore,
$$\begin{aligned}
A M & =\frac{A B}{2 \cos \angle P A M \cos \angl... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,657 |
65. Find the maximum value of the real number $\lambda$ such that point $P$ is inside the acute $\triangle ABC$, $\angle PAB = \angle PBC = \angle PCA$, and the rays $AP$, $BP$, $CP$ intersect the circumcircles of $\triangle PBC$, $\triangle PCA$, $\triangle PAB$ at $A_1$, $B_1$, $C_1$, respectively, so that $S_{\trian... | 65. Let $A A_{1} \cap B C=A_{2}$. From $\angle P A B=\angle P B C=\angle P A_{1} C$, we know $A B \parallel A_{1} C$. Therefore, $\triangle A_{1} A_{2} C \sim \triangle A A_{2} B$
$$\frac{S_{\triangle A_{1} B C}}{S_{\triangle A B C}}=\frac{A A_{2}}{A A_{2}}=\frac{C A_{2}}{B A_{2}}$$
Similarly, let $B_{2}=B B_{1} \cap ... | 3 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,659 |
66. Let the sides of $\triangle ABC$ be $a, b, c$, and the corresponding medians be $m_{a}, m_{b}, m_{c}$, and the angle bisectors be $w_{a}, w_{b}, w_{c}$. Suppose $w_{a} \cap m_{b}=P, w_{b} \cap m_{c}=Q, w_{c} \cap m_{a}=R$. Let the area of $\triangle PQR$ be $\delta$, and the area of $\triangle ABC$ be $F$. Find the... | 66. Let $a \geqslant b \geqslant c$, and let $G$ be the centroid of $\triangle ABC$. Then it is easy to see that $P$ lies on $BG$. Let $D$ be the midpoint of $BC$, and let $AP$ intersect $BC$ at $E$. By the Angle Bisector Theorem, we have $\frac{BE}{ED}=\frac{2c}{b-c}$.
Applying Menelaus' Theorem to $\triangle GBD$ an... | \frac{1}{6} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,660 |
67. Given a square $ABCD$ in the plane, find the minimum value of the ratio $\frac{OA+OC}{OB+OD}$. Here, $O$ is any point in the plane. (1993 St. Petersburg City Mathematical Selection Exam Problem) | 67. First, prove $\frac{O A+O C}{O B} \geqslant \frac{1}{\sqrt{2}}$. Squaring both sides of the inequality and eliminating the denominator, we get
$$2\left(O A^{2}+O C^{2}+2 O A \cdot O C\right) \geqslant O B^{2} \mp O D^{2}+2 O B \cdot O D$$
Since $O A^{2}+O C^{2}=O B^{2}+O D^{2}$, the above inequality simplifies to
... | \frac{1}{\sqrt{2}} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,661 |
68. Let $AB, CD$ be two perpendicular chords of a circle with center $O$ and radius $r$. The circle is divided into four parts by these chords, which are labeled in clockwise order as $X, Y, Z, W$. Find the maximum and minimum values of $\frac{A(X)+A(Z)}{A(Y)+A(W)}$, where $A(U)$ denotes the area of $U$. (1988 IMO Shor... | 68. Let's assume the center of the circle falls in the $Z$ as shown in figure (1). Then, as the chord $AB$ moves parallel, the shaded area in figure (2) is greater than the area without shading on its left, so $A(X)+A(Z)$ increases, while $A(Y)+A(W)$ decreases (note that the sum of the areas of $X, Y, Z, W$ is a consta... | \frac{\pi+2}{\pi-2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,662 |
69. The area of $\triangle A B C$ is $1, D, E$ are points on sides $A B, A C$ respectively, $B E, C D$ intersect at point $P$, and the area of quadrilateral $B C E D$ is twice the area of $\triangle P B C$. Find the maximum value of the area of $\triangle P D E$.
---
The area of $\triangle A B C$ is $1, D, E$ are poi... | 69.
$$\begin{array}{l}
\text { Let } \frac{A D}{A B}=x, \frac{A E}{A C}=y, \text { so } S_{\triangle A D E}=x y ; S_{B C E D}=1-x y ; S_{\triangle P B C}=\frac{1}{2}(1-x y) \\
\frac{S_{\triangle P D E}}{S_{\triangle P B C}}=\frac{P E \cdot P D}{P B \cdot P C}=\left(\frac{P E}{P B}\right) \cdot\left(\frac{P D}{P C}\righ... | 5 \sqrt{2}-7 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,663 |
70. Given $\triangle A B C$, find a point $P$ such that $A P+B P+C P$ is minimized. | 70. It is evident that $P$ cannot be outside $\triangle ABC$.
(1) If each interior angle is less than $120^{\circ}$, rotate $\triangle ABP$ counterclockwise by $60^{\circ}$ around point $B$ to the position of $\triangle A_{1}BQ$, as shown in Figure 70 (1). $\triangle BPQ$ is an equilateral triangle. Thus, $AP + BP + CP... | proof | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,664 |
71. In $\triangle A B C$, prove that $\frac{1}{\sin \frac{A}{2}}+\frac{1}{\sin \frac{B}{2}}+\frac{1}{\sin \frac{C}{2}} \geqslant \frac{1}{r} \cdot \sqrt{\frac{a^{2}+b^{2}+c^{2}+4 \sqrt{3} S}{2}}$ (where $r$ is the inradius of $\triangle A B C$). (2006 USA National Training Team Problem) | $$\begin{array}{l}
\text { 71. The inequality } \frac{1}{\sin \frac{A}{2}}+\frac{k}{\sin \frac{B}{2}}+\frac{1}{\sin \frac{C}{2}} \geqslant \frac{1}{r} \cdot \sqrt{\frac{a^{2}+b^{2}+c^{2}+4 \sqrt{3} S}{2}} \text { is equivalent to } \\
\frac{r}{\sin \frac{A}{2}}+\frac{r}{\sin \frac{B}{2}}+\frac{r}{\sin \frac{C}{2}} \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,665 |
72. In a convex quadrilateral $ABCD$, $P, Q, R, S$ are the midpoints of sides $BC, CD, DA, AB$, respectively. Prove: $4\left(A P^{2}+B Q^{2}+C R^{2}+D S^{2}\right) \leqslant 5\left(A B^{2}+B C^{2}+C D^{2}+D A^{2}\right) \cdot(2000$ Greece National | 72. In quadrilateral $ABCD$, $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}=\overrightarrow{0}$, so $\overrightarrow{AB}+\overrightarrow{CD}=-(\overrightarrow{BC}+\overrightarrow{DA})$. Also, $\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+\frac{1}{2} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,667 |
74. In $\triangle A B C$, define $l_{a}$ as the length of the line segment connecting the feet of the perpendiculars from the intersection of the angle bisector of $\angle A$ with $B C$ to the sides $A B$ and $A C$. Similarly, define $l_{b}, l_{c}$. Prove that: $\frac{l_{a} l_{b} l_{c}}{l^{3}} \leqslant \frac{1}{64}$, ... | 74. Let $a, b, c$ represent the side lengths opposite to vertices $A, B, C$ of $\triangle ABC$, respectively. Let $D$ be the intersection of $BC$ and the angle bisector of $\angle A$, and let $p = BD, q = CD$.
By the Angle Bisector Theorem, we have $bp = cq$. Combining this with $p + q = a$, we get
$$p = \frac{ac}{b+c... | \frac{l_a l_b l_c}{l^3} \leqslant \frac{1}{64} | Geometry | proof | Yes | Yes | inequalities | false | 733,669 |
76. In an acute triangle $\triangle ABC$, $A_{1}, B_{1}, C_{1}$ are the midpoints of $BC, CA, AB$ respectively, and $O$ is the center of the circumcircle. If the circumradius is 1, prove that $\frac{1}{O A_{1}}+\frac{1}{O B_{1}}+\frac{1}{O C_{1}} \geqslant 6$. (2007 Croatian Mathematical Olympiad) | 76. As shown in the figure, let $\angle C A B=\alpha, \angle A B C=\beta, \angle A C B=\gamma$, then $\angle B O A_{1}=\alpha$, so $\cos \alpha=O A_{1}$, similarly, $\cos \beta=O B_{1}, \cos \gamma=O C_{1}$,
By the arithmetic harmonic mean inequality, we have
$$\begin{array}{l}
\frac{1}{O A_{1}}+\frac{1}{O B_{1}}+\fra... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,671 |
77. In $\triangle ABC$, the lengths of the three sides are $a, b, c$, and the lengths of the corresponding angle bisectors are $w_{a}, w_{b}, w_{c}$. The circumradius of $\triangle ABC$ is $R$. Prove: $\frac{a^{2}+b^{2}}{w_{c}}+\frac{b^{2}+c^{2}}{w_{a}}+\frac{c^{2}+a^{2}}{w_{b}}>4 R$. (2007 India | 77. Since $w_{a}=\frac{2 b c \cos \frac{A}{2}}{b+c}, w_{b}=\frac{2 c a \cos \frac{B}{2}}{c+a}, w_{c}=\frac{2 a b \cos \frac{C}{2}}{a+b}$, therefore $\frac{a^{2}+b^{2}}{w_{c}}+$
$$\begin{array}{l}
\frac{b^{2}+c^{2}}{w_{a}}+\frac{c^{2}+a^{2}}{w_{b}}>4 R \text { is equivalent to } \\
\frac{\left(b^{2}+c^{2}\right)(b+c)}{4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,672 |
80. Let the area of $\triangle ABC$ be $T$, and $a, b, c$ be the lengths of sides $AC, BC, AB$ respectively. Let $x, y, z$ be the lengths of the three medians from $\angle A, \angle B, \angle C$ respectively. Prove the inequality: $\frac{a^{2}}{x}+\frac{b^{2}}{y}+\frac{c^{2}}{z} \geqslant 4 \sqrt{T \sqrt{3}}$.
$(1998$ | 80. By the median length formula of a triangle, we have
\[ x=\frac{1}{2} \sqrt{2 b^{2}+2 c^{2}-a^{2}}, \quad y=\frac{1}{2} \sqrt{2 c^{2}+2 a^{2}-b^{2}}, \]
\[ z=\frac{1}{2} \sqrt{2 a^{2}+2 b^{2}-c^{2}} \]
By the AM-GM inequality, we get
\[ \begin{array}{l}
\frac{a^{2}}{x}=\frac{2 a^{2}}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}=\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,675 |
81. Let the circumcenter of $\triangle ABC$ be $O$; the radius be $R$, and the radii of the incircles of $\triangle OBC$, $\triangle OCA$, and $\triangle OAB$ be $r_{1}$, $r_{2}$, and $r_{3}$, respectively. Prove: $\frac{1}{r_{1}}+\frac{k}{r_{2}}+\frac{1}{r_{3}} \geqslant \frac{4 \sqrt{3}+6}{R}$ (1998 Hungarian Math | 81. As shown in the figure, $\angle B O C=2 A, \triangle A B C$ has an circumradius $R$. By the Law of Sines, we have $B C=2 R \sin A$. Since the inradius of $\triangle O B C$ is $r_{1}$, by the area relationship of the triangle, we have
$$\begin{array}{l}
S_{\triangle O B C}=\frac{1}{2} O B \cdot O C \sin \angle B O C... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,676 |
44. Given $n$ real numbers $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, define $M_{1}=\frac{1}{n} \sum_{i=1}^{n} a_{i}, M_{2}=$ $\frac{2}{n(n-1)} \sum_{1 \leqslant i \leqslant j \leqslant n} a_{i} a_{j}, Q=\sqrt{M_{1}^{2}-M_{2}}$, prove: $a_{1} \leqslant M_{1}-Q \leqslant M_{1}+Q \leqslant a_{n}$, equality... | 44. Since $a_{\mathrm{F}} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, we have $M_{1}-a_{1} \geqslant 0$, so
$$a_{1} \leqslant M_{1}-Q \Leftrightarrow M_{1}-a_{1} \geqslant Q \Leftrightarrow\left(M_{1}-a_{1}\right)^{2} \geqslant M_{1}^{2}-M_{2} \Leftrightarrow a_{1}^{2}-2 a_{1} M_{1}+M_{2} \geqslant 0$$
Since fo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,677 |
83. As shown in the figure, $ABCD$ is a circumscribed quadrilateral of a circle with radius $r$, the points of tangency on sides $AB, BC, ED$, and $DA$ are $E, F, G, H$ respectively. The radii of the incircles of $\triangle EBF, \triangle FCG, \triangle GDH$, and $\triangle HAE$ are $r_{1}, r_{2}, r_{3}$, and $r_{4}$ r... | 83. As shown in the figure, let $\angle B A D=2 \alpha, \angle A B C=2 \beta, \angle B C D=2 \gamma, \angle C D A=2 \delta$, then $\alpha + \beta + \gamma + \delta = \pi$. Since the inradius of the incircle of $\triangle H A E$ is $r_{4}$, it is easy to get $A H = A E = r \cot \alpha, E H = 2 r \cos \alpha$, in the iso... | 2(2 - \sqrt{2}) r | Geometry | proof | Yes | Yes | inequalities | false | 733,679 |
85. $D, E$ are points on the sides $A B, A C$ of $\triangle A B C$, $D E$ is tangent to the incircle of $\triangle A B C$, and $D E \parallel B C$, prove that $: D E \leqslant \frac{1}{8}(A B+B C+C A) \cdot$ (1999 Italian Mathematical Olympiad) | 85 Let $P, Q, R$ be the points where the incircle of $\triangle ABC$ touches the sides $BC, CA, AB$ respectively. Since $DE \parallel BC$, we have $\triangle ADE \sim \triangle ABC$, thus,
$$\frac{AD + DE + EA}{AB + BC + CA} = \frac{DE}{BC} = \frac{DE}{a}$$
Because
$$AD + DE + EA = AR + AQ = b + c - a$$
Therefore,
$$... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,681 |
86. $O$ is the circumcenter of an acute triangle $ABC$, and the lines $CO, AO$, and $BO$ intersect the circumcircles of $\triangle AOB, \triangle BOC$, and $\triangle COA$ at the second points $A_1, B_1$, and $C_1$ different from $O$, respectively. Prove that: $\frac{AA_1}{OA_1} + \frac{BB_1}{OB_1} + \frac{CC_1}{OC_1} ... | 86. As shown in the figure, let the circumradius of $\triangle ABC$ be $R$, and the circumradius of $\triangle BOC$ be $R'$. It is easy to see that $\angle BAO = \angle ABO = 90^\circ - C$, $\angle CAO = \angle ACO = 90^\circ - B$, $\angle OBC = \angle OCB = \angle CA_1O = 90^\circ - A$, $\angle OA_2C = \angle ABC + \a... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,682 |
87. For any three non-collinear points $X, Y, Z$, let $R_{X Y Z}$ be the radius of the circumcircle of $\triangle X Y Z$. Let the incenter of $\triangle A B C$ be $I$, prove the inequality: $\frac{1}{R_{A B I}}+\frac{1}{R_{B C I}}+\frac{1}{R_{C A I}} \leqslant \frac{1}{B I}+\frac{1}{A I}+\frac{1}{C I}$ (2008 Germany | 87. Let the circumradius of $\triangle ABC$ be $R$, and the inradius be $r$. It is known that $AI = \frac{r}{\sin \frac{A}{2}}$. In $\triangle ABC$, $BC = 2R \sin A$, $BC = 2R_{BCl} \sin \left(\pi - \frac{B+C}{2}\right) = 2R_{BCl} \cos \frac{A}{2}$, so $R_{BCl} = 2R \sin \frac{A}{2}$. Therefore, the original inequality... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,683 |
89. For $\triangle A B C$, the circumradius is $R$, the perimeter is $P$, and the area is $K$. Determine the maximum value of $\frac{K P}{R^{3}}$. | 89. By the Law of Sines, $P=a+b+c=2 R(\sin A+\sin B+\sin C), K=\frac{a b \sin C}{2}=2 R^{2} \sin A \sin B \sin C$.
Therefore, by the AM-GM inequality,
$$\begin{aligned}
\frac{K P}{R^{3}}= & 4 \sin A \sin B \sin C(\sin A+\sin B+\sin C) \leqslant \\
& \frac{4}{27}(\sin A+\sin B+\sin C)^{4}
\end{aligned}$$
By Jensen's i... | \frac{27}{4} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,685 |
90. Prove: $\sqrt{\frac{A B_{1}}{A B}}+\sqrt{\frac{B C_{1}}{B C}}+\sqrt{\frac{C A_{1}}{C A}} \leqslant \frac{3}{\sqrt{2}}$, where $A_{1}, B_{1}, C_{1}$ are the points of tangency of the incircle of $\triangle A B C$ with the sides $B C, C A, A B$, respectively. (2009 Turkish National Team Selection Exam Problem) | 90. Let $x=A B_{1}, y=B C_{1}, z=C A_{1}$, the inequality is equivalent to $\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+$ $\sqrt{\frac{z}{z+x}} \leqslant \frac{3}{\sqrt{2}}$.
It is equivalent to proving
$$\begin{array}{l}
\sqrt{2 x(y+z)(z+x)}+\sqrt{2 y(z+x)(x+y)}+\sqrt{2 z(x+y)(y+z)} \leqslant \\
3 \sqrt{(x+y)(y+z)(z+x)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,686 |
91. A ray starting from vertex $A$ of $\triangle ABC$ intersects side $BC$ at $X$ and the circumcircle of $\triangle ABC$ at $Y$. Prove: $\frac{1}{AX} + \frac{1}{XY} \geqslant \frac{4}{BC}$. (2004 Baltic Way Mathematical Olympiad Problem) | 91. By the intersecting chords theorem, we have
$$\begin{array}{c}
B X \cdot X C=A X \cdot X Y \\
\frac{1}{A X}+\frac{1}{X Y} \geqslant \frac{4}{B C} \Leftrightarrow \frac{B C}{A X}+\frac{B C}{X Y} \geqslant 4 \Leftrightarrow \frac{B X}{A X}+\frac{X C}{A X}+\frac{B X}{X Y}+\frac{X C}{X Y} \geqslant 4
\end{array}$$
By ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,687 |
93. Let the circumradius of $\triangle ABC$ be $R$, and $G$ be the centroid of $\triangle ABC$. The extensions of $GA$, $GB$, and $GC$ intersect the circumcircle of $\triangle ABC$ at $D$, $E$, and $F$, respectively. Prove: $\frac{3}{R} \leqslant \frac{1}{GD} + \frac{1}{GE} + \frac{1}{GF} \leqslant \sqrt{3}\left(\frac{... | 93. Let the lines $A G, B G, C G$ intersect the sides $B C, C A, A B$ at $M, N, P$, respectively, and let $A M=m_{a}, B N=m_{b}, C P=m_{c}$. By the intersecting chords theorem, we have $A M \cdot M D = M B \cdot M C$. Since $M B = M C = \frac{a}{2}$, it follows that $M D = \frac{a^{2}}{4 m_{a}}$. Therefore, $G D = G M ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,690 |
94. In $\triangle A B C$, the three side lengths are $a, b, c$, $p=\frac{1}{2}(a+b+c)$, $R$ is the circumradius of $\triangle A B C$, $r$ is the inradius of $\triangle A B C$, and $l_{a}, l_{b}, l_{c}$ are the lengths of the angle bisectors from $A, B, C$, respectively. Prove: $l_{a} l_{b}+l_{b} l_{c}+l_{c} l_{a} \leqs... | 94. Since $S=\frac{1}{2}(b+c) l_{a} \sin \frac{A}{2}=\frac{1}{2} b c \sin A$, by the basic inequality we get
$$\begin{aligned}
l_{a}= & \frac{2 b c \cos \frac{A}{2}}{b+c}=\frac{2 b c}{b+c} \sqrt{\frac{1+\cos A}{2}}=\frac{2 b c}{b+c} \sqrt{\frac{1+\frac{b^{2}+c^{2}-a^{2}}{2 b c}}{2}}= \\
& \frac{\sqrt{b c\left[(b+c)^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,691 |
95. As shown in the figure, triangle $ABC$ is a right triangle, $\angle ACB=90^{\circ}$. $M_{1}, M_{2}$ are any two points inside $\triangle ABC$, $M$ is the midpoint of segment $M_{1} M_{2}$, lines $BM_{1}$, $BM_{2}$, $BM$ intersect side $AC$ at points $N_{1}$, $N_{2}$, $N$ respectively. Prove: $\frac{M_{1} N_{1}}{B M... | 95. Proof Let $H_{1}, H_{2}, H$ be the projections of $M_{1}, M_{2}, M$ on the line $B C$. Then $\frac{M_{1} N_{1}}{B M_{1}}=$ $\frac{H_{1} C}{B H_{1}}, \frac{M_{2} N_{2}}{B M_{2}}=\frac{H_{2} C}{B H_{2}}, \frac{M N}{B M}=\frac{H C}{B H}=\frac{H_{1} C+H_{2} C}{B H_{1}+B H_{2}}$. Without loss of generality, let $B C=1, ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,692 |
96. In a scalene $\triangle ABC$, the angle bisectors of $\angle BAC$ and $\angle ABC$ intersect the opposite sides at $D$ and $E$, respectively. Let $\angle BAC = \alpha$ and $\angle ABC = \beta$. Prove that the angle between line $DE$ and $AB$ does not exceed $\frac{|\alpha - \beta|}{3}$. (2009 Serbian Mathematical O... | 96. Let the sides opposite to $A, B, C$ in $\triangle ABC$ be $a, b, c$ respectively. By symmetry, assume $\alpha > \beta$. As shown in the figure, let $F$ be the intersection of line $DE$ and $AB$, and $\varphi$ be the angle between line $DE$ and $AB$. By the Angle Bisector Theorem, we have
$$\begin{array}{l}
\frac{B ... | \varphi<\frac{\alpha-\beta}{3} | Geometry | proof | Yes | Yes | inequalities | false | 733,693 |
97. In a right triangle $\triangle ABC$, where $C$ is the right angle vertex, the sides opposite to vertices $A, B, C$ are $a, b, c$, respectively. If the circle $K_{a}$ has its center on $a$ and is tangent to sides $b, c$; the circle $K_{b}$ has its center on $b$ and is tangent to sides $a, c$; and the radii of circle... | 97. Let $A^{\prime}$ be the point symmetric to $A$ with respect to $BC$, then $K_{a}$ is the incircle of $\triangle A^{\prime} A B$. Given that the perimeter of $\triangle A^{\prime} A B$ is $2(b+c)$ and the area is $a b$, we have $r_{a}=\frac{a b}{b+c}$. Similarly, $r_{b}=\frac{a b}{a+c}$.
Therefore,
$$\begin{array}{... | 1+\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,694 |
98. In $\triangle ABC$, take any points $K \in BC, L \in CA, M \in AB, N \in LM, R \in MK, F \in KL$. If $E_{1}, E_{2}, E_{3}, E_{4}, E_{5}, E_{6}$ and $E$ represent the areas of $\triangle AMR, \triangle CKR, \triangle BKF, \triangle ALF, \triangle BNM, \triangle CLN$ and $\triangle ABC$ respectively, prove: $E \geqsl... | 98. As shown in the figure, let $\frac{B K}{K C}=\frac{\lambda_{1}}{\mu_{1}}, \frac{C L}{L A}=\frac{\lambda_{2}}{\mu_{2}}, \frac{A M}{M B}=\frac{\lambda_{3}}{\mu_{3}}$, where $\lambda_{i}+\mu_{i}=1(i=1,2,3)$. Similarly, let $\frac{M N}{N L}=\frac{\lambda_{1}^{\prime}}{\mu_{1}^{\prime}}, \frac{L F}{F K}=\frac{\lambda_{2... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,695 |
99. Let $M$ be any point inside $\triangle ABC$, prove: $\min \{M A, M B, M C\} + M A + M B + M C < A B + B C + C A$. (40th IMO Shortlist) | Below is the proof of the original inequality. As shown in the figure, let $\triangle D E F$ be the triangle formed by the midpoints of the three sides of $\triangle A B C$, and it divides $\triangle A B C$ into four regions, each of which is covered by at least two of the convex quadrilaterals $A B D E, B C E F, C A F... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,696 |
100. Let $O$ be the circumcenter of acute-angled $\triangle ABC$. From $A$, draw the altitude to $BC$, with the foot of the perpendicular being $P$, and $\angle BCA \geq \angle ABC + 30^{\circ}$. Prove that $\angle CAB + \angle COP < 90^{\circ}$. (42nd IMO Problem) | 100. Let $\alpha=\angle C A B, \beta=\angle A B C, \gamma=\angle B C A, \delta=\angle C O P$.
Let $K, Q$ be the points symmetric to $A, P$ with respect to the perpendicular bisector of $B C$, and let $R$ be the circumradius of $\triangle A B C$. Then $O A=O B=O C=O K=R$. Since $K Q P A$ is a rectangle, $Q P=K A$, and ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,697 |
46. Given $a, b \in[1,3], a+b=4$, prove: $\sqrt{10} \leqslant \sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}<\frac{4 \sqrt{6}}{3}$. | 46. Let $y=\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}$, then
$$\begin{array}{c}
y^{2}=a+b+\frac{1}{a}+\frac{1}{b}+2 \sqrt{\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)}= \\
4+\frac{4}{a b}+2 \sqrt{a b+\frac{1}{a b}+\frac{b}{a}+\frac{a}{b}}= \\
4+\frac{4}{a b}+2 \sqrt{a b+\frac{1}{a b}+\frac{a^{2}+b^{2}}{a b}}= \\... | \sqrt{10} \leqslant \sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}<\frac{4 \sqrt{6}}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 733,699 |
102. In $\triangle ABC$, point $A_{1}$ is on side $BC$ internally, point $B_{1}$ is on side $CA$, and point $C_{1}$ is on side $AB$. If the three segments $AA_{1}$, $BB_{1}$, and $CC_{1}$ intersect at point $P$, and point $P$ does not coincide with $A$, prove that: $\frac{B_{1} C}{B_{1} A} + \frac{C_{1} B}{C_{1} A} \ge... | 102. Obviously, points $B_{1}, C_{1}$ are on the segments $A C, A B$ respectively,
$$\frac{A C_{1}}{B C_{1}}=\frac{S_{\triangle A P C_{1}}}{S_{\triangle B P C_{1}}}=\frac{S_{\triangle A C C_{1}}}{S_{\triangle B C C_{1}}}=\frac{S_{\triangle A C C_{1}}-S_{\triangle A P C 1}}{S_{\triangle B C C_{1}}-S_{\triangle B P C_{1}... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,700 |
103. In $\triangle ABC$, let $L=a+b+c$ (i.e., $L$ is the perimeter of $\triangle ABC$), $M$ is the sum of the lengths of the three medians, prove: $\frac{M}{L}>\frac{3}{4}$ (1994 Korean Mathematical Olympiad problem) | 103. As shown in the figure, let the three medians $A D, B E, C F$ intersect at the centroid $G$, then by the triangle inequality we have
$$A G+G B>A B, B G+G C>B C, G C+G A>C A$$
Adding these inequalities gives $2(A G+G B+G C)>L$, since $A G=\frac{2}{3} A D, B G=\frac{2}{3} B E, C G=\frac{2}{3} C F$ thus $\frac{M}{L}... | \frac{M}{L}>\frac{3}{4} | Geometry | proof | Yes | Yes | inequalities | false | 733,701 |
104, For any set $S$ of 5 points in the plane, satisfying that no three points in $S$ are collinear, let $M(S)$ and $m(S)$ be the maximum and minimum areas of triangles formed by any 3 points in $S$, respectively. Find the minimum value of $\frac{M(S)}{m(S)}$ $\quad($ Problem 43 of the 1 M O Preliminary Selection $)$ | 104. When these 5 points are the vertices of a regular pentagon, it is easy to know that $\frac{M(S)}{m(S)}$ equals the golden ratio $\tau=\frac{\sqrt{5}+1}{2}$.
Let the 5 points in $S$ be $A, B, C, D, E$, and the area of $\triangle A B C$ is $M(S)$. We prove that there exists a triangle whose area is less than or equ... | \frac{\sqrt{5}+1}{2} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,702 |
106. Let $A B C D$ be a tetrahedron, the sum of the lengths of its opposite edges is 1. Prove that $r_{A}+r_{B}+r_{C}+r_{D} \leqslant \frac{\sqrt{3}}{3}$, where $r_{A}, r_{B}, r_{C}, r_{D}$ are the radii of the incircles of the four triangular faces. Equality holds if and only if $A B C D$ is a regular tetrahedron (198... | 106. First, prove a lemma: In $\triangle ABC$, $3 \sqrt{3} r \leqslant p$, where $p$ is the semi-perimeter of $\triangle ABC$.
Under the substitution $a=y+z, b=z+x, c=x+y$, we have: the semi-perimeter $p=x+y+z$, the area $S=\sqrt{x y z(x+y+z)}$, the inradius $r=\frac{S}{p}=\sqrt{\frac{x y z}{x+y+z}}$, it suffices to p... | r_{A}+r_{B}+r_{C}+r_{D} \leqslant \frac{\sqrt{3}}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 733,704 |
107. Let the internal angle bisector of $\triangle A B C$ at vertex $A$ intersect side $B C$ at $A_{1}$ and the circumcircle of $\triangle A B C$ at $A_{2}$. Similarly, define $B_{1}, B_{2}, C_{1}, C_{2}$. Prove: $\frac{A_{1} A_{2}}{B A_{2}+A_{2} C}+\frac{B_{1} B_{2}}{C B_{2}+B_{2} A}+\frac{C_{1} C_{2}}{A C_{2}+C_{2} B... | 107. As shown in the figure, let the circumradius of $\triangle ABC$ be $R$, then by the Law of Sines, we have $B A_{2}=A_{2} C=2 R \sin \frac{A}{2}$. In $\triangle A_{1} A_{2} B$, by the Law of Sines, we get
$$\begin{aligned}
\frac{A_{1} A_{2}}{B A_{2}}= & \frac{\sin \angle A_{1} B A_{2}}{\sin \angle B A_{1} A_{2}}=\f... | \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2} | Geometry | proof | Yes | Yes | inequalities | false | 733,705 |
109: $\triangle A B C$ is any triangle, $M$ is any point inside $\triangle A B C$. The distances from $M$ to the sides $B C, C A, A B$ are denoted as $d_{a}, d_{b}, d_{c}$, respectively, and $S$ is the area of $\triangle A B C$. Prove: $a b d_{a} d_{b}+b c d_{b} d_{c}+c a d_{c} d_{a} \leqslant \frac{4}{3} S^{2}$ (1968 ... | 109. $\frac{1}{4}\left(a b d_{a} d_{b}+b c d_{b} d_{c}+c a d_{c} d_{a}\right)=S_{\triangle M B C} \cdot S_{\triangle M C A}+S_{\triangle M C A} S_{\triangle M A B}+S_{\triangle M A B}$ $S_{\triangle M C A} \leqslant \frac{1}{3}\left(S_{\triangle M B C}+S_{\triangle M C A}+S_{\triangle M A B}\right)^{2}=\frac{1}{3} S^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,707 |
110. In $\triangle A B C$, $\angle A=90^{\circ}$, the angle bisector of $\angle A$ intersects side $B C$ at point $D$, and the excenter of the excircle opposite to $B C$ is $I_{A}$. Prove: $\frac{A D^{-}}{D I_{A}} \leqslant \sqrt{2}-1$. (2004 Iran Mathematical Olympiad) | 110. Let $\angle A=x, A, D, I_{A}$ be collinear, by the Law of Sines we get
$$\begin{aligned}
\frac{A D}{D I_{A}}= & \frac{A D}{B D} \cdot \frac{B D}{D I_{A}}=\frac{\sin x}{\sin 45^{\circ}} \cdot \frac{\sin \left(45^{\circ}-\frac{x}{2}\right)}{\sin \left(90^{\circ}-\frac{x}{2}\right)}= \\
& \frac{2 \sin \left(45^{\circ... | \sqrt{2}-1 | Geometry | proof | Yes | Yes | inequalities | false | 733,708 |
111. The diagonals of quadrilateral $ABCD$ intersect at point $O$. Let the areas of $\triangle AOB$ and $\triangle COD$ be $S_{1}$ and $S_{2}$, and the area of quadrilateral $ABCD$ be $S$. Prove: $\sqrt{S_{1}} + \sqrt{S_{2}} \leq \sqrt{S}$. (1986 Swedish Mathematical Olympiad) | 111. As shown in the figure, let $O A=a, O B=b, O C=c, O D=d$, $\angle A O B=\angle C O D=\theta$,
then $S_{1}=\frac{1}{2} a b \sin \theta, S_{2}=\frac{1}{2} c d \sin \theta, S=$ $\frac{1}{2}(a+c)(b+d) \sin \theta$, by the Cauchy-Schwarz inequality we have $(a+c)(b+d) \geqslant(\sqrt{a b}+\sqrt{c d})^{2}$, i.e., $\sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,709 |
47. Let $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ be two sequences of positive numbers. Prove: $\left(\sum_{i \neq j} a_{i} b_{j}\right)^{2} \geqslant$ $\left(\sum_{i \neq j} a_{i} a_{j}\right)\left(\sum_{i \neq j} b_{i} b_{j}\right) \cdot(1998$ Yugoslav Mathematical Olympiad problem) | 47. Let $A=\sum_{i=1}^{n} a_{i}, B=\sum_{i=1}^{n} b_{i}$, the problem is equivalent to proving: $\left(A B-\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} \geqslant\left(A^{2}-\right.$ $\left.\sum_{i=1}^{n} a_{i}^{2}\right)\left(B^{2}-\sum_{i=1}^{n} b_{i}^{2}\right)$
Construct a quadratic function: $f(x)=\left(A^{2}-\sum_{i=1}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,710 |
112. For a tetrahedron \(ABCD\) with six edges of lengths \(a, b, c, d, e, f\), the areas of the four faces are \(S_{1}, S_{2}, S_{3}, S_{4}\), and the volume of the tetrahedron is \(V\), prove: \(2 \sqrt{S_{1} S_{2} S_{3} S_{4}} > 3 V \cdot \sqrt[6]{a b c d e f}\). | 112. As shown in the figure, in the tetrahedron $A_{1} A_{2} A_{3} A_{4}$, re-label $A_{i} A_{j}=a_{i j}, i, j \in\{1,2,3,4\}$. Let $S_{1}=S_{\Delta A_{1} A_{2} A_{3}}, S_{2}=S_{\Delta A_{1} A_{3} A_{4}}, S_{3}=S_{\Delta A_{1} A_{2} A_{4}}, S_{4}=S_{\Delta A_{2} A_{3} A_{4}}$, and denote the dihedral angle between the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,711 |
113. In $\triangle ABC$, the three internal angle bisectors $AD, BE, CF$ intersect the circumcircle at $P, Q, R$, respectively. Prove that $AP + BQ + CR \geq AB + BC + CA$. (26th Yugoslav Mathematical Olympiad) | 113. Let the angle bisector of $\angle A$ intersect the opposite side $BC$ at $D$. By the angle bisector property, we have
$$\frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}$$
and $BD + DC = a$. Therefore, $BD = \frac{ac}{b+c}$, $DC = \frac{ab}{b+c}$.
By the cosine rule, we get
$$\begin{array}{l}
AD^2 = AB^2 + BD^2 - 2 \cdot A... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,712 |
114. The incircle of $\triangle ABC$ touches the sides $BC$, $CA$, $AB$ at $A_1$, $B_1$, $C_1$ respectively. Let $I_1$, $I_2$, $I_3$ be the lengths of the shorter arcs $B_1C_1$, $C_1A_1$, $A_1B_1$ respectively. Denote the lengths of the sides $BC$, $CA$, $AB$ of $\triangle ABC$ as $a$, $b$, $c$ respectively. Prove that... | 114. Let the inradius of $\triangle ABC$ be $r$, it is easy to get $I_{1}=r(\pi-A), I_{2}=r(\pi-B), I_{3}=$ $r(\pi-C)$, thus $\frac{a}{I_{1}}+\frac{b}{I_{2}}+\frac{c}{I_{3}} \geqslant \frac{9 \sqrt{3}}{\pi}$ is equivalent to
$$\frac{a}{\pi-A}+\frac{b}{\pi-B}+\frac{c}{\pi-C} \geqslant \frac{9 \sqrt{3}}{\pi} r$$
By Cheb... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,713 |
116. In a convex quadrilateral $ABCD$, $N$ is the midpoint of side $BC$, and $\angle AND=135^{\circ}$. Prove: $AB + CD + \frac{1}{\sqrt{2}} BC \geqslant AD$. (2001 Baltic Way Mathematical Olympiad Problem) | 116. As shown in the figure, construct the symmetric points $B_{1}, C_{1}$ of points $B, C$ with respect to lines $A N$ and $N D$, respectively. Given that $\angle A N D = 135^{\circ}$, we have $\angle B N A + \angle C N D = 45^{\circ}$. Therefore, $\angle B_{1} N A + \angle C_{1} N D = 45^{\circ}$, which implies $\ang... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,715 |
117. The lateral surface area of a circular cone is $S$, and the volume is $V$. Prove: $\left(\frac{6 V}{\pi}\right)^{2} \leqslant\left(\frac{2 S}{\pi \sqrt{3}}\right)^{3}$, and determine when equality holds. (1966 IMO Shortlist) | 117. Let the radius of the base of a cone be $r$, and the height be $h$, then the slant height is $l=\sqrt{r^{2}+h^{2}}$, the lateral surface area is $S=$ $\pi r \sqrt{r^{2}+h^{2}}, V=\frac{1}{3} \pi r^{2} h$, the inequality $\left(\frac{6 V}{\pi}\right)^{2} \leqslant\left(\frac{2 S}{\pi \sqrt{3}}\right)^{3}$ is equiva... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,716 |
118. In $\triangle ABC$, the angle bisectors of $\angle A, \angle B$, and $\angle C$ intersect the circumcircle of $\triangle ABC$ at $A_{1}, B_{1}, C_{1}$, respectively. Prove that $A A_{1} + B B_{1} + C C_{1} > A B + B C + C A$. (1982 Australian Mathematical Olympiad) | 118. Applying Ptolemy's theorem to quadrilateral $A C A_{1} B$, we get $A A_{1} \cdot B C = A B \cdot A_{1} C + A C \cdot A_{1} B$. Let $A_{1} B = A_{1} C = x$. Noting that $2 x = A_{1} B + A_{1} C > B C$, we have $2 A A_{1} = 2 \frac{A B \cdot x + A C \cdot x}{B C} = (A B + A C) \cdot \frac{2 x}{B C} > A B + A C$, whi... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,717 |
120. Let the semi-perimeter of $\triangle A B C$ be $p$, the area be $S$, and the side length of the inscribed square $P Q R S$ be $x$, where $P, Q$ are on side $B C$, $R$ is on side $A C$, and $S$ is on side $A B$. Similarly, $y$ and $z$ are the side lengths of the other two inscribed squares, each with two points on ... | 120. Since $\triangle A S R \sim \triangle A B C$, we have $\frac{x}{a}=\frac{h_{a}-x}{h_{a}}$, i.e., $x=\frac{a h_{a}}{a+h_{a}}$. Similarly, $y=\frac{b h_{b}}{b+h_{b}}, z=\frac{c h_{c}}{c+h_{c}}$. Therefore, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,719 |
Example 2 Let $0<a_{i} \leqslant a, i=1,2, \cdots, n$, prove
(1) When $n=4$, we have
$$\frac{\sum_{i=1}^{4} a_{i}}{a}-\frac{a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+a_{4} a_{1}}{a^{2}} \leqslant 2 ;$$
(2) When $n=6$, we have
$$\frac{\sum_{i=1}^{6} a_{i}}{a}-\frac{a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{6} a_{1}}{a^{2}} \leqslant ... | Prove (1) Rewrite the inequality as $a_{1}\left(a-a_{2}\right)+a_{2}\left(a-a_{3}\right)+a_{3}\left(a-a_{4}\right)+$ $a_{4}\left(a-a_{1}\right) \leqslant 2 a^{2}$.
Each term on the left side is the area of one of the rectangles in Example 2's diagram. These 4 rectangles cover the square with side $a$ at most twice, so... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,722 |
Example 3 Let $a, b, c$ be the lengths of the three sides of a triangle with a perimeter not exceeding $2 \pi$. Prove: $\sin a, \sin b$, $\sin c$ can form the lengths of the three sides of a triangle. (2004 China National Training Team Selection Exam for IMO | Proof: From the given, we have $00, \sin b>0, \sin c>0$, $|\cos a||\cdot \sin \theta|+|\sin (c-\theta)| \geqslant|\sin (\theta+c-\theta)|=\sin c$ | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,724 |
4. Given that $a, b, c, A, B, C$ are all positive numbers, and $a+A=b+B=c+C=k$, prove: $a B+$ $b C+c A<k^{2}$. (21st All-Soviet Union Mathematical Olympiad problem) | 4. As shown in the figure, construct an equilateral triangle $P Q R$ with side length $k$. Points $D, E, F$ are on $Q R, R P, P Q$ respectively, with $Q D=A, D R=a, R E=B, E P=b, P F=C, F Q=c$. Then, from $S_{\triangle D R E}+S_{\triangle E P F}+$ $S_{\triangle F Q D}<S_{\triangle A B C}$, we get
$$\frac{1}{2} a B \sin... | a B+b C+c A<k^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 733,729 |
5. Prove that for any $x, y, z \in (\theta, 1)$, the inequality $x(1-y) + y(1-z) \pm z(1-x) < 1$ holds. (15th All-Russian Mathematical Olympiad Problem) | 5. As shown in the figure, construct an equilateral triangle $ABC$ with side length 1. Points $D, E, F$ are on $BC, CA, AB$ respectively, with $DC=x, BD=1-x, EA=y, CE=1-y, FB=z, AF=\mathrm{F}-z$. Then, from $S_{\triangle DCE} + S_{\triangle EAF} + S_{\triangle FBD} < S_{\triangle ABC}$, we get
$$\begin{array}{l}
\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,730 |
8. Given that $\alpha, \beta$ are acute angles, and $\alpha<\beta$, prove the inequality $\frac{\tan \alpha}{\alpha}<\frac{\tan \beta}{\beta}$. | 8. As shown in the figure, in the unit circle, take $\angle A O B=\alpha, \angle A O C=\beta$,
$$\begin{array}{l}
\frac{S_{\triangle O A B}}{S_{\text {sector } O A D}}<\frac{S_{\triangle O A B}}{S_{\triangle O M D}}=\left(\frac{O B}{O D}\right)^{2}=\frac{S_{\triangle O B C}}{S_{\triangle O D N}}<\frac{S_{\triangle O B ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,733 |
Example 9 Let $x, y, z$ all be positive real numbers, and $x+y+z=1$, find the minimum value of the ternary function $f(x, y, z)=$ $\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}$, and provide a proof. 2003 Hunan Province Mathematics Competition Test | Consider the function $g(t)=\frac{t}{1+t^{2}}$, we know that $g(t)$ is an odd function. Since when $t>0$, $\frac{1}{t}+t$ is decreasing in $(0,1)$, it is easy to see that $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in $(0,1)$. For $t_{1}, t_{2} \in(0,1)$ and $t_{1} \leqslant t_{2}$, we have
$$\left(t_{1}-t_{2}\right)\... | 0 | Algebra | proof | Yes | Yes | inequalities | false | 733,734 |
9. Let $x, y, z, t, u, v \in(0,1)$, prove the inequality: $x y z+u v(1-x)+(1-y)(1-$ $v) t+(1-z)(1-u)(1-t)<1$. (1996 Romanian Mathematical Olympiad) | 9. As shown in the figure, construct a regular tetrahedron $ABCD$ with edge length 1. Take points $E, F, G, P, M, N$ on $AB, AC, AD, BC, CD, BD$ respectively, and set $BP=x, BN=y, BE=z, CF=u, CM=v, DG=t$. Then
$$\begin{array}{l}
PC=1-x, ND=1-y, AE=1-z, AG=1-t, AF=1 \\
-u, MD=1-v . \\
\quad V_{B-ENP}+V_{C-FYP}+V_{D-MNG}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,735 |
10. Let $0<x_{1}<x_{2}<\cdots<x_{n}<\frac{\pi}{2}$, prove the inequality: $\sum_{k=1}^{n-1} \sin \left(2 x_{k}\right)-\sum_{k=1}^{n-1} \sin \left(x_{k}-\right.$ $\left.x_{k+1}\right)<\frac{\pi}{2}+\sum_{k=1}^{n-1} \sin \left(x_{k}+x_{k+1}\right) \cdot(1975$ IMO Shortlist $)$ | $$\begin{array}{l}
\text { 10. } \sum_{k=1}^{n-1} \sin \left(2 x_{k}\right)-\sum_{k=1}^{n-1} \sin \left(x_{k}-x_{k+1}\right)<\frac{\pi}{2}+\sum_{k=1}^{n-1} \sin \left(x_{k}+x_{k+1}\right) \Leftrightarrow \\
\sum_{k=1}^{n-1} \sin \left(2 x_{k}\right)-2 \sum_{k=1}^{n-1} \sin x_{k} \cos x_{k+1}<\frac{\pi}{2} \Leftrightarr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,736 |
13. Let real numbers $a_{i}, b_{i} (i=1,2, \cdots, n)$ satisfy $\sum_{i=1}^{n} a_{i}^{2}=\sum_{i=1}^{n} b_{i}^{2}=1, \sum_{i=1}^{n} a_{i} b_{i}=0$, prove: $\left(\sum_{i=1}^{n} a_{i}\right)^{2}+\left(\sum_{i=1}^{n} b_{i}\right)^{2} \leqslant n$. (2007 Romanian Mathematical Olympiad Problem) | $$\begin{array}{l}
\text { 13. Let } \vec{a}=\left(a_{1}, a_{2}, \cdots, a_{n}\right), \vec{b}=\left(b_{1}, b_{2}, \cdots, b_{n}\right), \vec{c}=(1,1, \cdots, 1), \\
\text { then } \vec{a}^{2}=\vec{b}^{2}=1, \vec{a} \cdot \vec{b}=0, \vec{c}^{2}=n. \\
\text { Let } \vec{d} = \vec{c}-(\vec{a} \cdot \vec{c}) \vec{a}-(\vec... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,739 |
14. Let $x, y \geqslant 0$, prove: $\sqrt{x^{2}-x+1}\left(\sqrt{y^{2}-y+1}+\sqrt{x^{2}+x+1}\right) \cdot \sqrt{y^{2}+y+1} \geqslant 2(x+y) \cdot(2010$ Kazakhstan Mathematical Olympiad problem $)$ | 14. Construct a quadrilateral $ABCD$ such that $AC=x+y$, $O$ is a point on segment $AC$ such that $OA=x$, $OB=OD=1$, $\angle AOB=\angle COD=60^{\circ}$. By the cosine rule, we get $AB=\sqrt{x^{2}-x+1}$, $BC=\sqrt{y^{2}+y+1}$, $CD=\sqrt{y^{2}-y+1}$, $DA=\sqrt{x^{2}+x+1}$. By the generalized Ptolemy's theorem, we have $A... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,740 |
Example 2 Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 3)$ be non-negative real numbers, and $x_{1}+x_{2}+\cdots+x_{n}=1$, prove that $x_{1}^{2} x_{2}$ $+x_{2}^{2} x_{3}+\cdots+x_{n-1}^{2} x_{n}+x_{n}^{2} x_{1} \leqslant \frac{4}{27}$. (1992 Taipei Mathematical Olympiad Problem) | Prove by mathematical induction on $n$. When $n \geqslant 3$, due to the cyclic symmetry of the inequality about $x_{1}, x_{2}, x_{3}$, we may assume $x_{1} \geqslant x_{2}, x_{1} \geqslant x_{3}$.
If $x_{2}<x_{3}$, then because
$$x_{1}^{2} x_{2}+x_{2}^{2} x_{3}+x_{3}^{2} x_{1}-\left(x_{1}^{2} x_{3}+x_{3}^{2} x_{2}+x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,742 |
Example 3 Let $r_{1}, r_{2}, \cdots, r_{n}$ be real numbers greater than or equal to 1, prove: $\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}+\cdots+$ $\frac{1}{r_{n}+1} \geqslant \frac{n}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1}$. (39th IMO Shortlist Problem) | Prove that when $n=1$, the inequality obviously holds. Below, we use mathematical induction to prove that the inequality holds for $n=2^{k} (k=1, 2, \cdots)$.
$$\text { When } k=1 \text {, we have } \frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}-\frac{2}{\sqrt{r_{1} r_{2}}+1}=\frac{\left(\sqrt{r_{1} r_{2}}-1\right)^{2}\left(\sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,743 |
Example 4 Given $x_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, satisfying $\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}=0$, prove: | $\left.\sum_{i=1}^{n} \frac{x_{i}}{i} \right\rvert\, \leqslant \frac{1}{2}-\frac{1}{2 n}$. (1989 National High School Mathematics League Test, Second Round) | We prove the strengthened proposition using mathematical induction: For $n(n \geqslant 2)$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$, if $\sum_{i=1}^{n}\left|x_{i}\right| \leqslant 1$ and $\sum_{i=1}^{n} x_{i}=0$, then $\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n}$.
(1) When $n=2$, $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,744 |
Example 10 Let $a, b, c, d$ be non-negative real numbers satisfying $a b+b c+c d+d a=1$, prove that: $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{d+a+b}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}$ (31st IMO Shortlist Problem) | Let $S=a+b+c+d$, then $S>0$. Construct the function $f(x)=\frac{x^{2}}{S-x}$. Since this function is increasing on $[0, S)$, for any $x \in[0, S)$, we have
$$-\left(x-\frac{S}{4}\right)\left[f(x)-f\left(\frac{S}{4}\right)\right] \geqslant 0$$
Therefore,
$$\frac{x^{3}}{S-x}-\frac{S x^{2}}{4(S-x)}-\frac{S x}{12}+\frac{S... | \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 733,745 |
Example 5 Let $\theta \in\left(0, \frac{\pi}{2}\right), n$ be a positive integer greater than 1, prove:
$$\left(\frac{1}{\sin ^{n} \theta}-\Gamma\right)\left(\frac{1}{\cos ^{n} \theta}-1\right) \geqslant 2^{n}-2 \sqrt{2^{n}}+1$$ | Prove that when $n=2$, both sides of equation (1) are equal, so the proposition holds for $n=2$. Assume the proposition holds for $n=k (k \geqslant 2)$, i.e., $\left(\frac{1}{\sin ^{k} \theta}-1\right)\left(\frac{1}{\cos ^{k} \theta}-1\right) \geqslant 2^{k}-2 \sqrt{2 k}+1$. Then for $n=k+1$,
$$\begin{array}{l}
\left(\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,746 |
Example 6 Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, and let $a=\min \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}$. Prove that:
$$\sum_{i=1}^{n} \frac{1+x_{j}}{1+x_{j+1}} \leqslant n+\frac{1}{(1+a)^{2}} \sum_{j=1}^{n}\left(x_{j}-a\right)^{2}$$
(Here, let $x_{n+1}=x_{1}$), and equality holds if and on... | Prove that when $n=1$, $a=x_{1}$, the inequality is written as $\frac{1+x_{1}}{1+x_{1}} \leqslant 1+\frac{1}{\left(1+x_{1}\right)^{2}}\left(x_{1}-x_{1}\right)$, which obviously holds and is an equality.
Assume the proposition holds for $n-1$, consider the case for $n$. Since the inequality is cyclically symmetric with... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,747 |
Example 7 Given $a>2,\left\{a_{n}\right\}$ is recursively defined as follows: $a_{0}=1, a_{1}=a, a_{n+1}=\left(\frac{a_{n}^{2}}{a_{n-1}^{2}}-2\right) a_{n}$. Prove: for any $n \cdot \mathbf{N}^{*}$, we have $\frac{1}{a_{0}}+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}<\frac{1}{2}\left(2+a-\sqrt{a^{2}-4}\right)$ (37th | $f^{(n)}(a)$.
Thus, $a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \cdots \frac{a_{1}}{a_{0}} \cdot a_{0}=f^{(n-1)}(a) \cdot f^{(n-2)}(a) \cdot \cdots \cdot f^{(0)}(a)$ (here we define $f^{(0)}(a)=a$)
We will use mathematical induction to prove $\frac{1}{a_{0}}+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,748 |
Example 8 If the sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies the conditions: $a_{1}=\frac{1}{2}, a_{k+1}=a_{k}+\frac{1}{n} a_{k}^{2},(k=0,1$, $2, \cdots)$, where $n$ is some fixed positive integer, prove: $1-\frac{1}{n}<a_{n}<1$. (1980 IMO of Five Countries | $$
\begin{aligned}
& \text{Proof: Since } a_{1}=a_{0}+\frac{1}{n} a_{0}^{2}=\frac{1}{2}+\frac{1}{4 n}=\frac{2 n+1}{4 n}, \text{ it is easy to prove that} \\
& \frac{n+1}{2 n+1}\frac{n+1}{2 n-k+2}+\frac{(n+1)^{2}}{n(2 n-k+2)^{2}}= \\
& \frac{n+1}{2 n-k+1}-\frac{n+1}{(2 n-k+1)(2 n-k+2)}+\frac{(n+1)^{2}}{n(2 n-k+2)^{2}}= ... | 1-\frac{1}{n}<a_{n}<1 | Algebra | proof | Yes | Yes | inequalities | false | 733,749 |
Example 9 Prove that for any $m, n \in \mathbf{N}$ and $x_{1}, x_{2}, \cdots, x_{n}, y_{1}, y_{2}, \cdots, y_{n} \in(0,1]$ satisfying $x_{\text {E }}+y_{i}=1, i=1,2, \cdots, n$, we have
$$\left(1-x_{1} x_{2} \cdots x_{n}\right)^{m}+\left(1-y_{1}^{m}\right)\left(1-y_{2}^{m}\right) \cdots\left(1-y_{n}^{m}\right) \geqslan... | Prove for $n \in \mathbf{N}$ using induction. First, because $\left(1-x_{1}\right)^{m}+\left(1-y_{1}^{m}\right)=y_{1}^{m}+(1-$ $\left.y_{1}^{m}\right)=1$, the conclusion is correct when $n=1$. Assume the conclusion holds for $n-1$, then
$$\begin{array}{l}
\left(1-x_{1} x_{2} \cdots x_{n}\right)^{m}+\left(1-y_{1}^{m}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,750 |
Example 11 Given the sequence $\left\{r_{n}\right\}$ satisfies $r_{1}=2, r_{n}=r_{1} r_{2} \cdots r_{n-1}+1, n=2,3, \cdots$, if positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<1$, prove:
$$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \leqslant \f... | Proof First, it is easy to prove by induction that the sequence $\left\{r_{n}\right\}$ has the following property:
$$\frac{1}{r_{1}}+\frac{1}{r_{2}}+\cdots+\frac{1}{r_{n}}=1-\frac{1}{r_{1} r_{2} \cdots r_{n}}$$
Below, we prove inequality (1) by induction.
When $n=1$, inequality (1) clearly holds.
Now assume that for $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,752 |
Example 12 From any $n(n \geqslant 2)$ given positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, each time take $k$ numbers to form a product. The $k$-th root of the arithmetic mean of all such products is called the $k$-th symmetric mean of these $n$ numbers, denoted as $A_{k}$, i.e. $\square$
$$A_{k}=\sqrt[k]{\frac{a_{1}... | From the lemma, when $k=1$, we have $\left(\sum_{n}^{1}\right)^{2} \geqslant\left(\sum_{n}^{2}\right)^{2}\left(\sum_{n}^{0}\right)^{0}$, which means $\sum_{n}^{1} \geqslant \sum_{n}^{2}$. Now, assume that $\sum_{n}^{1} \geqslant \sum_{n}^{2} \geqslant \cdots \geqslant \sum_{n}^{k} \cdot(2 \leqslant k<n)$ has already be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,753 |
1. Let real numbers $a_i, b_i (i=0,1, \cdots, 2 n)$ satisfy the following three conditions:
(1) For $i=0,1, \cdots, 2 n-1$, we have $a_{i}+a_{i+1} \geqslant 0$;
(2) For $j=0,1, \cdots, n-1$, we have $a_{2 j+1} \leqslant 0$;
(3) For any integers $p, q, 0 \leqslant p \leqslant q \leqslant n$, we have $\sum_{k=2 p}^{2 q} ... | Prove by mathematical induction: $\sum_{i=0}^{2 n}(-1)^{i} a_{i} b_{i} \geqslant 0$, and the equality holds if and only if $a_{0}=$ $a_{1}=\cdots=a_{2 n}=0$.
When $n=1$, by the conditions, $a_{0}+a_{1} \geqslant 0, a_{1}+a_{2} \geqslant 0, a_{1} \leqslant 0, b_{0}>0, b_{2}>0, b_{0}+$ $b_{1}+b_{2}>0$. Therefore,
$$\beg... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,754 |
1. Let $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ be a sequence of positive numbers, such that for all $n=0,1,2, \cdots$, we have $a_{n}^{2} \leqslant a_{n}-a_{n+1}$. Prove that for all $n=1,2, \cdots$, we have $a_{n}<\frac{1}{n+1}$.
(1965 Beijing Mathematical | 1. From the inequality $a_{0}^{2} \leqslant a_{0}-a_{1}$, we get $a_{1} \leqslant a_{0}-a_{0}^{2}=a_{0}\left(1-a_{0}\right) \leqslant$ $\left[\frac{a_{0}+\left(1-a_{0}\right)}{2}\right]^{2}=\frac{1}{4}<\frac{1}{2}$, so the inequality holds when $n=1$.
Assume that when $n=k$, the inequality holds, i.e., $a_{k}<\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,755 |
$\square$ Example 1 Given that $a, b, c$ are positive real numbers, prove that for any real numbers $x, y, z$, we have
$$\begin{aligned}
x^{2}+y^{2}+z^{2} \geqslant & 2 \sqrt{\frac{a b c}{(a+b)(b+c)(c+a)}}\left(\sqrt{\frac{a+b}{c}} x y+\sqrt{\frac{b+c}{a}} y z\right. \\
& \left.+\sqrt{\frac{c+a}{b}} z x\right) .
\end{a... | Prove the left side - right side of the above equation:
$$\begin{aligned}
= & {\left[\frac{b}{b+c} x^{2}+\frac{a}{c+a} y^{2}-2 \sqrt{\frac{a b}{(b+c)(c+a)}} x y\right] } \\
& +\left[\frac{c}{c+a} y^{2}+\frac{b}{a+b} z^{2}-2 \sqrt{\frac{b c}{(c+a)(a+b)}} y z\right] \\
& +\left[\frac{c}{b+c} x^{2}+\frac{a}{a+b} z^{2}-2 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,756 |
$$\square$$ Example 9 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, $\gamma=\alpha+\beta, \alpha \beta>0$, then
$$\begin{array}{l}
\quad \frac{1}{n}\left(a_{1}^{\gamma}+a_{2}^{\gamma}+\cdots+a_{n}^{\gamma}\right) \geqslant \frac{1}{n}\left(a_{1}^{\alpha}+a_{2}^{\alpha}+\cdots+a_{n}^{\alpha}\right) \cdot \... | Prove that because $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $\alpha, \beta$ have the same sign, we have
$$\begin{array}{c}
a_{1}^{\alpha+\beta}+a_{j}^{\alpha+\beta} \geqslant a_{1}^{\alpha} a_{j}^{\beta}+a_{1}^{\beta} a_{j}^{\alpha}(j=2,3, \cdots, n), \\
a_{2}^{\alpha+\beta}+a_{j}^{\alpha+\beta} \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,759 |
Example 10 If $a_{1}, a_{2}, \cdots, a_{n}$ are all positive numbers, $k$ is a positive integer, and let $a_{n+1}=$ $a_{1}$, then
$$\sum_{i=1}^{n} \frac{a_{i}^{k+1}}{a_{i}^{k}+a_{i}^{k-1} a_{i+1}+\cdots+a_{i} a_{i+1}^{k}+a_{i+1}^{k}} \geqslant \frac{1}{k+1} \sum_{i=1}^{n} a_{i}$$ | Proof: Let \( M=\sum_{i=1}^{n} \frac{a_{i}^{k+1}}{a_{i}^{k}+a_{i}^{k-1} a_{i+1}+\cdots+a_{i} a_{i+1}^{k}+a_{i+1}^{k}} \),
then \(\square\)
\[
\begin{array}{l}
N=\sum_{i=1}^{n} \frac{a_{i+1}^{k+1}}{a_{i}^{k}+a_{i}^{k-1} a_{i+1}+\cdots+a_{i} a_{i+1}^{k}+a_{i+1}^{k}} \\
\begin{array}{l}
M-N=\sum_{i=1}^{n} \frac{a_{i}^{k+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,760 |
$\square$ Example 11 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$, prove:
$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$. (46th IMO problem!) | To prove the original inequality is equivalent to:
$$\begin{aligned}
& \frac{x^{6}}{x^{5}+y^{2}+z^{6}}+\frac{y^{5}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}}{z^{5}+x^{2}+y^{2}} \\
\geqslant & \frac{x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{2}}{z^{5}+x^{2}+y^{2}}
\end{aligned}$$
From $\frac{a^{2}}{b} \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,761 |
■ Example 2 Given a tetrahedron $O-ABC$ with three lateral edges $OA, OB, OC$ mutually perpendicular, $P$ is a point within the base $ABC$, and $OP$ forms angles $\alpha, \beta, \gamma$ with the three lateral faces, respectively. Prove that $\frac{\pi}{2}<\alpha+\beta+\gamma \leqslant 3 \arcsin \frac{\sqrt{3}}{3}$. (20... | Given the problem, we have $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1$, and $\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$, so
$$\begin{aligned}
\sin ^{2} \alpha & =1-\left(\sin ^{2} \beta+\sin ^{2} \gamma\right)=\cos ^{2} \beta-\sin ^{2} \gamma \\
& =\frac{1}{2}(1+\cos 2 \beta)-\frac{1}{2}(1-\cos ... | \frac{\pi}{2}<\alpha+\beta+\gamma \leqslant 3 \arcsin \frac{\sqrt{3}}{3} | Geometry | proof | Yes | Yes | inequalities | false | 733,764 |
$\square$ Example 3 Given $5 n$ real numbers $r_{i}, s_{i}, t_{i}, u_{i}, v_{i}$ all greater than $1(1 \leqslant i \leqslant n)$, let $R=\left(\frac{1}{n} \sum_{i=1}^{n} r_{i}\right), S=\left(\frac{1}{n} \sum_{i=1}^{n} s_{i}\right), T=\left(\frac{1}{n} \sum_{i=1}^{n} t_{i}\right), U=\left(\frac{1}{n} \sum_{i=1}^{n} u_{... | Prove that the function $y=\ln \frac{\mathrm{e}^{x}+1}{\mathrm{e}^{x}-1}$ is a convex function $(x>0)$. For this, it suffices to prove that for real numbers $a, b>1$, we have
$$\left(\frac{a+1}{a-1}\right)\left(\frac{b+1}{b-1}\right) \geqslant\left(\frac{\sqrt{a b}+1}{\sqrt{a b}-1}\right)^{2}$$
This inequality is equi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,765 |
For example, $40<a, b, c<1$, and $a b+b c+c a=1$, prove: $\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}$ $+\frac{c}{1-c^{2}} \geqslant \frac{3 \sqrt{2}}{2}$. (2004 Singapore National Team Selection Exam Question) | Prove that because $a b+b c+c a=1$, we can set $a=\cot A, b=\cot B, c=\cot C$, where $A, B, C$ are the three interior angles of a triangle, and from $0<a, b, c<1$ we know $\frac{\pi}{4} < A, B, C < \frac{\pi}{2}$, thus
$$\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}}=-(\tan 2 A+\tan 2 B+\tan 2 C)$$
Therefore, i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,766 |
$\square$ Example $5 \alpha, \beta, \gamma$ are the three interior angles of a given triangle. Prove that $\csc ^{2} \frac{\alpha}{2}+$ $\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 12$, and find the condition for equality. (1994 National Mathematical Olympiad Problem) | Prove that by the AM-GM inequality,
$$\csc ^{2} \frac{\alpha}{2}+\csc ^{2} \frac{\beta}{2}+\csc ^{2} \frac{\gamma}{2} \geqslant 3 \sqrt[3]{\csc ^{2} \frac{\alpha}{2} \csc ^{2} \frac{\beta}{2} \csc ^{2} \frac{\gamma}{2}}$$
where equality holds if and only if $\alpha=\beta=\gamma$. Then, by the AM-GM inequality and the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,767 |
Example 6 Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive numbers $(n \geqslant 2)$, and $\sum_{i=1}^{n} x_{i}=1$, prove that: $\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{\sum_{i=1}^{n} \sqrt{x_{i}}}{\sqrt{n-1}}$. (4th CMO Problem) | Prove that the function $f(x)=\frac{x}{\sqrt{1-x}}(0<x<1)$ is a convex function. For $t_{1}, t_{2}>1$, and $x_{i}=1-\frac{1}{t_{i}^{2}}(i=1,2)$. Since
$$\frac{1}{2}\left(t_{1}+t_{2}\right) \geqslant \sqrt{t_{1} t_{2}}, \sqrt{t_{1}^{2}+t_{2}^{2}} \geqslant \sqrt{2 t_{1} t_{2}}$$
Thus, $\frac{1}{2}\left(t_{1}+t_{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,768 |
Example 13 Let $x, y, z$ be real numbers, $k_{1}, k_{2}, k_{3} \in\left(0, \frac{1}{2}\right)$ and $k_{1}+k_{2}+k_{3}=$ 1, prove: $k_{1} k_{2} k_{3}(x+y+z)^{2} \geqslant x y k_{3}\left(1-2 k_{3}\right)+y z k_{1}\left(1-2 k_{1}\right)+$ $z x k_{2}\left(1-2 k_{2}\right) .(1990$ China National Training Team Problem) | We first prove a lemma: In $\triangle ABC$, for any real numbers $x, y, z$, we have
$$x^{2}+y^{2}+z^{2} \geqslant 2 x y \cos C + 2 y z \cos A + 2 z x \cos B$$
In fact, since $\cos A = -\cos (B+C)$, we have
$$\begin{aligned}
& x^{2}+y^{2}+z^{2}-(2 x y \cos C + 2 y z \cos A + 2 z x \cos B) \\
= & (x - y \cos C - z \cos ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,769 |
$\square$ Example 7 Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1$, prove that: $\frac{x y}{\sqrt{x y+y z}}+$ $\frac{y z}{\sqrt{y z+x z}}+\frac{x z}{\sqrt{x z+x y}} \leqslant \frac{\sqrt{2}}{2} .(2006$ China National Training Team Exam Question) | Prove that since $\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=1$, by the generalized Jensen's inequality we have
$$\begin{array}{l}
\sum_{\mathrm{cyc}} \frac{x y}{\sqrt{x y+y z}}=\sum_{\mathrm{cyc}} \sqrt{\frac{x^{2} y}{x+z}}=\sum_{\mathrm{cyc}} \frac{x+y}{2} \sqrt{\frac{4 x^{2} y}{(x+y)^{2}(x+z)}} \\
\leqslant \sqrt{\su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,770 |
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