problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
32. Given an integer $n \geqslant 3$, real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\min _{1 \leqslant i<j \leqslant n}\left|a_{i}-a_{j}\right| \leqslant 1$, find the minimum value of $\sum_{k=1}^{n}$ $\left|a_{k}\right|^{3}$. (2009 China Mathematical Olympiad Problem) | 32. Let $a_{1}<a_{2}<\cdots<a_{n}$, then for $1 \leqslant k \leqslant n$, we have
$$\left|a_{k}\right|+\left|a_{n-k+1}\right| \geqslant\left|a_{n-k+1}-a_{k}\right| \geqslant|n+1-2 k|$$
Therefore,
$$\begin{array}{l}
\sum_{k=1}^{n}\left|a_{k}\right|^{3}= \frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|^{3}+\left|a_{n... | \frac{1}{32}\left(n^{2}-1\right)^{2} \text{ (when $n$ is odd) or } \frac{1}{32} n^{2}\left(n^{2}-2\right) \text{ (when $n$ is even)} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,489 |
33. Given the sequence $\left\{a_{n}\right\}$ defined as follows: $a_{1}=\frac{1}{2}, a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1}$, prove: $a_{1}+$ $a_{2}+\cdots+a_{n}<1$. (2003 Romanian National Training Team Selection Test) | 33. Let $b_{n}=\frac{1}{a_{n}}$, then $b_{n+1}=b_{n}^{2}-b_{n}+1$, thus $\frac{b_{n+1}-1}{b_{n}-1}=b_{n}$, i.e., $\frac{b_{n+1}-1}{b_{n}-1}=b_{n}$, $\frac{b_{n}-1}{b_{n-1}-1}=b_{n-1}, \frac{b_{n-1}-1}{b_{n-2}-1}=b_{n-2}, \cdots, \frac{b_{2}-1}{b_{1}-1}=b_{1}$, multiplying these gives $b_{n+1}-1=b_{1} b_{2} \cdots b_{n}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,490 |
34. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{3}, a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1}(n=1,2, \cdots)$, prove that: $\frac{1}{2}-$ $\frac{1}{3^{2^{n-1}}}<a_{1}+a_{2}+\cdots+a_{n}<\frac{1}{2}-\frac{1}{3^{2^{n}}}$. (2010 National High School Mathematics Competition $B$ Paper) | 34. From $a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1}$, we have
$$\frac{1}{a_{n+1}}=\frac{1}{a_{n}^{2}}-\frac{1}{a_{n}}+1, \frac{1}{a_{n+1}}-1=\frac{1}{a_{n}}\left(\frac{1}{a_{n}}-1\right)$$
Therefore,
$$\frac{a_{n+1}}{1-a_{n+1}}=\frac{a_{n}^{2}}{1-a_{n}}=\frac{a_{n}}{1-a_{n}}-a_{n}$$
That is,
$$a_{n}=\frac{a_{n}}{1-... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,491 |
35. Given the sequence $\left\{a_{n}\right\}$ defined as follows: $a_{1}=1, a_{2}=\frac{4}{3}, a_{n+1}=\sqrt{1+a_{n} a_{n-1}},(n \geqslant 2)$, prove that when $n \geqslant 2$, $a_{n}^{2}>a_{n-1}^{2}+\frac{1}{2}$, and $1+\sum_{i=1}^{n} \frac{1}{a_{i}}>2 a_{n}$. (2010 Indonesia Mathematical Olympiad) | 35. (1) Use mathematical induction. When $n=2$, the inequality obviously holds. Assume that when $n=k, k-1$, the inequality holds, i.e., $a_{k}^{2}>a_{k-1}^{2}+\frac{1}{2}, a_{k-1}^{2}>a_{k-2}^{2}+\frac{1}{2}$, then when $n=k+1$, by the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
a_{k+1}^{2}= & 1+a_{k} a_{k-1}>... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,492 |
36. Given a sequence $\left\{a_{n}\right\}$ where all terms are non-zero, the sum of its first $n$ terms is $S_{n}$, and for any $n \in \mathbf{N}^{*}$, we have $(1-p) S_{n}=p-p a_{n}$ (where $p$ is a constant greater than 1). Let $f(n)=\frac{1+C_{n}^{1} a_{1}+C_{n}^{2} a_{2}+\cdots+C_{n}^{n} a_{n}}{2^{n} S_{n}}$.
(1) ... | 36. (1) Notice that
$$(1-p) S_{n}=p-p a_{n}$$
Then
$$(1-p) S_{n+1}=p-p a_{n+1}$$
Subtracting equation (1) from equation (2), we get $(1-p) a_{n+1}=-p a_{n+1}+p a_{n}$. Thus, $a_{n+1}=p a_{n}$. Setting $n=1$ in equation (1), we get $a_{1}=p$. Therefore, $a_{n}=p^{n}$. Hence, $S_{n}=\frac{p^{n}-1}{p-1}$.
Notice that
$... | (2 n-1) f(n) \leqslant \sum_{k=1}^{2 n-1} f(k) \leqslant \frac{p+1}{p-1}\left[-\left(\frac{p+1}{2 p}\right)^{2 n-1}\right] | Algebra | proof | Yes | Yes | inequalities | false | 733,493 |
Example 1 Let $a, b, c$ be the lengths of the three sides of a triangle. If $a+b+c=1$, prove: $a^{2}+$ $b^{2}+c^{2}+4 a b c \leqslant \frac{1}{2}$. (1990 Italian Mathematical Olympiad) | Prove that the original inequality is equivalent to
$$\begin{array}{l}
(a+b+c)^{2}-2(a b+b c+c a)+4 a b c \leqslant \frac{1}{2} \Leftrightarrow \\
1-2(a b+b c+c a)+4 a b c \leqslant \frac{1}{2} \Leftrightarrow \\
a b+b c+c a-2 a b c \geqslant \frac{1}{4}
\end{array}$$
Since $a, b, c$ are the lengths of the three sides... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,494 |
28. Real numbers $u, v$ satisfy $\left(u+u^{2}+u^{3}+\cdots+u^{8}\right)+10 u^{9}=\left(v+v^{2}+v^{3}+\cdots+\right.$ $\left.v^{10}\right)+\mathrm{O} v^{11}=8$. Which is larger? Prove your conclusion. (18th US Mathematical Olympiad problem) | 28. First, from the equation $\left(u+u^{2}+u^{3}+\cdots+u^{8}\right)+10 u^{9}=\left(v+v^{2}+v^{3}+\cdots+\right.$ $\left.v^{10}\right)+10 v^{11}=8$ we have $u>0, v>0$.
Let
$$\begin{array}{l}
f(x)=\left(x+x^{2}+x^{3}+\cdots+x^{8}\right)+10 x^{9}-8 \\
g(x)=\left(x+x^{2}+x^{3}+\cdots+x^{10}\right)+10 x^{11}-8
\end{array}... | u<v | Algebra | proof | Yes | Yes | inequalities | false | 733,499 |
Example 6 Let real numbers $a, b, c$ be such that the sum of any two is greater than the third, then $\frac{2}{3}(a+b+c)\left(a^{2}+\right.$ $b^{2}+c^{2}$ ) $\geqslant a^{3}+b^{3}+c^{3}+3 a b c$. Equality holds if and only if $a=b=c$. (13th Putnam | It is not hard to verify that the above inequality is equivalent to
$$\begin{array}{l}
(-a+b+c)(a-b+c)(a+b-c) \geqslant \\
(3 a-b-c)(3 b-c-a)(3 c-a-b)
\end{array}$$
Moreover, at least two of $3 a-b-c, 3 b-c-a, 3 c-a-b$ are positive (otherwise, assume without loss of generality that $3 a-b-c \leqslant 0, 3 b-c-a \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,500 |
Example 7 In $\triangle A B C$, prove that: $\frac{\sqrt{\sin A \sin B}}{\sin \frac{C}{2}}+\frac{\sqrt{\sin B \sin C}}{\sin \frac{A}{2}}+\frac{\sqrt{\sin C \sin A}}{\sin \frac{B}{2}} \geqslant$ $3 \sqrt{3}$. | Prove that from the Law of Sines and the Law of Cosines, we have
$$\begin{aligned}
\frac{\sqrt{\sin A \sin B}}{\sin \frac{C}{2}}= & \frac{\sqrt{\sin A \sin B}}{\sin C} \cdot 2 \cos \frac{C}{2}=\frac{\sqrt{a b}}{c} \cdot \sqrt{2(1+\cos C)}= \\
& \frac{\sqrt{2 a b+2 a b \cos C}}{c}=\frac{\sqrt{(a+b)^{2}-c^{2}}}{c}= \\
& ... | 3 \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 733,501 |
Example 8 Prove that in an acute $\triangle A B C$, $\frac{a b c}{\sqrt{2\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}} \geqslant \frac{r}{2 R}$, where $r, R$ represent the inradius and circumradius of $\triangle A B C$, respectively. (2005 Jiangsu Mathematical Winter Camp) | To prove in $\triangle ABC$, $\frac{r}{R}=4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$, to prove $\frac{a b c}{\sqrt{2\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}} \geqslant \frac{r}{2 R}$, it suffices to prove
$$\frac{a b c}{\sqrt{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,502 |
F. Let $T$ be a triangle with perimeter 2, and let $a, b, c$ be the lengths of the sides of $T$. Prove: $abc + \frac{28}{27} \geqslant ab + bc + ca \geqslant abc + 1$. (19th Irish Mathematical Olympiad) | 1. Proof: From the given conditions, we know that $0 \leqslant a, b, c \leqslant 1, a+b+c=2$, thus $0 \leqslant(1-a)(1-b)(1-c) \leqslant\left(\frac{(1-a)+(1-b)+(1-c)}{3}\right)^{3}=\frac{1}{27}$.
Therefore, $0 \leqslant 1-a-b-c+a b+b c+c a-a b c \leqslant \frac{1}{27}$. Combining this with $a+b+c=2$, we have $1 \leqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,506 |
3. In $\triangle A B C$, prove that: $\frac{1}{(p-a)^{2}}+\frac{1}{(p-b)^{2}}+\frac{1}{(p-c)^{2}} \geqslant \frac{1}{r^{2}}$. (27th American Mathematical Competition Problem) | 3. $\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2} \geqslant \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$, and $\tan \frac{A}{2}=\frac{r}{p-a}, \tan \frac{B}{2}=\frac{r}{p-b}, \tan \frac{C}{2}=\frac{r}{p-c}$, so $\frac{1}{(p-a)^{2}}+\frac{1}... | \frac{1}{(p-a)^{2}}+\frac{1}{(p-b)^{2}}+\frac{1}{(p-c)^{2}} \geqslant \frac{1}{r^{2}} | Inequalities | proof | Yes | Yes | inequalities | false | 733,508 |
4. Prove: In any triangle, the inequality $\frac{\cos \alpha}{a^{3}}+\frac{\cos \beta}{b^{3}}+\frac{\cos \gamma}{c^{3}} \geqslant \frac{3}{2 a b c}$ holds. Here $a$, $b, c$ are the lengths of the sides of the triangle, and $\alpha, \beta, \gamma$ are the interior angles opposite to these sides, respectively. (2004 Croa... | 4. Apply the cosine theorem and the well-known inequality $x+\frac{1}{x} \geqslant 2$ (where $x>0$)
$$\begin{array}{l}
\frac{\cos \alpha}{a^{3}}+\frac{\cos \beta}{b^{3}}+\frac{\cos \gamma}{c^{3}}=\frac{b^{2}+c^{2}-a^{2}}{2 a^{3} b c}+\frac{c^{2}+a^{2}-b^{2}}{2 a b^{3} c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b c^{3}}= \\
\frac{... | \frac{3}{2 a b c} | Inequalities | proof | Yes | Yes | inequalities | false | 733,509 |
7. If $A, B, C$ are the three interior angles of a triangle, prove: $-2 \leqslant \sin 3 A+\sin 3 B+\sin 3 C \leqslant \frac{3 \sqrt{3}}{2}$. Determine when equality holds. (1981 10th United States of America Mathematical Olympiad) | 7. By the symmetry of the inequality, we can assume $A \geqslant B \geqslant C$, hence $C \leqslant 60^{\circ}, \sin 3 C \geqslant 0$, thus, $\sin 3 A+\sin 3 B+\sin 3 C \geqslant \sin 3 A+\sin 3 B \geqslant-2$. The equality cannot hold. Now we prove the right inequality.
Let $\alpha=\frac{3}{2}(B+C)$, then
$$\begin{ar... | \sin 3 A+\sin 3 B+\sin 3 C \leqslant \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 733,513 |
9. Let $A, B, C$ be the interior angles of an acute triangle, prove that: $\tan ^{n} A+\tan ^{n} B+\tan ^{n} C \geqslant 3^{\frac{n}{2}+1}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | 9. In $\triangle A B C$, since $\tan (A+B)=-\tan C$, i.e., $\frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C$, thus $\tan A+\tan B+\tan C=\tan A \tan B \tan C$. By the AM-GM inequality, we have
$$\tan A+\tan B+\tan C \geqslant 3(\tan A \tan B \tan C)^{\frac{1}{3}}$$
Solving this, we get
$$\tan A \tan B \tan C \geqslant 3... | null | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,515 |
10. For a triangle with side lengths $a, b, c$ and area $S$, the inequality $-6 a^{2}+10 b^{2}+ 123 c^{2} \geqslant 48 \sqrt{3} S$ is satisfied. (1992 "Friendship Cup" International Mathematics Competition) | 10. From the cosine theorem and the area formula of a triangle, we know that $a^{2}=b^{2}+c^{2}-2 b c \cos A, S=$ $\frac{1}{2} b c \sin A$.
Thus, the original inequality is equivalent to
$$\begin{array}{l}
-6\left(b^{2}+c^{2}-2 b c \cos A\right)+10 b^{2}+123 c^{2} \geqslant 48 \sqrt{3} \cdot \frac{1}{2} b c \sin A \Le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,516 |
11. In $\triangle A B C$, prove that: $\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2} \leqslant \frac{9 R r}{2 S}$. (where $S$ represents the area of $\triangle A B C$) (Strengthened version of a problem from the 26th IMO Shortlist) | 11. Let the three sides of $\triangle ABC$ be $a, b, c$, then it is easy to prove
$$\begin{array}{l}
\tan \frac{A}{2}=\frac{a^{2}-(b-c)^{2}}{4 S} \\
\tan \frac{B}{2}=\frac{b^{2}-(c-a)^{2}}{4 S} \\
\tan \frac{C}{2}=\frac{c^{2}-(a-b)^{2}}{4 S}
\end{array}$$
Thus, the original inequality becomes
$$a^{2}-(b-c)^{2}+b^{2}-(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,517 |
13. The inradius of $\triangle ABC$ is $r$, and the circumradius is $R$. Prove that $\sin \frac{A}{2} \sin \frac{B}{2} + \sin \frac{B}{2} \sin \frac{C}{2} + \sin \frac{C}{2} \sin \frac{A}{2} \leqslant \frac{5}{8} + \frac{r}{4 R}$. (29th IMO Shortlist) | 13. Since $a=r\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right)=r \cdot \frac{\sin \frac{B+C}{2}}{\sin \frac{B}{2} \sin \frac{C}{2}}=r \cdot \frac{\cos \frac{A}{2}}{\sin \frac{B}{2} \sin \frac{C}{2}}, a=$ $2 R \sin A$, thus $r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.
Therefore,
$$\begin{aligned}
\frac{r}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,519 |
Example 6 Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be real numbers. If they satisfy $\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}-1\right)\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}-1\right) \geq\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}-1\right)$, prove that $a_{1}^{2}+a_{2}^{2}+\cdots+a_... | Prove by contradiction, assume $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}<1$ and $b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}<1$, construct the quadratic function
$$\begin{array}{l}
f(x)=(x-1)-\sum_{k}\left(a_{k} x-b_{k}\right) \\
\left(1-\sum_{k=1}^{n} a_{k}^{2}\right) x^{2}-2\left(1-\sum_{k=1}^{n} a_{k} b_{k}\right) x+1-\sum_{k=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,521 |
30. Given that $a, b, c, d$ are positive numbers, and $a \leqslant b \leqslant c \leqslant d$, prove: $a^{b} b^{b} c^{d} d^{a} \geqslant b^{a} c^{b} d^{d} a^{d}$. | 30. Let the substitution $b=a x, c=a y, d=a z$, then by the conditions of the problem, we have $1 \leqslant x \leqslant y \leqslant z$. The inequality in the problem is
$$a^{a x}(a x)^{a y}(a y)^{a z}(a z)^{a} \geqslant(a x)^{a}(a y)^{a x}(a z)^{a y} a^{a z}$$
After canceling $a^{a} a^{a x} a^{a y} a^{a z}$, and takin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,522 |
15. Prove: For any triangle with sides of length $a, b, c$ and area $S$, we have $a b + b c + c a \geqslant 4 \sqrt{3} S$. (1977 IM Preliminary Question) | 15. In $\triangle A B C$, let $A B=c, B C=a, C A=b$, and $\angle B A C=\alpha$.
Consider the arc $\overparen{B A C}$ opposite to side $B C$ on the circumcircle of $\triangle A B C$. Since the midpoint $D$ of the arc is the farthest point from the chord $B C$, for the altitudes $A H=h$ and $D K$ of $\triangle A B C$ an... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,523 |
16. In $\triangle A B C$, prove the Garfunkel-Bankoff inequality $\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2}$
$$\frac{C}{2} \geqslant 2-8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$ | 16. Let $i, j, k$ be three unit vectors in a plane, and the angle between $j$ and $k$ is $\pi-A$, the angle between $k$ and $i$ is $\pi-B$, and the angle between $i$ and $j$ is $\pi-C$. Then $\left(i \tan \frac{A}{2}+j \tan \frac{B}{2}+k \tan \frac{C}{2}\right)^{2} \geqslant 0$.
Hence
$$\begin{array}{l}
\tan ^{2} \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,524 |
17. In $\triangle ABC$, prove that $(-a+b+c)(-a-b+c)+(a-b+c)(a+b-c)+$ $(a+b-c)(-a+b+c) \leqslant \sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c}) \cdot(2001$ Romanian National Team Problem) | 17. Let $a=y+z, b=z+x, c=x+y$.
$$\begin{array}{l}
(-a+b+c)(a-b+c)+(a-b+c)(a+b-c)+ \\
\quad(a+b-c)(-a+b+c) \leqslant \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})
\end{array}$$
This is equivalent to
$$4 x y+4 y z+4 z x \leqslant \sqrt{(y+z)(z+x)(x+y)}(\sqrt{y+z}+\sqrt{z+x}+\sqrt{x+y})$$
By the Cauchy-Schwarz inequality, we... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,525 |
18. In $\triangle A B C$, prove that $\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b} \leqslant \sqrt{a}+\sqrt{b}+$ $\sqrt{c}$. (1996 Asia Pacific Mathematical Olympiad Problem) | 18. Let $a=y+z, b=z+x, c=x+y \cdot \sqrt{a+b-c}+\sqrt{b+c-a}+$ $\sqrt{c+a-b} \leqslant \sqrt{a}+\sqrt{b}+\sqrt{c}$ is equivalent to
$$\begin{array}{l}
\sqrt{2 x}+\sqrt{2 y}+\sqrt{2 z} \leqslant \sqrt{y+z}+\sqrt{z+x}+\sqrt{x+y} \Leftrightarrow \\
2(x+y+z)+4 \sqrt{x y}+4 \sqrt{y z}+4 \sqrt{x z} \leqslant \\
2(x+y+z)+2 \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,526 |
19. Given that $a, b, c$ are the three sides of $\triangle ABC$, prove:
(1) The sides $\sqrt{a}, \sqrt{b}, \sqrt{c}$ can form a triangle;
(2) $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \leqslant a + b + c < 2(-\sqrt{ab} + \sqrt{bc} + \sqrt{ca})$.
Romanian Mathematical Olympiad Problem | 19. (1)=Since $a+b \rightarrow c$, therefore, $(\sqrt{a}+\sqrt{b})^{2}=a+b+2 \sqrt{a b}>c+2 \sqrt{a b}>$ $c$, hence, $\sqrt{a}+\sqrt{b}>\sqrt{c}$.
Similarly, we can prove $\sqrt{b}+\sqrt{c}>\sqrt{a}, \sqrt{c}+\sqrt{a}>\sqrt{b}$. Therefore, $\sqrt{a}, \sqrt{b}, \sqrt{c}$ can form the sides of a triangle.
(2) The left i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,527 |
22. In $\triangle A B C$, prove: $\frac{\sin ^{2} A}{a}+\frac{\sin ^{2} B}{b}+\frac{\sin ^{2} C}{c} \leqslant \frac{s^{2}}{a b c}$, where $s=\frac{a+b+c}{2}$. | 22. By the Law of Sines, we have
$$\begin{array}{l}
\frac{\sin ^{2} A}{a}+\frac{\sin ^{2} B}{b}+\frac{\sin ^{2} C}{c} \leqslant \frac{s^{2}}{a b c} \Leftrightarrow \\
a b c\left(\frac{\sin ^{2} A}{a}+\frac{\sin ^{2} B}{b}+\frac{\sin ^{2} C}{c}\right) \leqslant \frac{1}{4}(a+b+c)^{2} \Leftrightarrow \\
a b c\left(\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,530 |
24- (1)- In a non-obtuse triangle $ABC$, prove that: $\sin A + \sin B + \sin C > \cos A + \cos B + \cos C$, (1976 Yugoslav Mathematical Olympiad)
(2) $=$ In a non-obtuse triangle $ABC$, prove that: $1 < \sin A + \sin B + \sin C - (\cos A + \cos B + \cos C) \leqslant \frac{3 \sqrt{3} - 3}{2}$ | 24. (1) When $A, B, C \leqslant 90^{\circ}$, we have
$$\begin{array}{l}
\cos A+\cos B+\cos C=\cos \frac{A+B}{2} \cos \frac{A-B}{2}+ \\
\cos \frac{B+C}{2} \cos \frac{B-C}{2}+\cos \frac{C+A}{2} \cos \frac{C-A}{2}\cos \frac{A}{2}. \text{ Thus, }
$$\begin{array}{l}
\sin A+\sin B+\sin C-(\cos A+\cos B+\cos C)= \\
\sin A-\co... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,532 |
31. (1) Given that $a_{i}(i=1,2, \cdots, n)$ are real numbers, prove the inequality: $\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{a_{i} a_{j}}{i+j} \geqslant 0$. (1992 Polish Mathematical Olympiad Problem)
(2) Given that $a_{i}(i=1,2, \cdots, n)$ are real numbers, prove the inequality: $\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{a_{i... | 31. (1) Let $f(x)=\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j} x^{i+j-1}$. Then
$$x f(x)=\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} a_{j} x^{i+j}=\sum_{i=1}^{n} a_{i} x^{i} \sum_{j=1}^{n} a_{j} x^{j}=\left(\sum_{i=1}^{n} a_{i} x^{i}\right)^{2} \geqslant 0$$
Therefore, when $x \geqslant 0$, $f(x) \geqslant 0$. Thus,
$$\int_{0}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,533 |
25. Given that $a, b, c$ are the three sides of $\triangle ABC$, prove that $3 \leqslant \frac{a^{2}+b^{2}}{a b+c^{2}}+\frac{b^{2}+c^{2}}{b c+a^{2}}+\frac{c^{2}+a^{2}}{c a+b^{2}}<$
4. (2002 Romanian Olympiad Problem) | 25. By the mean inequality, we have
$$\begin{array}{l}
\frac{a^{2}+b^{2}}{a b+c^{2}}+\frac{b^{2}+c^{2}}{b c+a^{2}}+\frac{c^{2}+a^{2}}{c a+b^{2}} \geqslant \frac{a^{2}+b^{2}}{\frac{1}{2}\left(a^{2}+b^{2}\right)+c^{2}}+ \\
\frac{b^{2}+c^{2}}{\frac{1}{2}\left(b^{2}+c^{2}\right)+a^{2}}+\frac{c^{2}+a^{2}}{\frac{1}{2}\left(c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,534 |
26. (1) Given that $a, b, c$ are the three sides of $\triangle ABC$, prove: $2<\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \leq \frac{a^{3}+b^{3}+c^{3}}{a b c} \leqslant 3 \cdot(2001$ Austrian and Polish Mathematical Olympiad problem)
(2) Given that $a, b, c$ are the three sides of $\triangle ABC$, prove: $\frac{a}{b+c}+... | 26. (1) The left side is equivalent to $(b+c-a)(c+a-b)(a+b-c)>0$, the right side is equivalent to $(b+c-a)(c+a-b)(a+b-c) \leqslant a b c$ (Schur's inequality).
(2) It is equivalent to $a^{3}+b^{3}+c^{3}+3 a b c-a^{2} b-b^{2} a-a^{2} c-c^{2} a-a c^{2} \geqslant 0$ (Schur's inequality). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,535 |
29. The three sides of triangle $T_{1}$ are $a, b, c$, and its area is $P$. The three sides of triangle $T_{2}$ are $u, v, w$, and its area is $Q$. Prove the inequality: $16 P Q \leqslant a^{2}\left(-u^{2}+v^{2}+w^{2}\right)+b^{2}\left(u^{2}-v^{2}+\right.$ $\left.w^{2}\right)+c^{2}\left(u^{2}+v^{2}-w^{2}\right) \cdot(1... | 29. 16PQ $\leqslant a^{2}\left(-u^{2}+v^{2}+w^{2}\right)+b^{2}\left(u^{2}-v^{2}+w^{2}\right)+c^{2}\left(u^{2}+v^{2}-w^{2}\right) \Leftrightarrow$
$$\begin{array}{l}
16 P Q+2\left(a^{2} u^{2}+b^{2} v^{2}+c^{2} w^{2}\right) \leqslant \\
\left(a^{2}+b^{2}+c^{2}\right)\left(u^{2}+v^{2}+w^{2}\right)
\end{array}$$
By Heron'... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,538 |
30. Let $a, b, c$ be the three sides of $\triangle ABC$, and let $x, y, z$ be three numbers satisfying the equation $x+y+z=0$. Prove the inequality $a^{2} x y+b^{2} y z+c^{2} z x \leqslant 0$. (19th Lithuanian Mathematical Olympiad, 1988) | 30. Since from any three real numbers $x, y, z$, two can be chosen such that their product is not negative. For definiteness, let's assume $x y \geqslant 0$. Since $z=-(x+y)$, we have:
$$\begin{aligned}
a^{2} x y+b^{2} y z+c^{2} z x= & a^{2} x y-b^{2} y(x+y)-c^{2} x(x+y)= \\
& -c^{2} x^{2}-b^{2} y^{2}-\left(b^{2}+c^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,539 |
31 Find the smallest positive number $\lambda$, such that for any triangle with sides $a \geqslant \frac{b+c}{3}$, we have $a c + b c - c^{2} \leqslant \lambda\left(a^{2}+b^{2}+3 c^{2}+2 a b-4 b c\right) \cdot(1993$ China National Training Team Test Question $)$ | 31. $a^{2}+b^{2}+3 c^{2}+2 a b-4 b c=(a+b-c)^{2}+2 c^{2}+2 a c-2 b c=(a+b-$ $c)^{2}+2 c(a+c-b)$
Let
$$I=\frac{(a+b-c)^{2}+2 c(a+c-b)}{c(a+b-c)}=\frac{a+b-c}{2 c}+\frac{a+c-b}{a+b-c}$$
Since $a \geqslant \frac{b+c}{3}$, we have $a \geqslant \frac{1}{4}(a+b-c)+\frac{c}{2}$, thus, $a+c-b=2 a-(a+$ $b-c) \geqslant-\frac{1... | \frac{2 \sqrt{2}+1}{7} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,540 |
32. Let $a \leqslant b<c$ be the side lengths of a right triangle, find the largest constant $M$ such that $\frac{1}{a}+\frac{1}{b}+$ $\frac{1}{c} \geqslant \frac{M}{a+b+c}$. (1991 China National Training Team Test) | $$\begin{array}{l}
32 \text { Let } I=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+1+1=3+\frac{b}{a}+\frac{a}{b}+\frac{c}{a}+\frac{a}{c}+\frac{b}{c}+\frac{c}{b}\right. \\
\text { Since } \frac{b}{a}+\frac{a}{b} \geqslant 2, \frac{c}{a}+\frac{a}{c}=\frac{c}{a}+\frac{2 a}{c}-\frac{a}{c} \geqslant 2 \sqrt{2}-\frac{a}{c}, \frac{b}... | 5+3\sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,541 |
33. Given that $a, b, c$ are the three sides of $\triangle ABC$, and $p=\frac{1}{2}(a+b+c)$, prove the inequality:
$$a \sqrt{\frac{(p-b)(p-c)}{b c}}+b \sqrt{\frac{(p-c)(p-a)}{c a}}+c \sqrt{\frac{(p-a)(p-b)}{a b}} \geqslant p .$$ | 33. In $\triangle A B C$, $\sin \frac{A}{2}=\sqrt{\frac{1-\cos A}{2}}=\sqrt{\frac{(p-b)(p-c)}{b c}}$, by the Law of Sines we get
$$\frac{a}{b+c}=\frac{\sin A}{\sin B+\sin C}=\frac{2 \sin \frac{B+C}{2} \cos \frac{B+C}{2}}{2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}}=\frac{\cos \frac{B+C}{2}}{\cos \frac{B-C}{2}} \geqslant \c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,542 |
32. Let real numbers $a, b, c$ satisfy $a+b+c=3$, prove that: $\frac{1}{5 a^{2}-4 a+11}+\frac{1}{5 b^{2}-4 b+11}+$ $\frac{1}{5 c^{2}-4 c+11} \leqslant \frac{1}{4} .(2007$ China Western Mathematical Olympiad problem $)$ | 32. Let $f(x)=\frac{1}{5 x^{2}-4 x+11}$, then $f^{\prime}(x)=\frac{-2(5 x-2)}{\left(5 x^{2}-4 x+11\right)^{2}}$, so $f(x)$ is monotonically increasing on $\left(-\infty, \frac{2}{5}\right]$ and monotonically decreasing on $\left[\frac{2}{5},+\infty\right)$. Since $f^{\prime}(1)=-\frac{1}{24}$, the tangent line equation... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,544 |
35. Find all positive integers $k$ such that for any positive numbers $a, b, c$ satisfying the inequality $k(a b+b c+c a) > 5\left(a^{2}+b^{2}+c^{2}\right)$, there must exist a triangle with side lengths $a, b, c$. (2002 Year | 35. From $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geqslant 0$, we get $a^{2}+b^{2}+c^{2} \geqslant a b+b c+c a$, so, $k>5$, hence $k \geqslant 6$.
Since there does not exist a triangle with side lengths $1,1,2$, according to the problem, we have $k(1 \times 1+1 \times 2+1 \times 2) \leqslant 5\left(1^{2}+1^{2}+2^{2}\right)$, i... | 6 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,545 |
36. In an acute triangle $\triangle A B C$, prove: $\sin A+\sin B+\sin C>2$. | 36. When $A, B, C$ are $90^{\circ}, 90^{\circ}, 0^{\circ}$ respectively, i.e., $\triangle A B C$ is a degenerate triangle, $\sin A + \sin B + \sin C = 2$. Therefore, the problem becomes proving $\sin A + \sin B + \sin C > \sin 90^{\circ} + \sin 90^{\circ} + \sin 0^{\circ}$.
Below, the angles of $\triangle A B C$ are s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,546 |
38. $\alpha, \beta, \gamma$ are the three interior angles of $\triangle ABC$, prove: $\frac{1}{\sin \alpha}+\frac{1}{\sin \beta} \geqslant \frac{8}{3+2 \cos \gamma}$. And determine the condition for equality. (1996 Macedonian Mathematical Olympiad) | 38. It is not difficult to prove that $\sin \alpha \sin \beta=\cos ^{2} \frac{\alpha-\beta}{2}-\cos ^{2} \frac{\alpha+\beta}{2}$. In fact,
$$\begin{aligned}
\sin \alpha \sin \beta= & \frac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta)]= \\
& \frac{1}{2}\left[\left(2 \cos ^{2} \frac{\alpha-\beta}{2}-1\right)-\left(2 \co... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,548 |
39. In $\triangle A B C$, prove:
$$\frac{R}{2 r} \geqslant\left\{\frac{64 a^{2} b^{2} c^{2}}{\left[4 a^{2}-(b-c)^{2}\right]\left[4 b^{2}-(c-a)^{2}\right]\left[4 c^{2}-(a-b)^{2}\right]}\right\}^{2}$$ | 39. Let $a=y+z, b=z+x, c=x+y$, then $R=\frac{a b c}{4 S}, r=\frac{S}{p}$, where $S$ represents the area of the triangle, and $p$ is the semi-perimeter.
By Heron's formula $S^{2}=p(p-a)(p-b)(p-c)=x y z(x+y+z)$, so, $\frac{R}{r}=\frac{a b c p}{4 S^{2}}$.
$$\begin{array}{l}
\frac{R}{2 r} \geqslant\left(\frac{64 a^{2} b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,549 |
40. $I$ is the incenter of $\triangle ABC$, prove: $AI + BI + CI \leqslant 3R$, where $R$ is the circumradius of $\triangle ABC$. (2007 Moldova National Training Team Problem) | 40. Let $a=y+z, b=z+x, c=x+y \cdot p=\frac{a+b+c}{2}=x+y+z$,
$$\begin{aligned}
A I= & r \csc \frac{A}{2}=\frac{S}{p} \sqrt{\frac{2}{1-\cos A}}=\frac{2 S}{p} \sqrt{\frac{b c}{a^{2}-(b-c)^{2}}}= \\
& \frac{\sqrt{p(p-a)(p-b)(p-c)}}{p} \sqrt{\frac{b c}{(p-b)(p-c)}}= \\
& \sqrt{\frac{x(x+y)(x+z)}{x+y+z}}
\end{aligned}$$
Si... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,550 |
41. In $\triangle A B C$, prove: $\frac{3\left(a^{4}+b^{4}+c^{4}\right)}{\left(a^{2}+b^{2}+c^{2}\right)^{2}}+\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geqslant 2$. | 41. Let $a=y+z, b=z+x, c=x+y$, then
$$\begin{array}{l}
\frac{3\left(a^{4}+b^{4}+c^{4}\right)}{\left(a^{2}+b^{2}+c^{2}\right)^{2}}+\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geqslant 2 \Leftrightarrow \\
\left(x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x^{3} y+x y^{3}+y^{3} z+y z^{3}+z^{3} x+z x^{3} \geqslant\right. \\
3\left(x^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,551 |
42. In the right-angled $\triangle A B C$, find the largest positive real number $k$ such that the inequality $a^{3}+b^{3}+c^{3} \geqslant k(a+$ $b+c)^{3}$ holds. (2006 Iran Mathematical Olympiad) | 42. $a^{3}+b^{3}+c^{3} \geqslant k(a+b+c)^{3} \Leftrightarrow k \leqslant\left(\frac{a}{a+b+c}\right)^{3}+\left(\frac{b}{a+b+c}\right)^{3}+$ $\left(\frac{c}{a+b+c}\right)^{3}$
Let $x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$, then because $a^{2}+b^{2}=c^{2}$, so $x^{2}+y^{2}=z^{2}$
Also, because $x+y+z=1$... | \frac{3 \sqrt{2}-4}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,552 |
44. Given an acute triangle $\triangle ABC$ with side lengths $a, b, c$, and positive real numbers $x, y, z$ satisfying $\frac{a y z}{x}+\frac{b z x}{y}+\frac{c x y}{z}=P$, where $P$ is a given positive real number. Find the maximum value of $S=(b+c-a) x^{2}+(c+a-b) y^{2}+(a+b-c) z^{2}$, and determine the values of $x,... | 44. Since $\triangle A B C$ is an acute triangle, we have $a^{2}+b^{2}>c^{2}, b^{2}+c^{2}>a^{2}, c^{2}+a^{2}>b c^{2}$. Therefore, by the mean value inequality, we get
$$\begin{array}{l}
\left(b^{2}+c^{2}-a^{2}\right) x^{2}+\left(c^{2}+a^{2}-b^{2}\right) y^{2}+\left(a^{2}+b^{2}-c^{2}\right) z^{2} \leqslant \\
\frac{1}{2... | \frac{P^{2}}{a+b+c} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,554 |
33. Given that $a, b, c$ are positive numbers, and $a^{2}+b^{2}+c^{2}=1$, prove: $\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c} \geqslant \frac{3 \sqrt{3}+3}{2}$. | $$\begin{array}{l}
\text { 33. } f(x)=\frac{\sqrt{x}}{1-\sqrt{x}}, f(x) \text { at }\left(\frac{1}{3}, f\left(\frac{1}{3}\right)\right) \text { has the tangent line equation } y-\frac{\sqrt{3}+1}{2}= \\
\frac{6 \sqrt{3}+9}{4}\left(x-\frac{1}{3}\right) \\
\frac{\sqrt{x}}{1-\sqrt{x}}-\frac{\sqrt{3}+1}{2}-\frac{6 \sqrt{3}... | \frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c} \geqslant \frac{3 \sqrt{3}+3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 733,555 |
45. Let the semi-perimeter of $\triangle ABC$ be $p$, and $r_{a}, r_{b}, r_{c}$ be the radii of the excircles opposite to sides $BC, CA, AB$ respectively. Prove: $\frac{r_{a}}{\sin A}+\frac{r_{b}}{\sin B}+\frac{r_{c}}{\sin C} \geqslant 2 p$. (1993 Polish Mathematical Olympiad) | 45. In $\triangle A B C$, it is easy to get $r_{a}\left(\tan \frac{B}{2}+\tan \frac{C}{2}\right)=a$, and by the Law of Sines, $a=2 R \sin A$.
Therefore, $r_{a}=\frac{2 R \sin A \cos \frac{B}{2} \cos \frac{C}{2}}{\sin \left(\frac{B}{2}+\frac{C}{2}\right)}$, so $\frac{r_{a}}{\sin A}=\frac{2 R \cos \frac{B}{2} \cos \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,556 |
46. Let the area of $\triangle ABC$ be $S$, and the circumradius be $R$. Prove:
(1) $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \leqslant \frac{3 \sqrt{3}}{4 S}+\left(\frac{1}{a}-\frac{1}{b^{2}}\right)^{2}+\left(\frac{1}{b}-\frac{1}{c}\right)^{2}+\left(\frac{1}{c^{2}}-\frac{1}{a}\right)^{2}$;
(2) $\tan \frac{A}{2}... | 46. (1) Let $E=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}-\left(\frac{1}{a}-\frac{1}{b^{2}}\right)^{2}-\left(\frac{1}{b}-\frac{1}{c}\right)^{2}-\left(\frac{1}{c^{2}}-\frac{1}{a}\right)^{2}$, then
$$\begin{aligned}
E= & 2\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right)-\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,557 |
48. Given that $a, b, c$ are the three sides of $\triangle ABC$, prove:
(1) With $\sqrt{a}, \sqrt{b}, \sqrt{c}$ as the three sides, a $\triangle A_{1} B_{1} C_{1}$ can be formed;
(2) Let the area of $\triangle ABC$ be $S$, and the area of $\triangle A_{1} B_{1} C_{1}$ be $S_{1}$, prove the inequality: $S_{1}^{2} \geqsl... | 48. (1) Same as question 20;
(2) By the area formula of Qin Jiushao, we have $16 S^{2}=2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)$, so $16 S_{1}^{2}=2(a b+b c+c a)-\left(a^{2}+b^{2}+c^{2}\right)$, thus
$$\begin{array}{l}
\quad S_{1}^{2} \geqslant \frac{\sqrt{3} S}{4} \Leftrightarro... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,559 |
49. Given that $A, B, C$ are the three interior angles of $\triangle ABC$, prove: $\frac{\tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2}}{\sqrt{3}} \geqslant$ $\sqrt[6]{\tan ^{2} \frac{A}{2} \tan ^{2} \frac{B}{2} \tan ^{2} \frac{C}{2}}$. (2008 Northern China Mathematical Olympiad Invitational Competition Problem) | 49. In $\triangle A B C$,
$$\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$$
Therefore,
$$\left(\frac{\tan \frac{A}{2}+\tan \frac{B}{2}+\tan \frac{C}{2}}{\sqrt{3}}\right)^{2}=\frac{\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2}+2}{3}$$
Let $t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,560 |
50. Let the circumradius of $\triangle ABC$ be $R$, and the inradius be $r$. Prove: $\frac{\cos A}{\sin ^{2} A}+\frac{\cos B}{\sin ^{2} B}+$ $\frac{\cos C}{\sin ^{2} C} \geqslant \frac{R}{r} \cdot$ (2000 Beijing Mathematical Competition Problem) | 50. From the Law of Sines and the Law of Cosines, we have
\[
\frac{\cos A}{\sin ^{2} A}=\frac{\frac{b^{2}+c^{2}-a^{2}}{2 b c}}{\left(\frac{a}{2 R}\right)^{2}}=\frac{2 R^{2}}{a b c}\left(\frac{b^{2}+c^{2}-a^{2}}{a}\right)=\frac{2 R^{2}}{a b c}\left(\frac{b^{2}+c^{2}}{a}-a\right),
\]
symmetrically,
\[
\frac{\cos B}{\sin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,561 |
51. In $\triangle A B C$, the radii of the excircles opposite to sides $B C, C A, A B$ are $r_{a}, r_{b}, r_{c}$ respectively, and $2 p$ is the perimeter of $\triangle A B C$. Prove that: $\frac{r_{a}}{\sin A}+\frac{r_{b}}{\sin B}+\frac{r_{c}}{\sin C} \geqslant 2 p$. (1993 Poland-Austria Mathematical Olympiad) | 51. For $\triangle ABC$, the radius of the excircle opposite to side $BC$ is $r_{a}$, then
$$S_{\triangle ABC}=\frac{bc \sin A}{2}=\frac{(b+c-a) r_{a}}{2}$$
Therefore,
$$\begin{aligned}
r_{a}= & \frac{bc \sin A}{b+c-a}=\frac{2 R \sin B \sin C \sin A}{\sin B+\sin C-\sin A}= \\
& \frac{2 R \sin B \sin C \sin A}{2 \sin \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,562 |
52. Consider the inequality $a^{3}+b^{3}+c^{3}<k(a+b+c)(a b+b c+c a)$ in $\triangle A B C$, where $a, b, c$ are the side lengths of $\triangle A B C$, and $k$ is a real number. (1) Prove the inequality when $k=1$; (2) Find the largest real number $k$ for which the inequality holds. (2010 Albanian BMO Training Team Prob... | 52. (1) When $k=1$, $(a+b+c)(a b+b c+c a)-a^{3}+b^{3}+c^{3}>0$ is equivalent to
$$a^{2}(b+c-a)+b^{2}(c+a-b)+c^{2}(a+b-c)+3 a b c>0$$
(2) For problem (2), let $c=1, a=b=n$, then $k>\frac{2 n^{3}+1}{n(2 n+1)(n+2)} \rightarrow 1(n \rightarrow$ $+\infty)$, so the maximum value of $k$ is 1. | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 733,563 |
53. In an acute $\triangle ABC$, prove the inequality: $\sin ^{3} A \cos ^{2}(B-C)+\sin ^{3} B \cos ^{2}(C-A)+$ $\sin ^{3} C \cos ^{2}(A-B) \leqslant 3 \sin A \sin B \sin C$. And determine the condition for equality. (2002 China National Training Team Exam) | 53. We only need to prove the identity $\sin ^{3} A \cos (B-C)+\sin ^{3} B \cos (C-A)+$ $\sin ^{3} C \cos (A-B)=3 \sin A \sin B \sin C$.
In fact, $4 \sin ^{3} A \cos (B-C)=4 \sin ^{2} A \sin (B+C) \cos (B-C)=2 \sin ^{2} A$ $(\sin 2 B+\sin 2 C)=(1-\cos 2 A)(\sin 2 B+\sin 2 C)=\sin 2 B+\sin 2 C-\cos 2 A \sin 2 B-$ $\cos... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,564 |
54. In $\triangle A B C$, prove the inequality:
$$\left|\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}+\sqrt{\frac{b}{c}}-\sqrt{\frac{c}{b}}+\sqrt{\frac{c}{a}}-\sqrt{\frac{a}{c}}\right|<\frac{1}{10} .$$ | 54. Let $a \geqslant b \geqslant c$, and set $x=\sqrt{a}, y=\sqrt{b}, z=\sqrt{c}, t=\sqrt{b+c}$. Then $x \geqslant y \geqslant z>0$, $x=\sqrt{a} \leqslant \sqrt{b+c}=t, t=\sqrt{y^{2}+z^{2}}$. It is easy to see that
$$\left|\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}+\sqrt{\frac{b}{c}}-\sqrt{\frac{c}{b}}+\sqrt{\frac{c}{a}}-\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,565 |
34. Let $x, y, z$ be positive real numbers, and $x^{2}+y^{2}+z^{2}=3$, prove that: $\frac{x}{\sqrt{x^{2}+y+z}}+\frac{y}{\sqrt{y^{2}+z+x}}+$ $\frac{z}{\sqrt{z^{2}+x+y}} \leqslant \sqrt{3}$. (2008 Ukrainian Mathematical Olympiad Problem) | 34. Since $x^{2}+y^{2}+z^{2}=3$, by the mean value inequality we have $3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(x+$ $y+z)^{2}$.
Thus,
$$\frac{x}{\sqrt{x^{2}+y+z}} \leqslant \frac{x}{\sqrt{x^{2}+\sqrt{\frac{x^{2}+y^{2}+z^{2}}{3}}(y+z)}} \leqslant \frac{x}{\sqrt{x^{2}+\frac{x+y+z}{3}(y+z)}}$$
Since the degrees of the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,566 |
55. In $\triangle ABC$, consider the inequality: $a^{3}+b^{3}+c^{3}<k(a+b+c)(ab+bc+ca)$.
(1) When $k=1$, prove that the inequality holds;
(2) Find the smallest real number $k$ such that the inequality always holds. (2010 Albanian National Training Team) | 55. (1) When $k=1$, $(a+b+c)(a b+b c+c a)-\left(a^{3}+b^{3}+c^{3}\right)=a^{2}(b+$ $c-a)+b^{2}(c+a-b)+c^{2}(a+b-c)+3 a b c>0$, so $a^{3}+b^{3}+c^{3} & \frac{a^{3}+b^{3}+c^{3}}{(a+b+c)(a b+b c+c a)}= \\
& \frac{2\left(x^{3}+y^{3}+z^{3}\right)+3[x y(x+y)+y z(y+z)+z x(z+x)]}{2\left(x^{3}+y^{3}+z^{3}\right)+8[x y(x+y)+y z(... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 733,567 |
56. In $\triangle A B C$, $a+b+c=2$, prove: $\left|\frac{a^{3}}{b}+\frac{b^{3}}{c}+\frac{c^{3}}{a}-\frac{a^{3}}{c}-\frac{b^{3}}{a}-\frac{c^{3}}{b}\right|<3$. | 56. The inequality is equivalent to 1 $a^{4} c+b^{4} a+c^{4} b-a^{4} b-b^{4} c-c^{4} a l \leqslant 3 a b c \Leftrightarrow(a-b)$ $(b-c)(c-a)\left(a^{2}+b^{2}+c+a b+b c+c a\right) \mid \leqslant 3 a b \bar{c}$, let $a=y+z ; b=z+x$; $c=x+y$, then $x+y+z=1$, the inequality is equivalent to
$$\begin{array}{l}
(x-y)(y-z)(z-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,568 |
57. In $\triangle A B C$, prove: $\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{16}$. | 57. Let $a=\max \{a, b, c\}, a \geqslant b+c \geqslant \max b, c\}$, (when $a=b+c$, $\triangle A B C$ is a degenerate triangle, a line segment)
Because
$$\begin{aligned}
\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|= & \left|\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\r... | \left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right| < \frac{1}{16} | Inequalities | proof | Yes | Yes | inequalities | false | 733,569 |
58. In $\triangle A B C$, prove:
$$3 \sum_{c y c} a b(1+2 \cos C) \geqslant 2 \sum_{c y c} \sqrt{\left(c^{2}+a b(1+2 \cos C)\right)\left(b^{2}+c a(1+2 \cos B)\right)} \text {. }$$ | 58. From the cosine theorem $2 a b \cos C=a^{2}+b^{2}-c^{2}$, so $2 a b \cos C+2 b c \cos A+2 c a \cos B=a^{2}+b^{2}+c^{2}$, thus the original inequality is equivalent to $3\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\right) \geqslant$
$$\begin{array}{l}
2 \sum_{o c} \sqrt{\left(a^{2}+a b+b^{2}\right)\left(c^{2}+c a+a^{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,570 |
Example 1 Given that $D$ is any point on side $AB$ of $\triangle ABC$, point $E$ is any point on side $AC$, and connect $DE$. $F$ is any point on the line segment $DE$. Let $\frac{AD}{AB}=x$, $AE=y$, $\frac{DF}{DE}=z$, prove:
$$\begin{array}{l}
\quad(1) S_{\triangle B D F}=(1-x) y z S_{\triangle A B C}, \\
S_{\triangle... | Prove as shown in the figure, we have
$$\begin{aligned}
S_{\triangle B D F}=z S_{\triangle B D E}= & z(1-x) S_{\triangle A B E}=z(1-x) y S_{\triangle A B C} \\
S_{\triangle C E F}=(1-z) S_{\triangle C D E}= & (1-z)(1-y) S_{\triangle A A D}=(1-z)(1-y) x S_{\triangle A B C} \\
(2) \sqrt[3]{S_{\triangle B D F}}+\sqrt[3]{S... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,573 |
Example 3 Proof: The perimeter of any convex quadrilateral with an area of 1, plus the sum of its two diagonals, is not less than $4+2 \sqrt{2}$. (1985 Austrian and Polish Joint Mathematical Competition Problem) | Proof: Let $ABCD$ be any convex quadrilateral with an area of 1 (as shown in the figure). Then we have:
$$\begin{aligned}
1= & \frac{1}{2}(eg+gf+fh+he) \sin \alpha \leqslant \\
& \frac{1}{2}(eg+gf+fh+he)= \\
& \frac{1}{2}(e+f)(g+h) \leqslant \\
& \frac{1}{2}\left(\frac{e+f+g+h}{2}\right)^{2}
\end{aligned}$$
Thus, the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,575 |
Example 4 (Erdös - Mordell Inequality) Let $P$ be any point inside $\triangle ABC$, and let the distances from $P$ to the sides $BC$, $CA$, and $AB$ be $PB=p$, $PE=q$, and $PF=r$, respectively. Denote $PA=x$, $PB=y$, and $PC=z$. Then $x+y+z \geqslant 2(p+q+r)$.
- Equality holds if and only if $\triangle ABC$ is an equi... | Prove that, as shown in the figure, with the angle bisector of $\angle B$ as $A^{\prime} C^{\prime}$, and connecting $P A^{\prime}$ and $P G^{\prime}$, in $\triangle B A^{\prime} C^{\prime}$, it is easy to obtain
$$S_{\triangle B A^{\prime} P}+S_{\triangle B C^{\prime} P} \leqslant \frac{1}{2} B P \cdot A^{\prime} C^{\... | x+y+z \geqslant 2(p+q+r) | Inequalities | proof | Yes | Yes | inequalities | false | 733,576 |
Example 6 Let $ABCD$ be a convex quadrilateral with an inscribed circle, and each of its interior and exterior angles is not less than $60^{\circ}$. Prove: $\left.\frac{1}{3} \right\rvert\, AB^{3}-$ $AD^{3}|\leqslant| BC^{3}-CD^{3}|\leqslant 3| AB^{3}-AD^{3} \mid$. When does equality hold? (33rd United States of Americ... | Prove using the cosine theorem, we know
$$\begin{aligned}
B D^{2}= & A D^{2}+A B^{2}-2 A D \cdot A B \cos \angle D A B= \\
& C D^{2}+B C^{2}-2 C D \cdot B C \cos \angle D C B
\end{aligned}$$
From the given conditions, we know \(60^{\circ} \leqslant \angle D A B, \angle D C B \leqslant 120^{\circ}\), hence \(-\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,579 |
Example 8 Let circles $K$ and $K_{1}$ be concentric, with radii $R, R_{1}, R_{1}>R$, respectively. Quadrilateral $ABCD$ is inscribed in circle $K$, and quadrilateral $A_{1}B_{1}C_{1}D_{1}$ is inscribed in circle $K_{1}$. Points $A_{1}, B_{1}, C_{1}, D_{1}$ lie on the rays $CD, DA, \bar{A}B$, and $BC$, respectively. Pro... | To make writing convenient, let $AB=a, BC=b, CD=c, DA=d, AB_{1}=e; BC_{i}=f$, $CD_{1}=g, DA_{1}=h$, then
$$\begin{array}{c}
S_{\triangle AB_{1}C_{1}}=\frac{1}{2}(a+f) e \sin \angle B_{1}AC_{1} \\
S_{ABCD}=S_{\triangle ABD}+S_{\triangle BDC}=\frac{1}{2} ad \sin \angle DCB+\frac{F}{2} bc \sin \angle B_{1}AC_{1}
\end{arra... | \frac{S_{A_{1}B_{1}C_{1}D_{1}}}{S_{ABCD}} \geqslant \frac{R_{1}^{2}}{R^{2}} | Geometry | proof | Yes | Yes | inequalities | false | 733,581 |
Example 9 Given a quadrilateral $A_{1} A_{2} A_{3} A_{4}$ that has both an inscribed circle and a circumscribed circle, the inscribed circle touches the sides $A_{1} A_{2}$, $A_{2} A_{3}$, $A_{3} A_{4}$, $A_{4} A_{1}$ at points $B_{1}$, $B_{2}$, $B_{3}$, $B_{4}$ respectively. Prove: $\left(\frac{A_{1} A_{2}}{B_{1} B_{2... | Proof: As shown in the figure, let the radius of the incircle of quadrilateral \(A_{1} A_{2} A_{3} A_{4}\) be \(r\), and denote
\[
\begin{array}{l}
\angle A_{1} A_{2} A_{3}=2 \bar{\alpha} \\
\angle A_{2} A_{3} A_{4}=2 \beta \\
\angle A_{3} A_{4} A_{1}=2 \gamma \\
\angle A_{3} A_{4} A_{1} A_{2}=2 \theta
\end{array}
\]
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,582 |
For example, in $\triangle ABC$, the radius of the circumcircle $K$ is $R$, and the internal angle bisectors intersect the circle $K$ at $A_{1}, B_{1}, C_{1}$. Prove: $16 Q^{3} \geqslant 27 R^{4} P$. Here, $Q$ and $P$ are the areas of $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$, respectively. (30th | Proof: Let the internal angles of $\triangle ABC$ be $\alpha, \beta, \gamma$. Then
\[ P = \frac{1}{2} R^{2} (\sin 2 \alpha + \sin 2 \beta + \sin 2 \gamma). \]
Since the internal angles of $\triangle A_{1} B_{1} C_{1}$ are $\frac{\beta+\gamma}{2}, \frac{\gamma+\alpha}{2}, \frac{\alpha+\beta}{2}$, we have
\[ Q = \frac{... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,584 |
Example 13 Draw a line through point $P(a, b)(a>0, b>0)$ intersecting the $x$-axis and $y$-axis at points $M, N$ respectively. Try to find the range of values for $O M+O N-M N$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Let $\angle P M O=\theta\left(0-1\right)$ be monotonically decreasing on $(-1$, $\left.\sqrt{\frac{2 a}{b}}-1\right]$ and monotonically increasing on $\left[\sqrt{\frac{2 a}{b}}-1,+\infty\right)$. Considering that $0<t<1$, we discuss in three cases:
(1) When $1<\frac{2 a}{b}<4$, we have $\frac{a}{2}<b<2 a$, thus $0<\s... | not found | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,586 |
1. Given that $D$ is any point on side $A B$ of $\triangle A B C$ with area 1, $E$ is any point on side $A C$, and $F$ is any point on segment $D E$. Let $\frac{A D}{A B}=x, \frac{A E}{A C}=y, \frac{D F}{D E}=z$, and $y+z-x=\frac{1}{2}$. Try to find the maximum area of $\triangle B D F$. (2005 Hunan Province Mathematic... | 1. Connect $B E$, then the area of $\triangle B D F$
$$S_{\triangle B D F}=z S_{\triangle B D E}=z(1-x) S_{\triangle A B D}=z(1-x) y S_{\triangle A B C}=z(1-x) y$$
By the AM-GM inequality, we get $z(1-x) y \leqslant\left(\frac{z+(1-x)+y}{3}\right)^{3}=\frac{1}{8}$. Equality holds if and only if $z=1-x=y, y+z-x=\frac{1... | \frac{1}{8} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,587 |
36. Given that $a, b, c$ are real numbers, and $a+b+c=2, ab+bc+ca=1$, prove that: $\max \{a, b, c\} - \min \{a, b, c\} \leqslant \frac{2}{\sqrt{3}}$. (2008 Ukrainian Mathematical Olympiad Problem) | 36. From $a+b+c=2, ab+bc+ca=1$ eliminating $b$ we get $a^{2}+ac+c^{2}-2(a+c)+1=0$. Assuming $a \leqslant b \leqslant c$, let $c=a+x(x \geqslant 0)$, then
$$3 a^{2}+(3 x-4) a+(x-1)^{2}=0$$
Since $a$ is a real number, the discriminant
$$\Delta=(3 x-4)^{2}-12(x-1)^{2}=4-3 x^{2} \geqslant 0$$
Solving this, we get $x \leq... | \frac{2}{\sqrt{3}} | Inequalities | proof | Yes | Yes | inequalities | false | 733,588 |
4. If a triangle and a rectangle have equal perimeters and areas, they are called "twin." Prove: For a given triangle, there exists a "twin" rectangle that is not a square, and the ratio of the longer side to the shorter side is at least $\lambda-1+\sqrt{\lambda(\lambda-2)}$, where $\lambda=\frac{3 \sqrt{3}}{2}$. (2003... | 4. Let the side lengths of a rectangle be $x, y$, the semi-perimeter of a triangle be $p$, and the area be $S$. Then the necessary and sufficient condition for this triangle and rectangle to be "twin" is
$$\left\{\begin{array}{l}
x+y=p \\
x y=S
\end{array}\right.$$
The necessary and sufficient condition for the system... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,591 |
5. Given that $I$ is the incenter of $\triangle ABC$, and $AI, BI, CI$ intersect $BC, CA, AB$ at $A', B', C'$ respectively. Prove that $\frac{1}{4}<\frac{AI \cdot BI \cdot CI}{AA' \cdot BB' \cdot CC'} \leqslant \frac{8}{27}$. (32nd IMO Problem) | 5. Let $BC=a, CA=b, AB=c$, it is easy to know $\frac{BA^{\prime}}{A^{\prime}C}=\frac{c}{b^{\prime}}$, so $BA^{\prime}=\frac{ca}{b+c}, \frac{AI}{IA^{\prime}}=\frac{AB}{BA^{\prime}}=$ $\frac{b+c}{a}, \frac{\bar{A}I}{AA^{\prime}}=\frac{b+c}{a+b+c}$
Similarly, $\frac{BI}{BB^{\prime}}=\frac{c+a}{a+b+c}, \frac{CI}{CC^{\prim... | \frac{1}{4}<\frac{AI \cdot BI \cdot CI}{AA' \cdot BB' \cdot CC'} \leqslant \frac{8}{27} | Geometry | proof | Yes | Yes | inequalities | false | 733,592 |
6. Let $C_{1}, C_{2}$ be concentric circles, with the radius of $C_{2}$ being twice that of $C_{1}$. Quadrilateral $A_{1} A_{2} A_{3} A_{4}$ is inscribed in circle $C_{1}$. Extend $A_{4} A_{1}$ to intersect circle $C_{2}$ at $B_{1}$, extend $A_{1} A_{2}$ to intersect circle $C_{2}$ at $B_{2}$, extend $A_{2} A_{3}$ to i... | 6. Let the common center be $O$, connect $O A_{1}-O B_{1}$ and $O B_{2}$, in quadrilateral $O A_{1} B_{1} B_{2}$,
$$\begin{array}{c}
O B_{1} \cdot A_{1} B_{2} \leqslant O A_{1} \cdot B_{1} B_{2}+O B_{2} \cdot A_{1} B_{1} \\
O B_{1}=O B_{2}=2 O A_{1}
\end{array}$$
Thus,
$$2 A_{1} B_{2} \leqslant B_{1} B_{2}+2 A_{1} B_{... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,593 |
9. Let $\triangle ABC$ be an acute-angled triangle, and let $M, N, P$ be the feet of the perpendiculars from the centroid $G$ of $\triangle ABC$ to the sides $AB, BC$, and $CA$, respectively. Prove that $\frac{4}{27}<\frac{S_{\triangle MNP}}{S_{\triangle ABC}} \leqslant \frac{1}{4}$. (16th Balkan Mathematical Olympiad ... | 9. Let the altitudes from $A, B, C$ to the opposite sides $BC, CA, AB$ of $\triangle ABC$ be $h_a, h_b, h_c$ respectively. As shown in the figure, we know that $GN = \frac{1}{3} h_a$, $GP = \frac{1}{3} h_b$, $GM = \frac{1}{3} h_c$, and $\angle NGP = 180^\circ - \angle C$, $\angle MGP = 180^\circ - \angle A$, $\angle MG... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,596 |
10. Let the three sides of $\triangle ABC$ be $a, b, c$, and $P$ be any point inside $\triangle ABC$. $A', B', C'$ (different from $P$) are the intersections of lines $AP, BP, CP$ with the circumcircles of $\triangle BCP, \triangle CAP, \triangle ABP$, respectively. Let the perimeter of hexagon $AB'C A' B C'$ be $p$. P... | 10. It is known that $\angle B C A^{\prime}=\angle B P A^{\prime}=\angle A C^{\prime} B=$ $\angle A P B^{\prime}=\angle A C B^{\prime}$. Similarly, $\angle C A B^{\prime}=\angle C A^{\prime} B=$ $\angle B A C^{\prime}, \angle A B C^{\prime}=\angle A B^{\prime} C=\angle C B A^{\prime}$, then $\triangle A^{\prime} B C \s... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,597 |
11. Let the circle $C_{a}$ be the circle that is internally tangent to the circumcircle of $\triangle ABC$ and tangent to the sides $AB$ and $AC$, and let $r_{a}$ be the radius of circle $C_{a}$, and $r$ be the radius of the incircle of $\triangle ABC$. Similarly define $r_{b}$ and $r_{c}$. Prove: $r_{a} + r_{b} + r_{c... | 11. Let $O_{a}, O_{b}, O_{c}$ be the centers of circles $C_{a}, C_{b}, C_{c}$. Denote $M, N$ as the projections of circle $O_{a}$ on $A B, A C$, respectively. Then the incenter $I$ of $\triangle A B C$ is the midpoint of $M N$. Let $X, Y$ be the projections of $I$ on $A B, A C$, respectively. We have
$$\frac{r_{a}}{r}=... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,598 |
37. Given that $a, b, c$ are non-negative real numbers, and $a+b+c \leqslant 3$, find the minimum value of $\frac{a+1}{a(a+2)}+\frac{b+1}{b(b+2)}+$ $\frac{c+1}{c(c+2)}$. (2003 Austrian Mathematical Olympiad Problem) | 37. Let $x=a+1, y=b+1, z=c+1$, then the given conditions become $x, y, z \geqslant 1, x+y+z \leqslant 6$, and the problem becomes finding the minimum value of $\frac{x}{x^{2}-1}+\frac{y}{y^{2}-1}+\frac{z}{z^{2}-1}$.
When $x \geqslant 1$, it is easy to prove that $\frac{x}{x^{2}-1} \geqslant \frac{16-5 x}{9}$.
In fact,... | 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,599 |
12. Given a convex quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ are perpendicular and intersect at point $O$. Let the incenter of $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$ be $O_{1}, O_{2}, O_{3}, O_{4}$, respectively. Prove:
(1) The sum of the diameters of $\odot O_{1}, \odot O_{2}, \odot O_{3... | Then
$$d \leqslant a+b-\frac{a+b}{\sqrt{2}}=\frac{2-\sqrt{2}}{2}(a+b)$$
As shown in the figure, let the diameters of $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}$ be $d_{1}, d_{2}, d_{3}, d_{4}$, respectively, then
12. (1) As shown in the figure, let the lengths of the three sides of the right triangle be $a, b... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,600 |
13. In the tetrahedron $OABC$, the edges $OA$, $OB$, and $OC$ are pairwise perpendicular, $r$ is the radius of its inscribed sphere, and $H$ is the orthocenter of $\triangle ABC$. Prove: $OH \leqslant r(\sqrt{3}+1)$. (2003 Romanian Mathematical Olympiad) | 13. From $O C \perp O A, O C \perp O B$, we know $O C \perp$ plane $O A B$. Therefore, $O C \perp A B$.
Also, $C H \perp A B$, so, plane $O C H \perp A B, O H \perp A B$. Similarly, $O H E A C$. Thus, $O H \perp$ plane $A B C$.
Let $O A=a, O B=b, O C=c$, then it is not difficult to obtain
$$\begin{aligned}
\left(S_{\... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,601 |
14. Given $\triangle A B C:$ (1) If $M$ is any point in the plane, prove: $A M \cdot \sin A \leqslant B M \cdot \sin B + C M \cdot \sin C$
(2) Let points $A_{1}, B_{1}, C_{1}$ be on sides $B C, A C, A B$ respectively, and the interior angles of $\triangle A_{1} B_{1} C_{1}$ are $\alpha, \beta, \gamma$ respectively, pro... | 14. (1) In quadrilateral $A B M C$, applying the generalized Ptolemy's theorem (Ptolemy's inequality) we get
$$A M \cdot B C \leqslant B M \cdot A C + C M \cdot A B$$
In $\triangle A B C$, by the Law of Sines we have
$$A M \cdot 2 R \sin A \leqslant B M \cdot 2 R \sin B + C M \cdot 2 R \sin C$$
which simplifies to
$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,602 |
15. Let $D$ be a point inside an acute $\triangle ABC$. Prove that: $D A \cdot D B \cdot A B + D B \cdot D C \cdot B C + D C \cdot D A \cdot C A \geqslant A B \cdot B C \cdot C A$, equality holds if and only if $D$ is the orthocenter of $\triangle ABC$. (1998 CMO | $A D E F$ are all parallelograms.
Connecting $B F$ and $A E$, it is clear that $B C A F$ is also a parallelogram, thus, $\square$
$$A F=E D=B C, E F=A D, E B=C D, B F=A C$$
In quadrilaterals $A B E F$ and $A E B D$, by Ptolemy's inequality we get
$$\begin{array}{l}
A B \cdot E F+A F \cdot B E \geqslant A E \cdot B F ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,603 |
18. Let $M$ be a point inside $\triangle ABC$, and let $P, Q, R$ be the points of intersection of line $AM$ with $BC$, line $BM$ with $AC$, and line $CM$ with $AB$, respectively. Prove that: $\frac{A M}{M P} \cdot \frac{B M}{M Q} \cdot \frac{C M}{M R} \geqslant 8$. (1997 Macedonian Mathematical Olympiad | 18. From the area relationship of the triangle, we have
$$\begin{array}{l}
\frac{M P}{A P}+\frac{M Q}{B Q}+\frac{M R}{C R}=\frac{S_{\triangle M B C}}{S_{\triangle A B C}}+\frac{S_{\triangle M C A}}{S_{\triangle A B C}}+\frac{S_{\triangle M A B}}{S_{\triangle A B C}}= \\
\frac{S_{\triangle A B C}}{S_{\triangle A B C}}=1... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,606 |
19. Let $\triangle ABC$ be an equilateral triangle, and $P$ be a point inside it. The extensions of segments $AP, BP, CP$ intersect the opposite sides $BC, CA, AB$ at points $A_1, B_1, C_1$ respectively. Prove that: $A_1 B_1 \cdot B_1 C_1 \cdot C_1 A_1 \geqslant A_1 B \cdot B_1 C \cdot C_1 A = ($ 37th IMO Shortlist Pro... | 19. By the cosine theorem, $A_{1} B_{1}^{2}=A_{1} C^{2}+B_{1} C^{2}-A_{1} C \cdot B_{1} C \geqslant 2 A_{1} C \cdot B_{1} C - A_{1} C - B_{1} C = A_{1} C \cdot B_{1} C$. Similarly, $B_{1} C_{1}^{2} \geqslant B_{1} A_{1} \cdot C_{1} A$, $C_{1} A_{1}^{2} \geqslant C_{1} B \cdot A_{1} B$. By Ceva's theorem, we have
$$\fra... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,607 |
20- In $\triangle ABC$, the three angle bisectors of $\angle A, \angle B$, and $\angle C$ intersect the circumcircle of $\triangle ABC$ at points $A_{1}, B_{1}$, and $C_{1}$, respectively. Prove that $A A_{1} + B B_{1} + C C_{1} > AB + BC + CA$. (1982 Australian Math Olympiad) | 20. We prove $A \bar{A}_{1}>\frac{A B+A C}{2}$. In fact, by Ptolemy's theorem, we have $A A_{1} \cdot B C=A B \cdot A_{1} \bar{C}+A C \cdot A_{1} B$.
Noting that the inscribed angle $\angle B A A_{1}=\angle C A A_{1}$, therefore, $A_{1} B=$ $-A_{1} C=x$ Thus, from $2 x=A_{1} B+A_{1} C>B C$, we have $2 A A_{1}=$ $2 \fr... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,608 |
21. In trapezoid $ABCD$ with lower base $AB$ having two fixed points $M, N$, and a moving point $P$ on the upper base $CD$. Let $E=DN \cap AP$, $F=DN \cap MC$, $G=MC \cap PB$, and $DP=\lambda DC$. For what value of $\lambda$ is the area of quadrilateral PEFG maximized? (1988 National Training Team Selection Test) | 21. Take $DC$ as the unit length, i.e., $DC=1$, then $DP=\lambda, DC=1-\lambda$, let $AM=a, MN=b, NB=c$, and the height of the trapezoid be $h$.
Draw a line through $C$ parallel to $PB$ intersecting the extension of $AB$ at $Q$, as shown in the figure. Then $PCQB$ is a parallelogram, so $BQ=PC=1-\lambda$. Thus,
$$\beg... | \lambda = \frac{AN}{AN+MB} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,609 |
38. Given that $x, y, z$ are positive real numbers, and $x^{2}+y^{2}+z^{2}=1$, prove that $\frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}} \leqslant \frac{3 \sqrt{3}}{4} \cdot$ (1998 Bosnia and Herzegovina Mathematical Olympiad) | 38. Let $a=x^{2}, b=y^{2}, c=z^{2}$, the problem is equivalent to proving that under the condition $a+b+c=1$,
$$\frac{\sqrt{a}}{1+a}+
\frac{\sqrt{b}}{1+b}+\frac{\sqrt{c}}{1+c} \leqslant \frac{3 \sqrt{3}}{4}$$
Consider the function $y=f(x)=\frac{\sqrt{x}}{1+x}$ at the point $\left(\frac{1}{3}, \frac{\sqrt{3}}{4}\right... | \frac{3 \sqrt{3}}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 733,610 |
23. $A^{\prime}, B^{\prime}, C^{\prime}$ are three points on the sides $BC, CA, AB$ of $\triangle ABC$, respectively. Prove: $S_{\triangle ABC} \cdot S_{\triangle A^{\prime} B^{\prime} C^{\prime}} \geqslant 4 S_{\triangle AB^{\prime} C^{\prime}} \cdot S_{\triangle A^{\prime} B C^{\prime}} \cdot S_{\triangle A^{\prime} ... | $$\begin{array}{l}
\text { 23. Let } x=\frac{B A^{\prime}}{A^{\prime} C}, y=\frac{C B^{\prime}}{B^{\prime} A}, z=\frac{A C^{\prime}}{C B^{\prime},} \text { then } \frac{S_{\triangle A B^{\prime} C}}{S_{\triangle A B C}}=\frac{z}{(1+z)(1+y)}, \frac{S_{\triangle A^{\prime} B C^{\prime}}}{S_{\triangle A B C}}= \\
\frac{x}... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,612 |
24. $P$ is a point inside $\triangle ABC$, and the distances to $BC, CA, AB$ are $p, q, r$ respectively. $R$ is the circumradius of $\triangle ABC$. Prove that: $\frac{a^{2}+b^{2}+c^{2}}{18 \sqrt[3]{p q r}} \geqslant R$. (2005 Taiwan Training Team Problem) | 24. Obviously, $2 S=a p+b q+c r \geqslant 3 \sqrt[3]{a b c p q r}$, so, $\frac{a^{2}+b^{2}+c^{2}}{18 \sqrt[3]{p q r}} \geqslant$ $\frac{\left(a^{2}+b^{2}+c^{2}\right) \sqrt[3]{a b c}}{12 S}$, and $R=\frac{a b c}{4 S}$, so it is sufficient to prove $\left(a^{2}+b^{2}+c^{2}\right) \sqrt[3]{a b c} \geqslant 3 a b c$, whic... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,613 |
25. I is the incenter of $\triangle ABC$, prove: $I A^{2}+I B^{2}+I C^{2} \geqslant \frac{B C^{2}+C A^{2}+A B^{2}}{3}$. | $$\begin{array}{l}
25 I A=r \csc \frac{A}{2}, I B=r \csc \frac{B}{2}, I C=r \csc \frac{C}{2}, B C=r\left(\cot \frac{B}{2}+\cot \frac{C}{2}\right), C A= \\
r\left(\cot \frac{C}{2}+\cot \frac{A}{2}\right), A B=r\left(\cot \frac{A}{2}+\cot \frac{B}{2}\right), \\
I A^{2}+I B^{2}+I C^{2} \geqslant \frac{B C^{2}+C A^{2}+A B^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,614 |
27. Given that $\triangle A_{1} A_{2} A_{3}$ is an acute triangle, $O$ and $H$ are the circumcenter and orthocenter of $\triangle A_{1} A_{2} A_{3}$, respectively. For $1 \leqslant i \leqslant 3$, points $P_{i}$ and $Q_{i}$ lie on segments $O A_{i}$ and $A_{i+1} A_{i+2}$ (where $A_{i+3}=A_{i}$), respectively, such that... | 27. As shown in the figure, let the midpoint of $A_{2} A_{3}$ be $O_{1}$, then $O O_{1} \perp A_{2} A_{3}$, because $O$ is the circumcenter of $\triangle A_{1} A_{2} A_{3}$, so $\angle A_{2} O O_{1}=\angle A_{1}, \angle A_{2} A_{1} O=90^{\circ}-A_{3}, \angle H_{1} A_{1} O=90^{\circ}-A_{2}-\left(90^{\circ}-\right.$ $\le... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,616 |
28. In $\triangle A B C$, $C D$ is the internal angle bisector of $\angle C$, $S$ is the area of $\triangle A B C$, prove: $2 S\left(\frac{1}{A D}-\frac{1}{B D}\right) \leqslant A B$. (2004 Austrian-Polish Mathematical Olympiad Problem) | 28. Since $CD$ is the internal angle bisector of $\angle C$, by the Angle Bisector Theorem, $\frac{AD}{DB} = \frac{AC}{BC} = \frac{b}{a}$. Noting that $AD + DB = a$, we have:
$$AD = \frac{bc}{a+b}, \quad DB = \frac{ac}{a+b}$$
Let $S$ be the area of $\triangle ABC$, then $2S = ab \sin C$,
$$\begin{array}{l}
2S\left(\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,617 |
29. Let $P$ be any point on the plane of acute-angled $\triangle ABC$, and let $u, v, w$ be the distances from $P$ to $A, B, C$ respectively. Prove that: $u^{2} \tan A+v^{2} \tan B+w^{2} \tan C \geqslant 4 S$. Also, determine the condition for equality, where $S$ is the area of $\triangle ABC$. (1989 IMO Shortlist) | 29. Take the line through $B C$ as the $x$-axis, and the line through $A$ perpendicular to $BC$ as the $y$-axis, establishing a Cartesian coordinate system.
Let the coordinates of $A, B, C$ be $(0, a), (-b, 0), (c, 0)$ (where $a, b, c$ are all positive numbers), then $\tan B = \frac{a}{b}, \tan C = \frac{a}{c}, \tan A... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,618 |
30. Let $P$ be any point inside $\triangle ABC$, and let $R$ be the circumradius of $\triangle ABC$. Prove: $\frac{PA}{BC^2} + \frac{PB}{CA^2} + \frac{PC}{AB^2} \geqslant \frac{1}{R}$. (2001 USA National Training Team Selection Test) | 30. As shown in the figure, draw $P X \perp B C, P Y \perp C A$, $P Z \perp A B$, with $X, Y, Z$ being the feet of the perpendiculars. Extend $X P$, and draw perpendiculars from $Y, Z$ to the extended line, with the feet of the perpendiculars being $M, N$, respectively. Since $P Y \perp C A, P Z \perp A B$, points $A, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,619 |
39. Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$. (30th IMO Canadian Training Problem) | 39. When $x=y=z=\frac{\sqrt{3}}{3}$, $\frac{x}{1-x^{2}}=\frac{y}{1-y^{2}}=\frac{z}{1-z^{2}}=\frac{\sqrt{3}}{2}$, we conjecture that the minimum value is $\frac{3 \sqrt{3}}{2}$. Below, we prove that if $x, y, z$ are positive numbers, and $x^{2}+y^{2}+z^{2}=1$, then $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+$ $\frac{z}{1-z^{2... | \frac{3 \sqrt{3}}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,621 |
32. Let $K, L, M, N$ be the midpoints of the sides $AB, BC, CD, DA$ of a convex quadrilateral $ABCD$, respectively, and let $NL$ and $KM$ intersect at point $T$. Prove: $\frac{8}{3} S_{\text{quadrilateral } DNTM} < S_{\text{quadrilateral } ABCD} < 8 S_{\text{quadrilateral } DNTM}$. (2004 Belarus Math | 32. As shown in the figure, let $S_{\text {quadrilateral } \triangle \triangle T M}=x, S_{\text {quadrilateral } A B C D}=S$.
It is easy to see that $K L / / A C$, and $K L=\frac{1}{2} A C . N M / / A C$, and
$N M=\frac{1}{2} A C$. Therefore, $N M / / K L$ and $N M=K L$, so quadrilateral $K L M N$ is a parallelogram. ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,622 |
33. Given that $A B C D$ is a convex quadrilateral. Prove that $S_{\text {quadrilateral } A B C D} \leqslant \frac{A B^{2}+B C^{2}+C D^{2}+D A^{2}}{4}$. | 33. As shown in the figure, let $A B=a, B C=b, C D=c, D A=d$, then
$$2 S_{\text {quadrilateral } A B C D}=2 S_{\triangle A B C}+2 S_{\triangle A C D}=a b \sin B+c d \sin D$$
Therefore,
$$\begin{array}{l}
4\left(S_{\text {quadrilateral } A B C D}\right)^{2}=a^{2} b^{2} \sin ^{2} B+ \\
c^{2} d^{2} \sin ^{2} D+2 a b c d ... | S_{\text {quadrilateral } A B C D} \leqslant \frac{a^{2}+b^{2}+c^{2}+d^{2}}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 733,623 |
34. Given points $K, L, M, N$ are on the sides $AB, BC, CD, DA$ of a convex quadrilateral $ABCD$ respectively. Let $S_{1}=S_{\triangle A K N}, S_{2}=S_{\triangle B K L}, S_{3}=S_{\triangle C L M}, S_{4}=S_{\triangle D M N}, S=S_{\text {quadrilateral } A B C D}$. Prove: $\sqrt[3]{S_{1}}+\sqrt[3]{S_{2}}+$ $\sqrt[3]{S_{3}... | 34. As shown in the figure, let $\frac{A N}{A D}=\lambda_{1}, \frac{D M}{D C}=\lambda_{2}, \frac{C L}{C B}=\lambda_{3}, \frac{B K}{B A}=\lambda_{4}$,
then
$$\begin{array}{c}
S_{1}=S_{\triangle A K N}=\lambda_{1}\left(1-\lambda_{4}\right) S_{\triangle A B D} \\
S_{2}=S_{\triangle B K L}=\lambda_{4}\left(1-\lambda_{3}\r... | \sqrt[3]{S_{1}}+\sqrt[3]{S_{2}}+\sqrt[3]{S_{3}}+\sqrt[3]{S_{4}} \leqslant 2 \sqrt[3]{S} | Inequalities | proof | Yes | Yes | inequalities | false | 733,624 |
35. Let $ABCD$ be a trapezoid $(AB \parallel CD)$, $E$ a point on segment $AB$, and $F$ a point on $CD$. Line segments $CE$ and $BF$ intersect at point $H$, and line segments $ED$ and $AF$ intersect at point $G$. Prove that $S_{EHFG} \leqslant \frac{1}{4} S_{ABCD}$. If $ABCD$ is an arbitrary convex quadrilateral, does ... | 35. As shown in Figure (1), connect $E F$. In trapezoid $A E F D$, it is clear that
$$\sin \angle A G D=\sin \angle D G F=\sin \angle E G F=\sin \angle A G E$$
$$S_{\triangle A G D}=S_{\triangle A E D}-S_{\triangle A E G}=S_{\triangle A E F}-S_{\triangle A E G}=S_{\triangle E C F}(2)$$
From (1) and (2), we have.
$$\be... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,625 |
36. For any three points $P, Q, R$ in the plane, we define $m(P Q R)$ as the length of the shortest altitude of $\triangle P Q R$ (if $P, Q, R$ are collinear, let $m(P Q R)=0$). Let $A, B, C$ be points in the plane, and for any point $X$ in this plane, prove that: $m(A B C) \leqslant m(A B X)+m(A X C)+m(X B C)$. | 36. Suppose $A, B, C$ are not collinear. Extend $A B, B C, C A$ into lines, dividing the plane into $t$ parts, as shown in Figure (a), which are divided into three regions.
(1) Point $X$ is within region $I$, and let $l(P Q R)$ be the length of the longest side of $\triangle P Q R$.
Extend $A X$ to intersect $B C$ at ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,626 |
37. The side lengths of $ABC$ are $a, b, c$. Now extend $AB, AC$ by $a$ units, $BC, BA$ by $b$ units, and $CA, CB$ by $c$ units. Let the area of the convex hexagon formed by these six endpoints be $G$, and the area of $\triangle ABC$ be $F$. Prove that: $G \geqslant 13 F$. (1993 German Mathematical Olympiad Problem) | 37. It is known that $S_{\triangle A B_{2} C_{1}}=S_{\triangle C_{2} B_{1}}=S_{\triangle B C_{2} A_{1}}=S_{\triangle A B C}$, so
$$\begin{array}{l}
\frac{G}{F}=\frac{S_{A B C_{2} C_{1}}+S_{B C A_{2} A_{1}}+S_{A C B_{1} B_{2}}+4 F}{F}= \\
\frac{S_{\triangle A_{1} A_{2}}+S_{\triangle B B_{1} B_{2}}+S_{\triangle C C_{1} C... | 13 | Geometry | proof | Yes | Yes | inequalities | false | 733,627 |
38. In $\triangle A B C$, the medians on sides $A B$ and $A C$ are perpendicular to each other. Prove that: $\cot B+\cot C \geqslant \frac{2}{3}$. | 38. Let the medians $B B^{\prime}, C C^{\prime}$ intersect at $G$, and let $G B^{\prime}=m, G C^{\prime}=n$, then $B G=2 m, C G=2 n$. Therefore,
$$\cot B=\cot \left(\angle C B G+\angle G B C^{\{\prime}\right)=\frac{1-\frac{2 n}{2 m} \cdot \frac{n}{2 m}}{\frac{2 n}{2 m}+\frac{n}{2 m}}=\frac{2 m^{2}-n^{2}}{3 m n}$$
Simi... | \cot B+\cot C \geqslant \frac{2}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 733,628 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.