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114. Let $a, b, c \in \overline{\mathbf{R}^{-}}$, then
$$\sum \frac{\sqrt{b c+4 a(b+c)}}{b+c} \geqslant \frac{9}{2}$$
Equality holds if and only if $a=b=c$, or one of $a, b, c$ is zero and the other two are equal. | 114. Proof: Let
$$A=b c+4 a(b+c), B=c a+4 b(c+a), C=a b+4 c(a+b)$$
Square both sides of the original equation and rearrange to get
$$\sum b c \cdot \sum \frac{1}{(b+c)^{2}}+3 \sum \frac{a}{b+c}+2 \sum \frac{\sqrt{B C}}{(c+a)(a+b)} \geqslant \frac{81}{4}$$
Since
$$\sum b c \cdot \sum \frac{1}{(b+c)^{2}} \geqslant \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,423 |
115. (Recorded from Vasile Cirtoaje, editor of *Algebraic Inequalities. Old and New Methods*, § 3.4. Applications, Problem 16) Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, then
$$1+a+b+c \geqslant 2 \sqrt{1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$
Equality holds if and only if $a=b=c=1$. | 115. Proof: Let
$$s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c=1$$
Then the original expression, after squaring and simplifying, is equivalent to
$$s_{1}^{2}+2 s_{1}-4 s_{2}-3 \geqslant 0$$
From Example 14 in Chapter 1 "Proving Inequalities by Equivalent Transformation," it is easy to get
$$s_{2} \leqslant \frac{s_{1}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,424 |
116. Let $\lambda_{1}, \lambda_{2}, \lambda_{3} \in \mathbf{R}^{+}$, then
$$\begin{array}{l}
\sqrt{\frac{\lambda_{1}}{\lambda_{2}+\lambda_{3}}}+\sqrt{\frac{\lambda_{2}}{\lambda_{3}+\lambda_{1}}}+\sqrt{\frac{\lambda_{3}}{\lambda_{1}+\lambda_{2}}} \geqslant (6-3 \sqrt{3}) \sqrt{\frac{\lambda_{1} \lambda_{2} \lambda_{3}}... | 116. Proof: In $\triangle B_{1} B_{2} B_{3}$, let
$$\begin{aligned}
\lambda_{1} & =-\sin B_{1}+\sin B_{2}+\sin B_{3} \\
\lambda_{2} & =\sin B_{1}-\sin B_{2}+\sin B_{3} \\
\lambda_{3} & =\sin B_{1}+\sin B_{2}-\sin B_{3}
\end{aligned}$$
By using the sum-to-product identities, we get
$$\begin{array}{l}
\lambda_{1}=4 \sin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,425 |
117. (Recorded from Vasile Cirtoaje, editor, *Algebraic Inequalities. Old and New Methods*, § 3.4. Applications, Problem 10) Let \( x, y, z \in \overline{\mathbf{R}^{-}} \), then
$$\sqrt{1+\frac{48 a}{b+c}}+\sqrt{1+\frac{48 b}{c+a}}+\sqrt{1+\frac{48 c}{a+b}} \geqslant 15$$
with equality if and only if \( x=y=z \), or ... | 117. Proof: Let $x=-a+b+c, y=a-b+c, z=a+b-c, x+y+z=a+b+c=3$, then $a, b, c \in\left[0, \frac{3}{2}\right]$, and the original inequality can be transformed into
$$\sqrt{\frac{72}{a}-47}+\sqrt{\frac{72}{b}-47}+\sqrt{\frac{72}{c}-47} \geqslant 15$$
Therefore, to prove the original inequality, it suffices to prove that (1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,426 |
119. Let $a, b, c \in \overline{\mathbf{R}^{-}}$, then
$$\sum \frac{1}{4 b^{2}+4 c^{2}-b c} \geqslant \frac{9}{7 \sum a^{2}}$$
Equality holds if and only if $a=b=c$, or one of $a, b, c$ is zero and the other two are equal. | 119. By symmetry, we can assume \(a \geqslant b \geqslant c \geqslant 0\), then
$$\begin{array}{l}
\sum \frac{7\left(a^{2}+b^{2}+c^{2}\right)}{4 b^{2}+4 c^{2}-b c}-9= \\
{\left[2 \sum \frac{a^{2}+b^{2}+c^{2}}{b^{2}+c^{2}}-9\right]-\left[2 \sum \frac{a^{2}+b^{2}+c^{2}}{b^{2}+c^{2}}-\sum \frac{7\left(a^{2}+b^{2}+c^{2}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,428 |
121. (Original problem, 1987.08.24) Let
$$x, y, z \in[0,1], D=\left|\begin{array}{ccc}
0 & x & 1-x \\
1-y & 0 & y \\
z & 1-z & 0
\end{array}\right|$$
Then $x(1-y), y(1-z), z(1-x)$ must have one whose value is not greater than $D$. | 121. Proof: (1) When $y \geqslant z$
$$D-z(1-x)=(1-x-z)^{2}+(1-x)(x-y)+z(y-z) \geqslant 0$$
Thus, we have $\square$
$$z(1-x) \leqslant D$$
(2) When $y \leqslant z$, if $y+z \leqslant 1$, then
$$D-y(1-z)=(1-x-y)^{2}+(x-z)(1-x-y)+y(z-y) \geqslant 0$$
Thus, we have
$$y(1-z) \leqslant D$$
If $y+z \geqslant 1$, then
$$D-... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,430 |
Example 5 Let $a, b, c \in \mathbf{R}^{+}$, prove that
$$\frac{3 \sum a^{2}}{\sum a} \geqslant 2 \sum \frac{a^{2}}{a+b}$$
Equality in (5) holds if and only if $a=b=c$. | It is easy to prove $\sum \frac{a^{2}}{a+b}=\sum \frac{b^{2}}{a+b}$, so we have
$$2 \sum \frac{a^{2}}{a+b}=\sum \frac{a^{2}+b^{2}}{a+b}$$
Therefore, equation (5) is equivalent to
$$\frac{3 \sum a^{2}}{\sum a} \geqslant \sum \frac{a^{2}+b^{2}}{a+b}$$
Since equation (※) is symmetric with respect to $a, b, c$, we may as... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,431 |
122. ("Home of Math Olympiads" website, 2008. 04. 14, provided by polynasia) Let $a, b, c \in \mathbf{R}^{+}$, and $a b c=1$, then
$$\sum \frac{1}{a^{2}+2 b^{2}+3} \leqslant \frac{1}{2}$$ | 122. Proof: Since
$$a^{2}+2 b^{2}+3=\left(a^{2}+b^{2}\right)+\left(b^{2}+1\right)+2 \geqslant \frac{2}{c}+2 b+2$$
Therefore,
$$\frac{1}{a^{2}+2 b^{2}+3} \leqslant \frac{c}{2(1+c+b c)}$$
Similarly, we have
$$\begin{array}{l}
\frac{1}{b^{2}+2 c^{2}+3} \leqslant \frac{a}{2(1+a+c a)} \\
\frac{1}{c^{2}+2 a^{2}+3} \leqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,432 |
123. (Shen Yi, 2008.04. 10 proposed) Let $P$ be any point inside a non-obtuse $\triangle A B C$, and let the line $A P$ intersect the circumcircle of $\triangle A B C$ at another point $A_{1}$. From $A_{1}$, draw $A_{1} A_{2} \perp B C$ at $A_{2}$, and similarly define $B_{2}, C_{2}$. Then the area of $\triangle A_{2} ... | 123. Proof: (Chen Yi, Sichuan, 2008.04.19 provided) As shown in Figure 1, connect $A_{1} B, A_{1} C$. Let the circumradius of $\triangle A B C$ be $R$, and denote $\angle P A C, \angle P A B, \angle P B A, \angle P B C, \angle P C B, \angle P C A$ as $\alpha_{1}, \alpha_{2}, \beta_{1}, \beta_{2}, \gamma_{1}, \gamma_{2}... | proof | Geometry | proof | Yes | Yes | inequalities | false | 734,433 |
125. ("Oriental Hotline" Forum, 2008.04.14 provided) Let $O$ be any point inside $\triangle ABC$, and let the lines $AO, BO, CO$ intersect the sides $BC, CA, AB$ at $P, Q, R$ respectively. If $BC = a$ is the longest side, then $|OP| + |OQ| + |OR| < a$ | 125. Proof: Since $A P, B Q, C R$ pass through the same point $O$, by Ceva's theorem, we have $\frac{B P}{P C} \cdot \frac{C Q}{Q A} \cdot \frac{A R}{R B}=1$. Therefore, we can set $\frac{B P}{P C}=\frac{z}{y}, \frac{C Q}{Q A}=\frac{x}{z}, \frac{A R}{R B}=\frac{y}{x}, x, y, z \in \mathbf{R}^{+}$. Thus, we have $x \over... | proof | Geometry | proof | Yes | Yes | inequalities | false | 734,435 |
126. (2008, Serbian Mathematical Olympiad) Given that $a, b, c$ are positive numbers, and $a+b+c=1$, prove
$$\sum \frac{1}{bc + a + \frac{1}{a}} \leqslant \frac{27}{31}$$ | 126. Simplify: Original expression $\Leftrightarrow \sum \frac{9 a^{2}+9 a b c+9-31 a}{a^{2}+a b c+1} \geqslant 0$.
Assume without loss of generality that $a \geqslant b \geqslant c$, it is clear that $9(a+b)<31$, etc., so it is easy to prove
$$\begin{array}{c}
9 a^{2}+9 a b c+9-31 a \leqslant 9 b^{2}+9 a b c+9-31 b \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,436 |
131. For positive numbers $x, y, z$, prove that
$$\frac{(y+z)(z+x)(x+y)}{x y z}+\frac{2 \sqrt{2}(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}} \geqslant 8+6 \sqrt{2}$$
Equality holds if and only if $x=\sqrt{2}, y=z=1$; or $y=\sqrt{2}, z=x=1$; or $z=\sqrt{2}, x=y=1$. | 131. Proof: Let $s_{1}=\sum x=1, s_{2}=\sum y z, s_{3}=x y z$. After transformation, the original inequality is equivalent to
$$s_{1}-2 s_{2}^{2}-(9+4 \sqrt{2}) s_{3}+(18+12 \sqrt{2}) s_{2} s_{3} \geqslant 0$$
Now we prove inequality (1). Applying Corollary 3 of Theorem 2 in the appendix of Chapter 4 "Application of B... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,441 |
Example 6 (Original Problem, 1992.03.01) In $\triangle A B C$, $B C=a, C A=b, A B=c$, the circumradius is $R$, and the inradius is $r$, then
$$1-\frac{2 r}{R} \geqslant\left|\frac{(a-b)(b-c)(a-c)}{a b c}\right|$$
Equality in (6) holds if and only if $\triangle A B C$ is an equilateral triangle. | Assume $a \geqslant b \geqslant c$, and $a=c+\alpha+\beta, b=c+\alpha, \alpha, \beta \in \mathbf{R}^{-}$, then we have $c \geqslant \beta$. Since $\frac{2 r}{R}=\frac{(-a+b+c)(a-b+c)(a+b-c)}{a b c}$, and
$$\begin{array}{l}
\quad a b c-\prod(-a+b+c)-(a-b)(b-c)(a-c)= \\
\left(\alpha^{2}+\alpha \beta+\beta^{2}\right) c+(2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,443 |
134. (Original problem, 2008.04.05) Let $a, b, c \in \mathbf{R}^{+}$, and $a b c \leqslant m$, then
$$\sum \frac{1}{m+a^{2}(b+c)} \leqslant \frac{3}{m+2 a b c}$$
Equality holds if and only if $a b c=m$, or $a=b=c$. | $$\begin{array}{l}
\frac{3}{m+2 a b c}-\sum \frac{1}{m+a^{2}(b+c)}= \\
\sum\left[\frac{1}{m+2 a b c}-\frac{1}{m+a^{2}(b+c)}\right]= \\
\frac{1}{m+2 a b c} \sum \frac{a^{2}(b+c)-2 a b c}{m+a^{2}(b+c)}= \\
\frac{1}{m+2 a b c} \sum \frac{a[c(a-b)-b(c-a)]}{m+a^{2}(b+c)}= \\
\frac{1}{m+2 a b c} \sum a(b-c)\left[\frac{b}{m+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,446 |
135. (Original problem, 2008. 12. 21) Let $x_{i} \in \overline{\mathbf{R}^{-}}, i=1,2, \cdots, n$, and denote $s_{1}=\sum_{i=1}^{n} x_{i}, s_{2}=$ $\sum_{1<i<j<n} x_{i} x_{j}, s_{3}=\sum_{1<i<j<k<n} x_{i} x_{j} x_{k}$, then
$$\sum_{i=1}^{n} \sqrt{\frac{x_{i}}{s_{1}-x_{i}}} \geqslant \sqrt{\frac{s_{1}^{3}}{s_{1} s_{2}-3... | 135. Proof: By applying the Cauchy-Schwarz inequality twice, we have
$$\sum_{i=1}^{n} \sqrt{\frac{x_{i}}{s_{1}-x_{i}}} \geqslant \frac{s_{1}^{2}}{\sum_{i=1}^{n} \sqrt{x_{i}^{3}\left(s_{1}-x_{i}\right)}}$$
Also,
$$\sum_{i=1}^{n} \sqrt{x_{i}^{3}\left(s_{1}-x_{i}\right)} \leqslant \sqrt{\sum_{i=1}^{n} x_{i} \cdot \sum_{i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,447 |
136. Let $a, b, c, d \in \mathbf{R}$, then
$$a^{2} c^{2}+a^{2} d^{2}+b^{2} d^{2}+b^{2} c^{2}+4 a^{2} b^{2} \geqslant 4 a^{2} b d+4 a b^{2} c$$ | 136. Proof: If $c=0$, the original equation obviously holds; if $c \neq 0$, then the left side - right side $=$
$$\begin{array}{l}
\left(c^{2}+d^{2}+4 b^{2}-4 b d\right) a^{2}-4 b^{2} c \cdot a+\left(b^{2} c^{2}+b^{2} d^{2}\right)= \\
\left(c^{2}+d^{2}+4 b^{2}-4 b d\right)\left(a-\frac{2 b^{2} c}{c^{2}+d^{2}+4 b^{2}-4 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,448 |
137. For positive numbers $a, b, c, d$ and $k \geqslant 0$, we have
$$\frac{a}{b+k d}+\frac{b}{c+k a}+\frac{c}{d+k b}+\frac{d}{a+k c} \geqslant \frac{4}{1+k}$$
Equality holds if and only if $a=b=c=d$, or $k=1$, and $a+c=b+d$. | 137. Simplified Proof
$$\begin{array}{l}
\left(\frac{a}{b+k d}+\frac{b}{c+k a}+\frac{c}{d+k b}+\frac{d}{a+k c}\right) \cdot \\
{[a(b+k d)+b(c+k a)+c(d+k b)+d(a+k c)] \geqslant} \\
(a+b+c+d)^{2}
\end{array}$$
Thus, it suffices to prove
$$\frac{(a+b+c+d)^{2}}{a(b+k d)+b(c+k a)+c(d+k b)+d(a+k c)} \geqslant \frac{4}{1+k}$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,449 |
139. (Original problem, 2007. 12.21) Let $x, y, z \in \mathbf{R}^{-}$, then
$$\left(x^{2}-y^{2}-x y\right)^{2}+\left(y^{2}-z^{2}-y z\right)^{2}+\left(z^{2}-x^{2}-z x\right)^{2} \geqslant y^{2} z^{2}+z^{2} x^{2}+x^{2} y^{2}$$
Equality holds if and only if $x=y=z$. | 139. Simplify: The original expression is equivalent to
$$\sum\left(x^{2}-y^{2}\right)^{2} \geqslant 2 \sum x y\left(x^{2}-y^{2}\right)$$
When $x \geqslant y \geqslant z$, we have
$$\sum x y\left(x^{2}-y^{2}\right)=(x+y+z)(x-y)(y-z)(x-z) \geqslant 0$$
Therefore, to prove (1), it suffices to prove that (1) holds when ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,451 |
140. (Original problem, 2008.08.03) Given positive numbers $a, b, c$ satisfying $\sum bc=1$, prove that
$$\sum \sqrt{a+a^{4}}>\sqrt{(\sqrt{3}+1) \sum a}$$ | 140. Proof: Since
$$\sum \sqrt{a+a^{4}} \geqslant \sqrt{\left(\sum \sqrt{a}\right)^{2}+\left(\sum a^{2}\right)^{2}}$$
it suffices to prove that
$$\left(\sum \sqrt{a}\right)^{2}+\left(\sum a^{2}\right)^{2} \geqslant(\sqrt{3}+1) \sum a$$
If we let $\sqrt{a}=x, \sqrt{b}=y, \sqrt{c}=z$, we need to prove
$$\begin{array}{l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,452 |
141. Let $x, y, z \in \mathbf{R}^{-}$, and satisfy $x+y+z=2$, then
$$\sum \sqrt{9-8 y z} \geqslant 7$$
Equality holds if and only if $x=y=z=\frac{2}{3}$, or one of $x, y, z$ is zero and the other two are equal to 1. | 141. Simplified Proof: For the convenience of proof, we make the following transformation: Let $x=2a, y=2b, z=2c$. At this point, the original proposition is equivalent to:
Let $a, b, c \in \overline{\mathbf{R}^{-}}$, and satisfy $a+b+c=1$, then
$$\sum \sqrt{9-32bc} \geqslant 7$$
Equality holds if and only if $a=b=c=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,453 |
Example 7 (Self-created problem, 2000.03.02) In $\triangle A B C$, the lengths of the three sides are $a, b, c$. Prove that
$$\sum a \sum \frac{1}{a} \geqslant 9+\left(\frac{a-c}{b}\right)^{2}$$
Equality in (7) holds if and only if $\triangle A B C$ is an equilateral triangle. | Prove that when $a \geqslant b \geqslant c$, it is easy to verify that $\frac{a-c}{b} \geqslant \frac{a-b}{c}, \frac{a-c}{b} \geqslant \frac{b-c}{a}$. Therefore, we only need to prove that inequality (7) holds when $a \geqslant b \geqslant c$. Let $a=c+\alpha+\beta, b=c+\alpha$, then $c \geqslant \beta, \alpha$, $\beta... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,454 |
142. (Original problem, 2008. 08. 30) Let $a_{1}, b_{1}, a_{2}, b_{2} \in \mathbf{R}$, and $a_{1}^{2}-b_{1}^{2}=a_{2}^{2}-b_{2}^{2}=m^{2}$, then $\square$
$$\frac{\left(a_{1}+b_{2}\right)\left(a_{2}+b_{1}\right)-m^{2}}{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right)-m^{2}} \geqslant \frac{\left(a_{1}+b_{1}+a_{2}+b_{2}... | 142. Proof: Let $a_{1}+b_{1}=u, a_{2}+b_{2}=v$, then from $a_{1}^{2}-b_{1}^{2}=a_{2}^{2}-b_{2}^{2}=m^{2}$, we have $\frac{m^{2}}{u}=a_{1}-b_{1}, \frac{m^{2}}{v}=a_{2}-b_{2}$, thus
$$\begin{array}{l}
\left(a_{1}+b_{2}\right)\left(a_{2}+b_{1}\right)-m^{2}= \\
{\left[\frac{1}{2}\left(u+\frac{m^{2}}{u}\right)+\frac{1}{2}\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,455 |
143. ("Home of Math Olympiad" website, 2009.01.19 "ppppqqqq" provided) Let \( x, y, z \in \mathbf{R}^{-} \), and \( x^{2}+y^{2}+z^{2}=1 \), then
$$\sum \frac{1}{1-y z} \leqslant \frac{4 \sum x \sum y z}{(y+z)(z+x)(x+y)}$$
Equality holds if and only if \( x=y=z=\frac{\sqrt{3}}{3} \), or one of \( x, y, z \) is zero, an... | 143. Simplify: Homogenize the original expression to get
$$\sum \frac{x^{2}+y^{2}+z^{2}}{x^{2}+y^{2}+z^{2}-yz} \leqslant \frac{4 \sum x \sum yz}{(y+z)(z+x)(x+y)}$$
Let $s_{1}=\sum x=1, s_{2}=\sum yz, s_{3}=xyz$, and after removing the denominators and simplifying, we get
$$\begin{aligned}
\text { Equation (1) } \Leftr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,456 |
145. Let $\triangle A B C$ have side lengths $a, b, c$, with $\max \{a, b, c\}=a$, and $\angle A \geqslant \frac{2 \pi}{3}$. Let $P$ be any point inside or on the boundary of $\triangle A B C$, and let $P A=R_{1}, P B=R_{2}, P C=R_{3}$. Then
$$\left(R_{2}+R_{3}\right)\left(R_{3}+R_{1}\right)\left(R_{1}+R_{2}\right) \ge... | 145. The author proved the conjecture of Teacher Liu Jian on April 24, 2009, as follows:
(1) When $R_{2} R_{3} \geqslant b c$
$$\begin{array}{l}
\left(R_{2}+R_{3}\right)\left(R_{3}+R_{1}\right)\left(R_{1}+R_{2}\right)= \\
R_{2} R_{3}\left(R_{1}+R_{2}+R_{3}\right)+R_{1}\left(R_{2}^{2}+R_{3}^{2}+R_{2} R_{3}+R_{3} R_{1}+R... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,458 |
1. Let $a, b, c, d$ be real numbers such that $a^{2}+b^{2}+c^{2}+d^{2}=4$. Prove that $a^{3}+b^{3}+c^{3}+d^{3} \leqslant 8$ | 1. (2007.01.15) To prove: It suffices to show that $\left(\sum a^{2}\right)^{3} \geqslant\left(\sum a^{3}\right)^{2}$. This inequality can be easily derived from the following: $\left(\sum a^{2}\right)^{3}=\left(\sum a^{2}\right)^{2} \cdot \sum a^{2} \geqslant \sum a^{4} \cdot \sum a^{2} \geqslant\left(\sum a^{3}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,459 |
2. If $a, b, c$ are non-negative numbers, then
$$a^{3}+b^{3}+c^{3}-3 a b c \geqslant\left(\frac{b+c}{2}-a\right)^{3}$$ | $$\begin{array}{l}
a^{3}+b^{3}+c^{3}-3 a b c=\frac{1}{2}(a+b+c)\left[(b-a)^{2}+(c-a)^{2}+(a-b)^{2}\right] \geqslant \\
\qquad \quad \frac{1}{2}(a+b+c)\left[(b-a)^{2}+(c-a)^{2}\right] \geqslant \\
\frac{1}{4}(a+b+c)(b-a+c-a)^{2} \geqslant \\
2\left(\frac{b+c}{2}-a\right)^{3}
\end{array}$$
Prove the following inequality... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,460 |
4. Let $a, b, c$ be non-negative numbers such that $a^{3}+b^{3}+c^{3}=3$. Prove that
$$a^{4} b^{4}+b^{4} c^{4}+c^{4} a^{4} \leqslant 3$$ | 4. (2007.02.26) Prove briefly: If $x, y, z \in \mathbf{R}^{-}$, then $x y z \geqslant \prod(-x+y+z)$, i.e., $\sum x^{3}+3 x y z \geqslant \sum y z(y+z)$. The following will use this inequality. Since
$$\begin{aligned}
9 \sum b^{4} c^{4}= & 9 \sum\left(b^{3} c^{3} \cdot b c\right) \leqslant 3 \sum b^{3} c^{3}\left(b^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,462 |
5. If $a, b, c$ are non-negative numbers, then
$$a^{2}+b^{2}+c^{2}+2 a b c+1 \geqslant 2(a b+b c+c a)$$ | 5. (2007.01.16) Proof: Among $1-a, 1-b, 1-c$, there must be two that are both not greater than zero or both not less than zero. Without loss of generality, assume they are $1-b, 1-c$, thus $(1-b)(1-c) \geqslant 0$. Therefore,
$$\begin{array}{l}
a^{2}+b^{2}+c^{2}+2 a b c+1-2(a b+b c+c a)= \\
(a-1)^{2}+(b-c)^{2}+2 a(1-b)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,463 |
Example 8 (Self-created problem, 2000.07.21) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in \mathbf{R}, \max \left\{a_{1}, a_{2}, a_{3}, a_{4}\right.$, $\left.a_{5}\right\}=a_{1}, \min \left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}=a_{5}$, then
$$5 \sum a_{1}^{2}-\left(\sum a_{1}\right)^{2} \leqslant 6\left(a_{1}-a_{5}\r... | Prove: Without loss of generality, let $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant a_{4} \geqslant a_{5}, a_{1}=a_{5}+\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}, a_{2}=$
$$\begin{array}{l}
a_{5}+\alpha_{1}+\alpha_{2}+\alpha_{3}, a_{3}=a_{5}+\alpha_{1}+\alpha_{2}, a_{4}=a_{5}+\alpha_{1}, \alpha_{1}, \alpha_{2}, \a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,464 |
6. If $a, b, c$ are distinct real numbers, then
$$\frac{a^{2}}{(b-c)^{2}}+\frac{b^{2}}{(c-a)^{2}}+\frac{c^{2}}{(a-b)^{2}} \geqslant 2$$ | 6. (2007.01.16) Proof: By symmetry, without loss of generality, assume $a > b > c$, then it is easy to prove
$$\frac{a}{b-c} > \frac{a-c}{b}$$
Thus,
$$\begin{array}{r}
\frac{a^{2}}{(b-c)^{2}} + \frac{b^{2}}{(c-a)^{2}} + \frac{c^{2}}{(a-b)^{2}} \geqslant \frac{a^{2}}{(b-c)^{2}} + \frac{b^{2}}{(c-a)^{2}} \geqslant \\
\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,465 |
7. If $a, b, c$ are non-negative numbers, then
$$\left(a^{2}-b c\right) \sqrt{b+c}+\left(b^{2}-c a\right) \sqrt{c+a}+\left(c^{2}-c a\right) \sqrt{a+b} \geqslant 0$$ | 7. (2007.01.21) Brief proof: Let $b+c=x^{2}, c+a=y^{2}, a+b=z^{2}, x, y, z \in \overline{\mathbf{R}^{-}}$, then the original expression becomes
$$\begin{array}{l}
\sum\left[\left(-x^{2}+y^{2}+z^{2}\right)^{2}-\left(x^{2}-y^{2}+z^{2}\right)\left(x^{2}+y^{2}-z^{2}\right)\right] x \geqslant 0 \Leftrightarrow \\
\sum\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,466 |
8. If $a, b, c, d$ are non-negative numbers, then
$$\frac{a-b}{a+2b+c}+\frac{b-c}{b+2c+d}+\frac{c-d}{c+2d+a}+\frac{d-a}{d+2a+b} \geqslant 0$$ | 8. (2007.01.11) Simplify
$$\begin{array}{c}
\text { Original } \Leftrightarrow \sum\left(\frac{a-b}{a+2 b+c}+\frac{1}{2}\right) \geqslant 2 \Leftrightarrow \\
\sum \frac{3 a+c}{a+2 b+c} \geqslant 4
\end{array}$$
Now we prove equation (1). Applying the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{[(3 a+c)(a+2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,467 |
12. Let $a, b, c$ be positive numbers and let
$$E(a, b, c)=a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$$
Prove that
(a) $(a+b+c) E(a, b, c) \geqslant a b(a-b)^{2}+b c(b-c)^{2}+c a(c-a)^{2}$;
(b) $2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) E(a, b, c) \geqslant(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$. | 12. (2007.01.07) Simplify: $E(a, b, c)=\frac{1}{2} \sum(-a+b+c)(b-c)^{2}$.
(a) Without loss of generality, assume $a \geqslant b \geqslant c$, then
$$\begin{aligned}
\text { Original } \Leftrightarrow & \frac{1}{2} \sum(a+b+c)(-a+b+c)(b-c)^{2} \geqslant \sum b c(b-c)^{2} \Leftrightarrow \\
& \sum\left(-a^{2}+b^{2}+c^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,471 |
13. Let $a, b, c$ and $x, y, z$ be real numbers such that $a+x \geqslant b+y \geqslant c+z \geqslant 0$ and $a+b+c=x+y+z$. Prove that
$$a y+b x \geqslant a c+x z$$ | 13. (2008.08. 29 , excerpted and translated from the original book) Simplified proof
$$\begin{aligned}
a y+b x= & a c-x z=a(y-c)+x(b-z)= \\
& a(a+b-x-z)+x(b-z)= \\
& a(a-x)+(a+x)(b-z)= \\
& \frac{1}{2}(a-x)^{2}+\frac{1}{2}\left(a^{2}-x^{2}\right)+(a+x)(b-z)= \\
& \frac{1}{2}(a-x)^{2}+\frac{1}{2}(a+x)(a+2 b-x-2 z)= \\
&... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,472 |
14. Let \(a, b, c \in \left[\frac{1}{3}, 3\right]\). Prove that
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac{7}{5}$$ | 14. (2007.02.10) Simplified proof: The original equation, after removing the denominator and rearranging, is equivalent to
$$(b+c)(c+a)(a+b) \geqslant 5(c-b)(c-a)(b-a)$$
From equation (1), we know that we only need to prove that equation (1) holds when $c \geqslant b \geqslant a$ to establish the original proposition.... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,473 |
Lemma 2 Given $a_{1}, a_{2}, \cdots, a_{n}$ are non-negative real numbers, and $\max _{1<i<n}\left\{a_{i}\right\}=a_{1}, \min _{1<i<n}\left\{a_{i}\right\}=a_{n}$, then for $t=1,2, \cdots, n-1$, the following inequality holds, that is
$$\sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2} \leqslant \min \{t, n-t\} \cdot\left(... | (1) If $t \leqslant\left[\frac{n}{2}\right]$, then $t \leqslant n-t$. Classify all expressions of the form $\left(a_{i}-a_{i+t}\right)^{2}$ into the following $t$ categories based on the remainder of $i$ modulo $t$: $A_{r}=\left\{\left(a_{i}-a_{i+t}\right)^{2} \mid i=r(\bmod t), 1 \leqslant i \leqslant n-t\right\}$ (wh... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,475 |
20. Let $a, b, c$ be non-negative numbers such that $a b+b c+c a=3$. Prove that
$$\frac{1}{a^{2}+2}+\frac{1}{b^{2}+2}+\frac{1}{c^{2}+2} \leqslant 1$$ | 20. (2007.01.17) Proof 1: The original inequality is equivalent to
$$\sum b^{2} c^{2}+a^{2} b^{2} c^{2} \geqslant 4$$
Let $b c=x, c a=y, a b=z, x, y, z \in \overline{\mathbf{R}^{-}}$, then $x+y+z=3$, thus we need to prove
$$\frac{\sum x}{3} \cdot \sum x^{2}+x y z \geqslant 4 \cdot\left(\frac{\sum x}{3}\right)^{3}$$
T... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,480 |
21. Let $a, b, c$ be non-negative real numbers such that $a b+b c+c a=3$. Prove that
$$\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1} \leqslant \frac{3}{2}$$ | 21. (2007.01.17) Simplify and prove: the original inequality is equivalent to
$$3+\sum a^{2} \geqslant \sum b^{2} c^{2}+3 a^{2} b^{2} c^{2}$$
Let $b c=x, c a=y, a b=z, x, y, z \in \overline{\mathbf{R}^{-}}$, then $x+y+z=3$, thus we need to prove
$$\begin{aligned}
3 \cdot\left(\frac{\sum x}{3}\right)^{3}+\sum \frac{y z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,481 |
Example 9 (Original Problem, 1991.08.09) In $\triangle A B C$, the lengths of the three sides are $a, b, c$, then
$$a b^{2}+b c^{2}+c a^{2} \geqslant 3 a b c+2|(b-c)(c-a)(a-b)|$$
Equality in (10) holds if and only if $\triangle A B C$ is an equilateral triangle. | To prove that if $a \geqslant b \geqslant c$, then
$$a^{2} b+b^{2} c+c^{2} a-\left(a b^{2}+b c^{2}+c a^{2}\right)=(a-b)(b-c)(a-c) \geqslant 0$$
Therefore, to prove inequality (10), we only need to prove it under the condition $a \geqslant b \geqslant c$.
$$\begin{array}{l}
\text { Let } a=c+\alpha+\beta, b=c+\alpha \t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,486 |
28. If $a, b, c, d$ are positive real numbers, then
$$\left(\frac{a}{a+b}\right)^{2}+\left(\frac{b}{b+c}\right)^{2}+\left(\frac{c}{c+d}\right)^{2}+\left(\frac{d}{d+a}\right)^{2} \geqslant 1$$ | 28. (2007.01.20) Simplify to prove: the original expression is equivalent to
$$\begin{array}{c}
\frac{1}{\left(1+\frac{b}{a}\right)^{2}}+\frac{1}{\left(1+\frac{c}{b}\right)^{2}}+\frac{1}{\left(1+\frac{d}{c}\right)^{2}}+\frac{1}{\left(1+\frac{a}{d}\right)^{2}} \geqslant 1 \\
\text { Let } \frac{b}{a}=x, \frac{c}{b}=y, \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,489 |
30. Let $a, b, c$ be non-negative numbers, no two of them are zero. Then
$$\frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}} \geqslant \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$ | 30. (2007.01.30) Brief proof: By symmetry, without loss of generality, assume $a \geqslant b \geqslant c$, then
$$\begin{array}{l}
\sum \frac{a^{2}}{b^{2}+c^{2}}-\sum \frac{a}{b+c}=\sum \frac{a b(a-b)+a c(a-c)}{\left(b^{2}+c^{2}\right)(b+c)}= \\
{\left[\frac{a b+a c}{\left(b^{2}+c^{2}\right)(b+c)}-\frac{b a}{\left(c^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,491 |
31. If $a, b, c$ are non-negative numbers, then
$$2\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \geqslant(a+1)(b+1)(c+1)(a b c+1)$$ | 31. (2007.01. 24) Proof: We will prove a stronger inequality:
$$\begin{array}{l}
2\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \geqslant \\
(a+1)(b+1)(c+1) \cdot \sqrt[3]{\left(a^{3}+1\right)\left(b^{3}+1\right)\left(c^{3}+1\right)}
\end{array}$$
To prove inequality (1), we first prove:
$$\begin{aligne... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,492 |
32. If $a, b, c, d$ are non-negative numbers, then
$$3\left(1-a+a^{2}\right)\left(1-b+b^{2}\right)\left(1-c+c^{2}\right) \geqslant 1+a b c+a^{2} b^{2} c^{2}$$ | 32. (2007.01.21) Proof: Since
$$3\left(1-a+a^{2}\right)^{3}-\left(1+a^{3}+a^{6}\right)=(1-a)^{4}\left(2-a+2 a^{2}\right) \geqslant 0$$
Similarly, there are two other inequalities, so
$$\begin{array}{l}
{\left[3\left(1-a+a^{2}\right)\left(1-b+b^{2}\right)\left(1-c+c^{2}\right)\right]^{3} \geqslant} \\
\left(1+a^{3}+a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,493 |
33. If $a, b, c$ are non-negative numbers, then
$$\left(1-a+a^{2}\right)\left(1-b+b^{2}\right)\left(1-c+c^{2}\right)\left(1-d+d^{2}\right) \geqslant\left(\frac{1+a b c d}{2}\right)^{2}$$ | 33. (2007.01.21) Proof: Since
$$\begin{array}{c}
2\left(1-a+a^{2}\right)^{2}-\left(1+a^{4}\right)=(1-a)^{4} \geqslant 0 \\
2\left(1-a+a^{2}\right)^{2} \geqslant\left(1+a^{4}\right)
\end{array}$$
Similarly,
$$\begin{array}{l}
2\left(1-b+b^{2}\right)^{2} \geqslant\left(1+b^{4}\right) \\
2\left(1-c+c^{2}\right)^{2} \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,494 |
34. If $a, b, c$ are non-negative numbers, then
$$\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geqslant(a b+b c+c a)^{3}$$ | 34. (2007.01.21) Simplify and prove
$$\begin{aligned}
\text { LHS }= & \left(a b+b^{2}+a^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(a^{2}+c^{2}+a c\right) \geqslant \\
& (a c+a b+b c)^{3}
\end{aligned}$$
(Apply Hölder's inequality).
Note: For the generalization of this problem, refer to Example 22 in Chapter 4 "Provin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,495 |
35. Let $a, b, c, d$ be positive numbers such that $a b c d=1$. Prove that
$$\begin{array}{l}
\frac{1}{1+a b+b c+c a}+\frac{1}{1+b c+c d+d b}+\frac{1}{1+c d+d a+a c}+ \\
\frac{1}{1+d a+a b+b d} \leqslant 1
\end{array}$$ | 35. (2007.02.01) Simplify
$$\begin{array}{l}
\frac{1}{1+a b+b c+c a} \leqslant \frac{\sqrt{d}}{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}} \Leftrightarrow \\
\sqrt{d}+\sqrt{a^{2} b^{2} d}+\sqrt{b^{2} c^{2} d}+\sqrt{c^{2} a^{2} d} \geqslant \sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \Leftrightarrow \\
\sqrt{\frac{a b}{c}}+\sqrt{\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,496 |
Example 10 (Original problem, 1991.08.08) $\triangle A B C$ is a non-obtuse triangle, with side lengths $a, b, c$, circumradius $R$, and inradius $r$, then
$$1-\frac{2 r}{R} \geqslant \frac{3|(a-b)(b-c)(a-c)|}{a b c}$$
Equality in (11) holds if and only if $\triangle A B C$ is an equilateral triangle. | Given $a \geqslant b \geqslant c$, and $a=c+\alpha+\beta, b=c+\alpha, c \geqslant \sqrt{2 \alpha \beta+2 \beta^{2}}+\beta, \alpha, \beta \in \overline{\mathbf{R}^{-}}$, then
$$\begin{aligned}
\text { Equation (11) } \Leftrightarrow \frac{\left(\alpha^{2}+\alpha \beta+\beta^{2}\right) c+\left(2 \alpha \beta^{2}+\beta^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,497 |
36. If $a, b, c$ and $x, y, z$ are real numbers, then
$$4\left(a^{2}+x^{2}\right)\left(b^{2}+y^{2}\right)\left(c^{2}+z^{2}\right) \geqslant 3(b c x+c a y+a b z)^{2}$$ | 36. (2007.01.30) Simplify and prove: Let
Since
$$\begin{aligned}
f(x)= & 4\left(a^{2}+x^{2}\right)\left(b^{2}+y^{2}\right)\left(c^{2}+z^{2}\right)-3(b c x+c a y+a b z)^{2}= \\
& {\left[4\left(b^{2}+y^{2}\right)\left(c^{2}+z^{2}\right)-3 b^{2} c^{2}\right] x^{2}-6 a b c(c y+b z) x+} \\
& {\left[4 a^{2} \cdot\left(b^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,498 |
37. If $a \geqslant b \geqslant c \geqslant d \geqslant e$, then
$$(a+b+c+d+e)^{2} \geqslant 8(a c+b d+c e)$$
For $e>0$, determine when equality occurs. | 37. (2007.01.22) Proof: Let
$$\begin{aligned}
f(e)= & (a+b+c+d+e)^{2}-8(a c+b d+c e)= \\
& e^{2}+2(a+b+d-3 c) e+(a+b+c+d)^{2}-8(a c+b d)
\end{aligned}$$
Its discriminant is
$$\begin{aligned}
\Delta= & 4(a+b+d-3 c)^{2}-4(a+b+c+d)^{2}+32(a c+b d)= \\
& 8(c-b)(c-d) \leqslant 0
\end{aligned}$$
(Note that $a \geqslant b \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,499 |
38. If $a, b, c, d$ are real numbers, then
$$6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+(a+b+c+d)^{2} \geqslant 12(a b+b c+c d)$$ | 38. (2007.01.30) Simplify and prove: Let
$$\begin{aligned}
f(b)= & 6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+(a+b+c+d)^{2}-12(a b+b c+c d)= \\
& 7 b^{2}-2 b(5 a+5 c-d)+6\left(a^{2}+c^{2}+d^{2}\right)+(a+c+d)^{2}-12 c d
\end{aligned}$$
Because
$$\begin{aligned}
\Delta= & 4\left[(5 a+5 c-d)^{2}-42\left(a^{2}+c^{2}+d^{2}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,500 |
39. If $a, b, c$ are positive numbers, then
$$\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)} \geqslant 1+\sqrt{1+\sqrt{\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)}}$$ | 39. (2007.02.01) Proof: Let $s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c$, then the original inequality is equivalent to
$$\begin{array}{l}
\left(\sqrt{\sum a \cdot \sum \frac{1}{a}}-1\right)^{2} \geqslant 1+\sqrt{\sum a^{2} \cdot \sum \frac{1}{a^{2}}} \Leftrightarrow \\
\sum a \cdot \sum \frac{1}{a} \geqslant 2 \sqrt{\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,501 |
40. If $a, b, c, d$ are positive numbers, then
$$5+\sqrt{2\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right)-2} \geqslant(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ | 40. (2007.02.01) Simplify: The original equation is equivalent to
$$2 \sum a^{2} \cdot \sum \frac{1}{a^{2}}-2 \geqslant\left(\sum a \cdot \sum \frac{1}{a}\right)^{2}-10 \sum a \cdot \sum \frac{1}{a}+25$$
Let $s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c$, then
$$\begin{array}{l}
\text { Equation (1) } \Leftrightarrow \fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,502 |
43. Let \(a, b, c\) and \(x, y, z\) be positive real numbers such that
$$(a+b+c)(x+y+z)=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)=4$$
Prove that
$$a b c x y z<\frac{1}{36}$$ | 43. (2007.01.30) Proof: Since
$$(a+b+c)(x+y+z)=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)=4$$
Therefore,
$$\begin{array}{l}
\left(\sum a\right)^{2} \cdot\left(\sum x\right)^{2}=4 \sum a^{2} \cdot \sum x^{2} \Leftrightarrow \\
\left(\sum a^{2}+2 \sum b c\right)\left(\sum x^{2}+2 \sum y z\right)=4 \sum... | a b c x y z < \frac{1}{36} | Inequalities | proof | Yes | Yes | inequalities | false | 734,504 |
45. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\frac{1}{a^{2}+b c}+\frac{1}{b^{2}+c a}+\frac{1}{c^{2}+a b} \geqslant \frac{3}{a b+b c+c a}$$ | 45. (2007.02.01) Simplify: The original expression is equivalent to
$$\sum b c \cdot \sum\left(b^{2}+c a\right)\left(c^{2}+a b\right)-3 \prod\left(a^{2}+b c\right) \geqslant 0$$
By symmetry, assume $c$ is the smallest, then
$$\begin{array}{l}
\sum b c \sum\left(b^{2}+c a\right)\left(c^{2}+a b\right)-3 \prod\left(a^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,506 |
41. If $a, b, c, d$ are positive numbers, then
$$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \geqslant 0$$ | 41. (2007.01.21) Simplify and prove: The original expression is equivalent to
$$\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b} \geqslant 4$$
The left side of (※) $=\left(\frac{a+c}{b+c}+\frac{c+a}{d+a}\right)+\left(\frac{b+d}{c+d}+\frac{d+b}{a+b}\right)=$
$$\begin{array}{l}
\frac{(a+c)(d+a)+(b+c)(c+a)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,507 |
Example 11 (Original Problem, 1991.08.15) For $\triangle ABC$ with side lengths $a, b, c$, then
$$a b^{2}+b c^{2}+c a^{2} \geqslant 3 a b c+\left(1-\frac{2}{\sqrt{27}} b(a-c)^{2}\right.$$
Equality in (13) holds if and only if $a=b=c$ or the sides of $\triangle ABC$ are in the ratio $(\sqrt{3}+1): \sqrt{3}: 1$. | Prove that if $a \geqslant b \geqslant c$, then $a^{2} b+b^{2} c+c^{2} a \geqslant a b^{2}+b c^{2}+c a^{2}$, and
$$b(a-c)^{2} \geqslant a(b-c)^{2}, b(a-c)^{2} \geqslant c(a-b)^{2}$$
Therefore, to prove equation (13), it is sufficient to prove it under the condition $a \geqslant b \geqslant c$.
Let $a=c+\alpha+\beta, b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,508 |
49. Let $a, b, c$ be non-negative numbers such that $a b+b c+c a=3$. Prove that
$$a^{3}+b^{3}+c^{3}+7 a b c \geqslant 10$$ | 49. (2007.02.04) Proof: Let $s_{1}=\sum a, s_{2}=\sum b c=3, s_{3}=a b c$, then the original inequality is equivalent to
$$s_{1}^{3}-3 s_{1} s_{2}+10 s_{3} \geqslant 10$$
which is
$$s_{3} \geqslant \frac{10-s_{1}^{3}+9 s_{1}}{10}$$
We will prove this in two cases.
(1) If $3 \leqslant s_{1} \leqslant \frac{-3+\sqrt{12... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,512 |
50. If $a, b, c$ are positive numbers such that $a b c=1$, then
$$(a+b)(b+c)(c+a)+7 \geqslant 5(a+b+c)$$ | 50. (2007.01.30) Proof: From problem 55 of this chapter, we have
$$\sum b c \geqslant \frac{9 \sum a}{6+\sum a}$$
Thus,
$$\begin{array}{l}
\quad(a+b)(b+c)(c+a)+7-5 \sum a= \sum a \cdot \sum b c+6-5 \sum a \geqslant \\
\sum a \cdot \frac{9 \sum a}{6+\sum a}+6-5 \sum a= \\
\frac{4\left(\sum a-3\right)^{2}}{6+\sum a} \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,513 |
51. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\begin{array}{l}
\frac{a^{3}}{\left(2 a^{2}+b^{2}\right)\left(2 a^{2}+c^{2}\right)}+\frac{b^{3}}{\left(2 b^{2}+c^{2}\right)\left(2 b^{2}+a^{2}\right)}+\frac{c^{3}}{\left(2 c^{2}+a^{2}\right)\left(2 c^{2}+b^{2}\right)} \leqslant \\
\frac{1... | 51. (2007.02.17) Simplify and prove
$$\begin{array}{c}
\sum \frac{a^{3}}{\left(2 a^{2}+b^{2}\right)\left(2 a^{2}+c^{2}\right)}=\sum \frac{a^{3}}{\left(a^{2}+a^{2}+b^{2}\right)\left(a^{2}+c^{2}+a^{2}\right)} \leqslant \\
\sum \frac{a^{3}}{\left(a^{2}+a c+a b\right)^{2}}=\sum \frac{a}{(a+b+c)^{2}}=\frac{1}{a+b+c}
\end{ar... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,514 |
52. Let \(a, b, c\) be non-negative numbers such that \(a+b+c \geqslant 3\). Prove that
$$\frac{1}{a^{2}+b+c}+\frac{1}{a+b^{2}+c}+\frac{1}{a+b+c^{2}} \leqslant 1$$ | 52. (2007.02.20) Simplify
$$\begin{aligned}
\sum \frac{1}{a^{2}+b+c}-1= & \sum \frac{1+b+c}{\left(a^{2}+b+c\right)(1+b+c)}-1 \leqslant \\
& \sum \frac{1+b+c}{(a+b+c)^{2}}-1=\frac{3+2 \sum a}{\left(\sum a\right)^{2}}-1= \\
& \frac{\left(1+\sum a\right)\left(3-\sum a\right)}{\left(\sum a\right)^{2}} \leqslant 0
\end{alig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,515 |
53. Let $a, b, c$ be non-negative numbers such that $a b+b c+c a=3$. If $r \geqslant 1$, then
$$\frac{1}{r+a^{2}+b^{2}}+\frac{1}{r+b^{2}+c^{2}}+\frac{1}{r+c^{2}+a^{2}} \leqslant \frac{3}{r+2}$$ | 53. (2007.02.06) Simplify: The original expression, after removing the denominator and simplifying, is equivalent to
$$\begin{array}{l}
-6 r^{2}+\left(2 r^{2}-8 r\right) \sum a^{2}+2(r-1) \sum\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)+ \\
3 \prod\left(b^{2}+c^{2}\right) \geqslant 0
\end{array}$$
Notice that $\su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,516 |
Example 12 In $\triangle A B C$, the lengths of the three sides are $a, b, c$, then
$$a^{2}\left(\frac{b}{c}-1\right)+b^{2}\left(\frac{c}{a}-1\right)+c^{2}\left(\frac{a}{b}-1\right) \geqslant 0$$
Equality in (14) holds if and only if $\triangle A B C$ is an equilateral triangle. | To prove that
$$\begin{array}{l}
\frac{a^{2} b}{c}+\frac{b^{2} c}{a}+\frac{c^{2} a}{b}-\left(\frac{a b^{2}}{c}+\frac{b c^{2}}{a}+\frac{c a^{2}}{b}\right)= \\
\frac{1}{a b c}\left[\left(\frac{a^{3} b^{2}}{c}+\frac{b^{3} c^{2}}{a}+\frac{c^{3} a^{2}}{b}\right)-\left(\frac{a^{2} b^{3}}{c}+\frac{b^{2} c^{3}}{a}+\frac{c^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,519 |
56. If $a, b, c$ are real numbers, then
$$2(1+a b c)+\sqrt{2\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)} \geqslant(1+a)(1+b)(1+c)$$ | $$\begin{array}{l}
2(1+a b c)+\sqrt{2\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)}= \\
2(1+a b c)+\sqrt{2\left[\left(\sum b c-1\right)^{2}+\left(\sum a-a b c\right)^{2}\right]} \geqslant \\
2(1+a b c)+\sum b c-1+\sum a-a b c= \\
1+\sum a+\sum b c+a b c= \\
(1+a)(1+b)(1+c)
\end{array}$$
Translate the ab... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,520 |
57. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\frac{a(b+c)}{a^{2}+b c}+\frac{b(c+a)}{b^{2}+c a}+\frac{c(a+b)}{c^{2}+a b} \geqslant 2$$ | 57. (2007.01.29) Proof: We will prove a stronger form of the inequality:
$$\sum \frac{a(b+c)}{a^{2}+b c} \geqslant 2+\frac{8 a^{2} b^{2} c^{2}}{\prod\left(a^{2}+b c\right)}$$
When one of $a, b, c$ is zero, assume without loss of generality that $c=0$, then inequality (1) becomes
$$\frac{b}{a}+\frac{a}{b} \geqslant 2$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,521 |
58. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\sqrt{\frac{a(b+c)}{a^{2}+b c}}+\sqrt{\frac{b(c+a)}{b^{2}+c a}}+\sqrt{\frac{c(a+b)}{c^{2}+a b}} \geqslant 2$$ | 58. (2007.02.20) Simplify
$$\begin{aligned}
\sum \sqrt{\frac{a(b+c)}{a^{2}+b c}}= & \sum \frac{a(b+c)}{\sqrt{(a b+a c)\left(a^{2}+b c\right)}} \geqslant \sum \frac{2 a(b+c)}{a b+a c+a^{2}+b c}= \\
& \sum \frac{2 a(b+c)}{(a+b)(a+c)}=\frac{2 \sum a(b+c)^{2}}{(a+b)(b+c)(a+c)}= \\
& \frac{2(a+b)(b+c)(a+c)+8 a b c}{(a+b)(b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,522 |
59. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b} \geqslant \frac{a}{a^{2}+b c}+\frac{b}{b^{2}+c a}+\frac{c}{c^{2}+a b}$$ | 59. (2007.02.05) Simplify: By symmetry, without loss of generality, assume $a \geqslant b \geqslant c$, then
$$\begin{aligned}
\text { LHS }- \text { RHS }= & \frac{(a-b)(a-c)}{(b+c)\left(a^{2}+b c\right)}-\frac{(a-b)(b-c)}{(c+a)\left(b^{2}+c a\right)}+\frac{(a-c)(b-c)}{(a+b)\left(c^{2}+a b\right)}= \\
& \frac{(a-b)\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,523 |
60. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b} \geqslant \frac{2 a}{3 a^{2}+b c}+\frac{2 b}{3 b^{2}+c a}+\frac{2 c}{3 c^{2}+a b}$$ | 60. (2007.02.09) Brief proof: By symmetry, without loss of generality, assume \(a \geqslant b \geqslant c\), then
$$\begin{aligned}
\sum \frac{1}{b+c}-\sum \frac{2 a}{3 a^{2}+b c}= & \sum \frac{(a-b)(a-c)+a(a-b)+a(a-c)}{(b+c)\left(3 a^{2}+b c\right)}= \\
& {\left[\frac{(a-b)(a-c)}{(b+c)\left(3 a^{2}+b c\right)}-\frac{(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,524 |
61. Let \(a, b, c\) be positive numbers such that \(a^{2}+b^{2}+c^{2}=3\). Prove that
$$5(a+b+c)+\frac{3}{a b c} \geqslant 18$$ | 61. (2007.01.23) Simplify
$$\begin{aligned}
5 \sum a+\frac{3}{a b c}= & \sum a+\sum a+\sum a+\sum a+\sum a+\frac{\sum a^{2}}{a b c} \geqslant \\
& 6 \cdot \sqrt[6]{\frac{\left(\sum a\right)^{5} \cdot \sum a^{2}}{a b c}}
\end{aligned}$$
Thus, it suffices to prove
$$6 \cdot \sqrt[6]{\frac{\left(\sum a\right)^{5} \cdot \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,525 |
62. Let $a, b, c$ be non-negative numbers such that $a+b+c=3$. Prove that
$$\frac{1}{6-a b}+\frac{1}{6-b c}+\frac{1}{6-c a} \leqslant \frac{3}{5}$$ | 62. (2007.01.24) Simplify: Note that $a+b+c=3$, after removing the denominator and simplifying the original expression, we get its equivalent form
$$\begin{array}{c}
36-16 \sum b c+13 a b c-a^{2} b^{2} c^{2} \geqslant 0 \\
a+b+c=3
\end{array}$$
Since
there must be one number among $a, b, c$ that is no greater than 1. ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,526 |
63. Let \( n \geqslant 4 \) and let \( a_{1}, a_{2}, \cdots, a_{n} \) be real numbers such that
\[ a_{1} + a_{2} + \cdots + a_{n} \geqslant n \text{ and } a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \geqslant n^{2} \]
Prove that
\[ \max \left\{a_{1}, a_{2}, \cdots, a_{n}\right\} \geqslant 2 \] | 63. (2007.01.30) Proof: Since
$$\sum_{i=1}^{n} a_{i} \geqslant n, \quad n \geqslant \sum_{i=1}^{n}\left(2-a_{i}\right)$$
and since
$$\sum_{i=1}^{n} a_{i}^{2} \geqslant n^{2} \quad n \geqslant 4$$
it follows that
$$\sum\left(2-a_{1}\right)^{2} \geqslant n^{2}-4 n+4 \sum_{i=1}^{n}\left(2-a_{i}\right) \geqslant n \cdot ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,527 |
64. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{13}{6}-\frac{2(a b+b c+c a)}{3\left(a^{2}+b^{2}+c^{2}\right)}$$ | 64. (2007.01.28) Proof: Without loss of generality, assume $a \geqslant b \geqslant c$, then it is easy to prove
$$(a+c)(a-c)^{2} \geqslant(b+c)(b-c)^{2}$$
Thus,
$$\begin{aligned}
\sum \frac{a}{b+c}-\frac{13}{6}+\frac{2 \sum b c}{3 \sum a^{2}}= & \left(\sum \frac{a}{b+c}-\frac{3}{2}\right)-\frac{2}{3}\left(1-\frac{\su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,528 |
Example 13 (USA, Pham Kim Hung) Let $a, b, c$ be the lengths of the sides of a triangle, then
$$2 \sum \frac{a^{2}}{b} \geqslant \sum a+\sum \frac{b^{2}}{a}$$
Equality in (15) holds if and only if $\triangle A B C$ is an equilateral triangle. | To prove that when $a \geqslant b \geqslant c$, it is easy to show that
$$\frac{b^{2}}{a}+\frac{c^{2}}{b}+\frac{a^{2}}{c} \geqslant \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}$$
Therefore, inequality (15) only needs to be proven under the condition $a \geqslant b \geqslant c$. In this case,
$$\begin{array}{l}
2 \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,530 |
66. Let \(a, b, c\) be non-negative numbers such that
\[
(a+b)(b+c)(c+a)=2
\]
Prove that
\[
\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \leqslant 1
\] | $$\begin{array}{l}
\Pi(b+c)^{2}-4 \Pi\left(a^{2}+b c\right)=\left[(a+b) c^{2}+(a+b)^{2} c+a b(a+b)\right]^{2}- \\
{\left[4 a b c^{4}+4\left(a^{3}+b^{3}\right) c^{3}+8 a^{2} b^{2} c^{2}+4 a b\left(a^{3}+b^{3}\right) c+4 a^{3} b^{3}\right]=} \\
(a-b)^{2} c^{4}+2(a+b)\left[2 a b-(a-b)^{2}\right] c^{3}+\left[(a+b)^{4}+2 a ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,531 |
1. Let $a, b, c$ be positive numbers such that $a b c=1$. Prove that
$$\sqrt{\frac{a+b}{b+1}}+\sqrt{\frac{b+c}{c+1}}+\sqrt{\frac{c+a}{a+1}} \geqslant 3$$ | 1. (2007.02.28) Prove briefly: $\sum \sqrt{\frac{a+b}{b+1}} \geqslant 3 \sqrt[6]{\frac{\prod(a+b)}{\prod(b+1)}}$, it is easy to prove $\Pi(b+c) \geqslant$ $\Pi(b+1)$, in fact, we have
$$\Pi(b+c)-\Pi(b+1)=\left(\sum a-3\right)\left(\sum b c-1\right)+2\left(\sum b c-3\right) \geqslant 0$$
Therefore, the original inequali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,532 |
2. Let $a, b, c$ be positive numbers such that $a b c=1$. Prove that
$$\sqrt{\frac{a}{b+3}}+\sqrt{\frac{b}{c+3}}+\sqrt{\frac{c}{a+3}} \geqslant \frac{3}{2}$$ | 2. (2007.03.27) Proof: Let \( a = \frac{y}{x}, b = \frac{z}{y}, c = \frac{x}{z}, x, y, z \in \mathbf{R}^{+} \), then the original inequality is equivalent to
$$\sum \frac{y}{\sqrt{z x + 3 x y}} \geqslant \frac{3}{2}$$
By the Cauchy-Schwarz inequality, we have
$$\sum y \sqrt{z x + 3 x y} \cdot \sum \frac{y}{\sqrt{z x +... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,533 |
3. Let $a, b, c$ be non-negative numbers such that $a+b+c=3$. Prove that
$$\frac{5-3 b c}{1+a}+\frac{5-3 c a}{1+b}+\frac{5-3 a b}{1+c} \geqslant a b+b c+c a$$ | 3. (2007.03.30) Simplify: Let $s_{1}=a+b+c=3, s_{2}=bc+ca+ab, s_{3}=abc$, then
$$\begin{aligned}
\sum \frac{5-3bc}{1+a}-\sum bc= & \frac{45-7s_{2}-3s_{2}^{2}+27s_{3}}{4+s_{2}+s_{3}}-s_{1}= \\
& \frac{45-11s_{2}-4s_{2}^{2}-s_{2}s_{3}+27s_{3}}{4+s_{2}+s_{3}}
\end{aligned}$$
Therefore, it suffices to prove
$$45-11s_{2}-4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,534 |
5. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\sqrt{\frac{a}{4 a+5 b}}+\sqrt{\frac{b}{4 b+5 c}}+\sqrt{\frac{c}{4 c+5 a}} \leqslant 1$$ | 5. (2007.03.27) Brief proof: The original proposition, after equivalent transformation, becomes
$$\sum \frac{1}{\sqrt{1+\frac{5}{4} \cdot \frac{b}{a}}} \leqslant 2$$
Let $x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}, x, y, z \in \mathbf{R}^{+}$, and $x y z=1$, then the above expression is equivalent to
$$\sum \frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,536 |
7. Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be real numbers. Prove that
$$\begin{array}{l}
a_{1} b_{1}+\cdots+a_{n} b_{n}+\sqrt{\left(a_{1}^{2}+\cdots+a_{n}^{2}\right)\left(b_{1}^{2}+\cdots+b_{n}^{2}\right)} \geqslant \\
\frac{2}{n}\left(a_{1}+\cdots+a_{n}\right)\left(b_{1}+\cdots+b_{n}\right... | 7. (2007.08.28, Excerpt from the original book) Proof: Let $b=\frac{1}{n} \sum b_{1}, x_{i}=2 b-b_{i}, i=1$, $2, \cdots, n$, then the original inequality is equivalent to
$$\sqrt{\sum a_{1}^{2} \cdot \sum b_{1}^{2}} \geqslant \sum a_{1}\left(2 b-b_{1}\right)$$
Furthermore,
$$\sum_{i=1}^{n} x_{i}^{2}=\sum_{i=1}^{n}\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,538 |
8. Let $k$ and $n$ be positive integers with $k<n$, and let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers such that $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. Prove that
$$\left(a_{1}+a_{2}+\cdots a_{n}\right)^{2} \geqslant n\left(a_{1} a_{k+1}+a_{2} a_{k+2}+\cdots+a_{n} a_{k}\right)$$
in the following c... | 8. (2007.08.29, Excerpt from the original book) Proof: (a) We only need to prove
$$\left(a_{1}+a_{2}+\cdots+a_{2 k}\right)^{2} \geqslant 4 k\left(a_{1} a_{k+1}+a_{2} a_{k+2}+\cdots+a_{k} a_{2 k}\right)$$
Let $x \in \mathbf{R}$, and $a_{k} \leqslant x \leqslant a_{k+1}$, we have
$$\left(x-a_{1}\right)\left(a_{k+1}-x\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,539 |
10. If $a, b, c$ are non-negative numbers, then
$$9\left(a^{4}+1\right)\left(b^{4}+1\right)\left(c^{4}+1\right) \geqslant 8\left(a^{2} b^{2} c^{2}+a b c+1\right)^{2}$$ | 10. (2007.02.03) Proof: Since
$$\begin{array}{l}
9\left(1+a^{4}\right)^{3}-8\left(1+a^{3}+a^{6}\right)^{2}=1-16 a^{3}+27 a^{4}-24 a^{6}+27 a^{8}-16 a^{9}+a^{12}= \\
(1-a)^{4}\left(1+4 a+10 a^{2}+4 a^{3}-2 a^{4}+4 a^{5}+10 a^{6}+4 a^{7}+a^{8}\right) \geqslant \\
0 \\
\text { Therefore, } \quad 9\left(1+a^{4}\right)^{3} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,542 |
12. Let $a, b, c$ be non-negative numbers, no two of which are zero. Then
$$\frac{1}{a^{2}+a b+b^{2}}+\frac{1}{b^{2}+b c+c^{2}}+\frac{1}{c^{2}+c a+a^{2}} \geqslant \frac{9}{(a+b+c)^{2}}$$ | 12. (2007.02.25) Proof: Take a point $P$ on the plane, and draw line segments $P A=a$, $P B=b$, $P C=c$, such that the angles from ray $P A$ to $P B$, $P B$ to $P C$, and $P C$ to $P A$ are all $120^{\circ}$. Connect $A$, $B$, and $C$ to form $\triangle A B C$, and denote $B C=a^{\prime}$, $C A=b^{\prime}$, $A B=c^{\pr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,544 |
14. Let $a, b, c$ be positive numbers, no two of which are zero. If $n$ is a positive integer, then
$$\frac{2 a^{n}-b^{n}-c^{n}}{b^{2}-b c+c^{2}}+\frac{2 b^{n}-c^{n}-a^{n}}{c^{2}-c a+a^{2}}+\frac{2 c^{n}-a^{n}-b^{n}}{a^{2}-a b+b^{2}} \geqslant 0$$ | 14. (2007.04.06) Simplify
$$\begin{array}{l}
\sum \frac{2 a^{n}-b^{n}-c^{n}}{b^{2}-b c+c^{2}}=\sum \frac{\left(a^{n}-b^{n}\right)-\left(c^{n}-a^{n}\right)}{b^{2}-b c+c^{2}}= \\
\sum\left(\frac{1}{b^{2}-b c+c^{2}}-\frac{1}{c^{2}-c a+a^{2}}\right)\left(a^{n}-b^{n}\right)= \\
\sum \frac{(a+b-c)(a-b)\left(a^{n}-b^{n}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,546 |
15. Let $0 \leqslant a \leqslant b$ and let $a_{1}, a_{2}, \cdots, a_{n} \in[a, b]$. Prove that $a_{1}+a_{2}+\cdots+a_{n}-n \sqrt[n]{a_{1} a_{n} \cdots a_{n}} \leqslant(n-1)(\sqrt{b}-\sqrt{a})^{2}$ | 15. (2007.08.29, Excerpt from the original book) Proof: First, we will prove that when $a_{1}, a_{2}, \cdots, a_{n} \in \{a, b\}$, the left-hand side of the equation is the largest. To prove this, consider $a_{2}, \cdots, a_{n}$ as fixed, and use proof by contradiction. If
$$f\left(a_{1}\right)=a_{1}+a_{2}+\cdots+a_{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,547 |
17. Let $a, b, c$ and $x, y, z$ be positive numbers such that $x+y+z=a+b+c$. Prove that
$$\frac{x(3 x+a)}{b c}+\frac{y(3 y+a)}{c a}+\frac{z(3 z+a)}{a b} \geqslant 12$$ | 17. (2007.03.30) Simplify: After eliminating the denominators and rearranging, it is sufficient to prove
$$3\left(a x^{2}+b y^{2}+c z^{2}\right)+\left(a^{2} x+b^{2} y+c^{2} z\right) \geqslant 12 a b c$$
From the previous problem (16), we have
$$3\left(a x^{2}+b y^{2}+c z^{2}\right)+3 x y z \geqslant 12 a b c$$
Also, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,549 |
18. Let \(a, b, c\) be positive numbers such that \(a^{2}+b^{2}+c^{2}=3\). Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{9}{a+b+c}$$ | 18. (2007.02.26) Proof: Since $\left(\sum \frac{a}{b}\right) \cdot \sum ab \geqslant \left(\sum a\right)^{2}$, it suffices to prove that
$$\frac{\left(\sum a\right)^{2}}{\sum bc} \geqslant \frac{9}{\sum a}$$
$$\left(\sum a\right)^{3} \geqslant 9 \sum bc$$
and
$$\begin{array}{l}
\left(\sum a\right)^{6} - 27\left(\sum b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,550 |
19. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers such that $a_{1} a_{2} \cdots a_{n}=1$. Prove that
$$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}+\frac{4 n}{n+a_{1}+a_{2}+\cdots+a_{n}} \geqslant n+2$$ | 19. (2007.08. 28 , Excerpt from the original book) Proof: Since
$$\frac{1}{n} \sum \frac{1}{a_{1}} \geqslant \sqrt[n-1]{\frac{\sum \frac{1}{a_{1} a_{2} \cdots a_{n-1}}}{n}}=\sqrt[n-1]{\frac{\sum a_{1}}{n}}$$
and let
$$\sqrt[n-1]{\frac{\sum a_{1}}{n}}=a$$
Therefore, it is sufficient to prove
$$\begin{array}{l}
n a+\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,551 |
Example 6 (Self-created problem, $1988,10,13$) Let the side lengths of two convex quadrilaterals in the same plane be $a$, $b, c, d$ and $a^{\prime}, b^{\prime}, c^{\prime}, d^{\prime}$, and their areas be $\Delta$ and $\Delta^{\prime}$, respectively. Then,
$$a a^{\prime}+b b^{\prime}+c c^{\prime}+d d^{\prime} \geqslan... | We know that, among convex quadrilaterals with given side lengths, the area is maximized when the quadrilateral is cyclic (inscribed in a circle) (explanation attached). Therefore, we only need to prove the case where both convex quadrilaterals are cyclic. The area of a cyclic convex quadrilateral with sides $a, b, c, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,552 |
20. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers such that $a_{1} a_{2} \cdots a_{n}=1$. Prove that
$$a_{1}+a_{2}+\cdots+a_{n}-n+1 \geqslant \sqrt[n-1]{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-n+1}$$ | 20. (2007.08. 28 , Excerpt from the original book) Proof: Let \( a = \frac{1}{n} \sum a_{1} \), then from \( a_{1} a_{2} \cdots a_{n} = 1 \) we know \( a \geqslant 1 \), and
$$a = \frac{\sum a_{1}}{n} \geqslant \sqrt[n-1]{\frac{\sum a_{1} a_{2} \cdots a_{n-1}}{n}} = \sqrt[n-1]{\frac{\sum \frac{1}{a_{1}}}{n}}$$
Thus,
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,554 |
22. Let $a, b, c$ be positive real numbers such that $a b c \geqslant 1$. Prove that
(a) $a^{\frac{a}{b}} b^{\frac{b}{c}} c^{\frac{c}{a}} \geqslant 1$;
(b) $a^{\frac{6}{b}} b^{\frac{6}{c}} c^{c} \geqslant 1$ | 22. (2007.08.29, Excerpt from the original book) Proof: (a) Using the substitution method \(x=\frac{a}{r}, y=\frac{b}{r}\) and \(z=\frac{c}{r}\), where \(r=\sqrt[3]{a b c} \geqslant 1\), we have \(x y z=1\) and
$$a^{\frac{a}{b}} b^{\frac{b}{c}} c^{\frac{c}{a}}=x^{\frac{x}{y}} y^{\frac{y}{z}} z^{\frac{z}{x}} r^{\frac{a}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,556 |
23. Let \(a, b, c, d\) be non-negative numbers. Prove that
\[4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)+15(a b c+b c d+c d a+d a b) \geqslant(a+b+c+d)^{3}\] | 23. (2007.02.27) Simplify: Let
$$\begin{array}{c}
s_{1}=a+b+c+d \\
s_{2}=a b+a c+a d+b c+b d+c d \\
s_{3}=b c d+a c d+a b d+a b c \\
s_{4}=a b c d
\end{array}$$
The left side - right side $=4\left(s_{1}^{3}-3 s_{1} s_{2}+3 s_{3}\right)+15 s_{3}-s_{1}^{3}=3\left(s_{1}^{3}-4 s_{1} s_{2}+9 s_{3}\right)$
$$\frac{3}{s_{1}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,557 |
24. Let $a, b, c$ be positive numbers such that $(a+b-c)\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)=4$. Prove that
$$\left(a^{4}+b^{4}+c^{4}\right)\left(\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}\right) \geqslant 2304$$ | 24. (2007.08. 28 , Excerpt from the original book) Proof: Since
$$\begin{array}{l}
4=(a+b-c)\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)=(a+b)\left(\frac{1}{a}+\frac{1}{b}\right)- \\
{\left[(a+b) \frac{1}{c}+\left(\frac{1}{a}+\frac{1}{b}\right) c\right]+1 \leqslant } \\
(a+b)\left(\frac{1}{a}+\frac{1}{b}\right)-2 ... | 2304 | Inequalities | proof | Yes | Yes | inequalities | false | 734,558 |
25. Let $a, b, c$ be positive numbers. Prove that
$$\frac{1}{a^{2}+2 b c}+\frac{1}{b^{2}+2 c a}+\frac{1}{c^{2}+2 a b}>\frac{2}{a b+b c+c a}$$ | 25. (2007.04.06) To prove the original inequality, it suffices to show that the left-hand side minus the right-hand side after clearing the denominators is not less than zero. Let $s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c$, then
$$\begin{array}{l}
\sum b c \cdot \sum\left(b^{2}+2 c a\right)\left(c^{2}+2 a b\right)-2 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,559 |
26. Let $a, b, c$ be non-negative numbers, no two of which are zero. Prove that
$$\frac{a(b+c)}{a^{2}+2 b c}+\frac{b(c+a)}{b^{2}+2 c a}+\frac{c(a+b)}{c^{2}+2 a b}>1+\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}$$ | 26. (2007.03.30) Simplify
$$\begin{aligned}
\sum \frac{a(b+c)}{a^{2}+2 b c}-1-\frac{\sum b c}{\sum a^{2}}= & \sum\left[\frac{a(b+c)}{a^{2}+2 b c}-\frac{a(b+c)}{\sum a^{2}}\right]+ \\
& \frac{2 \sum b c}{\sum a^{2}}-1-\frac{\sum b c}{\sum a^{2}}= \\
& \sum \frac{a(b+c)(b-c)^{2}}{\left(a^{2}+2 b c\right) \sum a^{2}}-\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,560 |
27. Let $a, b, c$ be non-negative numbers, no two of which are zero. Then
$$\frac{(b+c)^{2}}{a^{2}+b c}+\frac{(c+a)^{2}}{b^{2}+c a}+\frac{(a+b)^{2}}{c^{2}+a b} \geqslant 6$$ | $$\begin{array}{l}
2 a b(a b+a c) \geqslant\left(a^{2}+c^{2}\right)\left(b^{2}+c^{2}\right) \\
2 a^{2}\left(b^{2}+b c\right) \geqslant\left(a^{2}+c^{2}\right)\left(b^{2}+c^{2}\right)
\end{array}$$
This inequality is obviously true, so inequality (2) holds. Therefore,
$$\begin{array}{l}
\sum \frac{a(b+c)\left(-a^{2}+b^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,561 |
28. Let $a, b, c$ be non-negative numbers, no two of which are zero. Then
$$\frac{b+c}{2 a^{2}+b c}+\frac{c+a}{2 b^{2}+c a}+\frac{a+b}{2 c^{2}+a b} \geqslant \frac{6}{a+b+c}$$ | 28. (2007.04.06) Simplify: The original expression only needs to prove that the left side minus the right side after removing the denominator is not less than zero. Let
$$s_{1}=\sum a, s_{2}=\sum b c, s_{3}=a b c$$
Then
$$\begin{array}{l}
\sum a \cdot \sum\left(2 b^{2}+c a\right)\left(2 c^{2}+a b\right)(b+c)-6 \prod\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,562 |
Example 1 Let $a_{1}, a_{2}, a_{3} \in [0,1]$, prove that
$$\begin{array}{l}
\frac{a_{1}}{a_{2}+a_{3}+1}+\frac{a_{2}}{a_{3}+a_{1}+1}+\frac{a_{3}}{a_{1}+a_{2}+1}+ \\
\left(1-a_{1}\right)\left(1-a_{2}\right)\left(1-a_{3}\right) \leqslant 1
\end{array}$$
Equality in (1) holds if and only if two of $a_{1}, a_{2}, a_{3}$ a... | The left side of equation (1) is a symmetric expression in $a_{1}, a_{2}, a_{3}$. Without loss of generality, assume $0 \leqslant a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant 1$, then
the left side of equation (1) $\leqslant \frac{a_{1}}{a_{1}+a_{2}+1}+\frac{a_{2}}{a_{1}+a_{2}+1}+\frac{a_{3}}{a_{1}+a_{2}+1}+\left(1-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,564 |
30. Let $a, b, c$ be non-negative numbers, no two of which are zero. Then
$$\frac{a^{2}-b c}{\sqrt{a^{2}+b c}}+\frac{b^{2}-c a}{\sqrt{b^{2}+c a}}+\frac{c^{2}-a b}{\sqrt{c^{2}+a b}} \geqslant 0$$ | 30. (2007.05.01, provided by Chen Ji, see http://www. mathlinks. ro/ Forum/ viewtopic. php?t $=80731$ ) Simplify and prove: Let $D_{a}=\sqrt{a^{2}+k b c}, D_{b}=\sqrt{b^{2}+k c a}, D_{c}=\sqrt{c^{2}+k a b}, k \geqslant \frac{2}{3}$, then we can prove a more general form of the original inequality:
$$\sum \frac{a^{2}-b ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,565 |
31. If $a, b, c$ are non-negative numbers, then
$$\left(a^{2}-b c\right) \sqrt{a^{2}+4 b c}+\left(b^{2}-c a\right) \sqrt{b^{2}+4 c a}+\left(c^{2}-a b\right) \sqrt{c^{2}+4 a b} \geqslant 0$$ | 31. (2007.08.28, Excerpt from the original book) Proof: By symmetry, let $a \geqslant b \geqslant c$, and denote
$$\begin{array}{c}
x=\left(a^{2}-b c\right)(b+c), y=\left(b^{2}-c a\right)(c+a), z=\left(c^{2}-a b\right)(a+b) \\
A=\frac{\sqrt{a^{2}+4 b c}}{b+c}, B=\frac{\sqrt{b^{2}+4 c a}}{c+a}, C=\frac{\sqrt{c^{2}+4 a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,566 |
32. If $a, b, c$ are positive numbers, then
$$\frac{a^{2}-b c}{\sqrt{8 a^{2}+(b+c)^{2}}}+\frac{b^{2}-c a}{\sqrt{8 b^{2}+(c+a)^{2}}}+\frac{c^{2}-a b}{\sqrt{8 c^{2}+(a+b)^{2}}} \geqslant 0$$ | 32. (2007.05.02) Simplify: Let
Then
$$\begin{array}{l}
\text { LHS }=\frac{1}{2} \sum \frac{(a-b)(c+a)-(c-a)(a+b)}{D_{a}}= \\
\frac{1}{2} \sum\left(\frac{a+b}{D_{b}}-\frac{c+a}{D_{c}}\right)(b-c)= \\
\frac{1}{2} \sum \frac{\left[(a+b) D_{c}-(c+a) D_{b}\right](b-c)}{D_{b} D_{c}}= \\
\frac{1}{2} \sum \frac{\left\{(a+b)^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,567 |
33. If $a, b, c$ are non-negative numbers, then
$$\sqrt{a^{2}+b c}+\sqrt{b^{2}+c a}+\sqrt{c^{2}+a b} \leqslant \frac{3}{2}(a+b+c)$$ | 33. (2006.11.05) Proof: Without loss of generality, assume \( c \geqslant b \geqslant a \geqslant 0 \), then
$$\sqrt{c^{2}+a b} \leqslant \sqrt{c^{2}+a c} \leqslant c+\frac{a}{2}$$
Additionally, we prove
$$\sqrt{a^{2}+b c}+\sqrt{b^{2}+c a} \leqslant \frac{c}{2}+a+\frac{3 b}{2}$$
Since \(\left(\sqrt{a^{2}+b c}+\sqrt{b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,568 |
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