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79. For any positive real numbers $a, b, c$, we have $\frac{a^{3}}{b^{2}-b c+c^{2}}+\frac{b^{3}}{c^{2}-c a+a^{2}}+\frac{c^{3}}{a^{2}-a b+b^{2}} \geqslant$ $a+b+c$. (2006 Balkan Mathematical Olympiad) | 79. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\frac{a^{3}}{b^{2}-b c+c^{2}}+\frac{b^{3}}{c^{2}-c a+a^{2}}+\frac{c^{3}}{a^{2}-a b+b^{2}}= \\
\frac{a^{4}}{a\left(b^{2}-b c+c^{2}\right)}+\frac{b^{4}}{b\left(c^{2}-c a+a^{2}\right)}+\frac{c^{4}}{c\left(a^{2}-a b+b^{2}\right)} \geqslant \\
\frac{\left(a^{2}+b^{2}+c^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,504 |
80. Prove that for any real numbers $x_{0}, x_{1}, x_{2}, \cdots, x_{2 n}$ the inequality holds:
$$\sum_{k=0}^{2 n} x_{k}^{2} \geqslant \frac{1}{2 n+1}\left(\sum_{k=0}^{2 n} x_{k}\right)^{2}+\frac{3}{n(n+1)(2 n+1)}\left(\sum_{k=0}^{2 n}(k-n) x_{k}\right)^{2}$$ | 80. Replace $x_{k}$ in the original inequality with $x_{k}-u$, where $u$ is a constant, and the inequality remains unchanged. This means the inequality is equivalent to proving $\square$
$$\begin{array}{l}
\sum_{k=0}^{2 n}\left(x_{k}-u\right)^{2} \geqslant \frac{1}{2 n+1}\left[\sum_{k=0}^{2 n}\left(x_{k}-u\right)\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,505 |
81. Let $x, y, z$ be positive real numbers, and $x y z=1$, prove:
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4}$$
(39th IMO Shortlist Problem) | 81. By Cauchy-Schwarz inequality,
$$\begin{aligned}
3\left(x^{2}+y^{2}+z^{2}\right) \geqslant & (x+y+z)^{2}=(x+y+z)(x+y+z) \geqslant \\
& 3 \sqrt[3]{x y z}(x+y+z)= \\
& 3(x+y+z)
\end{aligned}$$
Therefore,
$$\begin{array}{l}
\text { Hence } x^{2}+y^{2}+z^{2} \geqslant x+y+z \\
2\left[\left(x^{2}+y^{2}+z^{2}\right)-(x y... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,506 |
27. Given that $a, b, c$ are non-negative real numbers, prove: $\frac{1}{3}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \leqslant$ $a^{2}+b^{2}+c^{2}-3 \sqrt[3]{a^{2} b^{2} c^{2}} \leqslant(a-b)^{2}+(b-c)^{2}+(c-a)^{2}$. (2005 Ireland Math | 27. By the AM-GM inequality, we have $a^{2}+b^{2}+c^{2}-3 \sqrt[3]{a^{2} b^{2} c^{2}} \geqslant 0$, then
$$\begin{array}{l}
\frac{1}{3}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]= \\
\frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)-\frac{2}{3}(a b+b c+c a) \leqslant \\
\frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)-\frac{2}{3} \cdo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,507 |
82. Given $x, y>0$, and $x^{3}+y^{4} \leqslant x^{2}+y^{3}$, prove: $x^{3}+y^{3} \leqslant 2$. (2002 Hungarian Mathematical Olympiad problem) | 82. By the Cauchy-Schwarz inequality,
$$\left(x^{2}+y^{3}\right)\left(x+y^{2}\right) \geqslant\left(x^{3}+y^{4}\right)\left(x+y^{2}\right) \geqslant\left(x^{2}+y^{3}\right)^{2}$$
Therefore,
$$x+y^{2} \geqslant x^{2}+y^{3}$$
Similarly, by (1) and the Cauchy-Schwarz inequality,
$$(1+y)\left(x+y^{2}\right) \geqslant(1+y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,508 |
84. (1) Let $a, b, c$ and $x, y, z$ be real numbers, prove:
$$\begin{array}{l}
a x+b y+c z+\sqrt{\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)} \geqslant \\
\frac{2}{3}(a+b+c)(x+y+z)
\end{array}$$
(1989 Ukrainian Mathematical Olympiad Problem) | 84. (1) We only need to prove:
$$\begin{array}{l}
\sqrt{\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)} \geqslant \\
\frac{1}{3}[a(2 y+2 z-x)+b(2 z+2 x-y)+c(2 x+2 y-z)]
\end{array}$$
By the transformation of the Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
\frac{1}{3}[a(2 y+2 z-x)+b(2 z+2 x-y)+c(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,510 |
(2) Let $a_{1}, a_{2}, \cdots, a_{n}$ and $x_{1}, x_{2}, \cdots, x_{n}$ be $2n$ real numbers. Prove:
$$\sum_{i=1}^{n} a_{i} x_{i}+\sqrt{\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} x_{i}^{2}\right)} \geqslant \frac{2}{n}\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} x_{i}\right)$$ | (2) Let $X=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}$, the original inequality is equivalent to
$$\begin{array}{c}
\sqrt{\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(-\sum_{i=1}^{n} x_{i}^{2}\right)} \geqslant \sum_{i=1}^{n} a_{i}\left(2 X-x_{i}\right) \\
\sum_{i=1}^{n}\left(2 X-x_{i}\right)^{2}=\sum_{i=1}^{n}\left(4 X^{2}-4 X ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,511 |
87. Let $x, y, z$ be non-negative numbers, and $x^{2}+y^{2}+z^{2}=3$, prove:
$$\frac{x}{\sqrt{x^{2}+y+z}}+\frac{y}{\sqrt{y^{2}+z+x}}+\frac{z}{\sqrt{z^{2}+x+y}} \leqslant \sqrt{3}$$
(2008 Ukrainian Mathematical Olympiad Problem) | 86. After squaring both sides, we only need to prove
$$\begin{array}{l}
2 \sqrt{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)}+2 \sqrt{\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}+2 \sqrt{\left(c^{2}+a^{2}\right)\left(a^{2}+b^{2}\right)} \geqslant \\
2\left(a^{2}+b^{2}+c^{2}\right)+3(a b+b c+c a)
\end{array}$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,513 |
88. Let \(a, b, c \in \left(\frac{1}{\sqrt{6}}, +\infty\right)\), and \(a^{2} + b^{2} + c^{2} = 1\), prove:
$$\frac{1+a^{2}}{\sqrt{2 a^{2}+3 a b-c^{2}}}+\frac{1+b^{2}}{\sqrt{2 b^{2}+3 b c-a^{2}}}+\frac{1+c^{2}}{\sqrt{2 c^{2}+3 c a-b^{2}}} \geqslant 2(a+b+c)$$
(2007 Croatian National Training Team Problem) | 88. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(\sqrt{2 a^{2}+3 a b-c^{2}}+\sqrt{2 b^{2}+3 b c-a^{2}}+\sqrt{2 c^{2}+3 c a-b^{2}}\right) \\
\left(\frac{a^{2}}{\sqrt{2 a^{2}+3 a b-c^{2}}}+\frac{b^{2}}{\sqrt{2 b^{2}+3 b c-a^{2}}}+\frac{c^{2}}{\sqrt{2 c^{2}+3 c a-b^{2}}}\right) \geqslant(a+b+c)^{2} \\
\left(\sqr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,514 |
$89-$ Let $a, b, c, d, e$ be non-negative real numbers, and $a+b=c+d+e$, find the largest positive real number $\bar{T}$, such that the inequality $\sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}} \geqslant T(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e})^{2}$ holds. (2007 Iran Mathematical Olympiad) | 89. Taking $a=b=3, c=d=e=2$, we get $T \leqslant \frac{\sqrt{30}}{6(\sqrt{3}+\sqrt{2})^{2}}$.
Next, we prove
$$\sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}} \geqslant \frac{\sqrt{30}}{6(\sqrt{3}+\sqrt{2})^{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}+\sqrt{e})^{2}$$
Let $X=a+b=c+d+e$, by the Cauchy-Schwarz inequality we have $a^{2... | \frac{\sqrt{30}}{6(\sqrt{3}+\sqrt{2})^{2}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,515 |
90. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, $S=\sum_{i=1}^{n} a_{i}$. Prove:-
$$(2 S+n)\left(-2 S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right) \geqslant 9\left(\sqrt{a_{1} a_{2}}+\sqrt{a_{2} a_{3}}+\cdots+\sqrt{a_{n} a_{1}}\right)^{2}$$
(2007 Baltic Way Mathematical Competition) | 90. Since
$$\begin{aligned}
2 S+n=\left(a_{1}+a_{2}+\cdots+a_{n}\right)+\left(a_{2}+\right. & \left.a_{3}+\cdots+a_{n}+a_{1}\right)+1+1+\cdots+1 \\
2 S+a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}= & \left(a_{2}+a_{3}+\cdots+a_{n}+a_{1}\right)+ \\
& \left(a_{1}+a_{2}+\cdots+a_{n}\right)+ \\
& \left(a_{1} a_{2}+a_{2} a_{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,516 |
92. Let $x, y, z$ be positive real numbers, and $x+y+z+xyz=4$, prove
$$\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \geqslant \frac{\sqrt{2}}{2}(x+y+z)$$
(2007 BMO (Balkan Mathematical Olympiad) Problem) | $$\begin{array}{l}
92. First, use proof by contradiction to prove x+y+z \geqslant x y+y z+z x. Assume x+y+z4(x+y+z)-(x+y+z)^{2}> \\
(x+y+z)[4-(x+y+z)]=x y z(x+y+z)
\end{array}$$
Thus, \( x+y+z<3 \). By the AM-GM inequality, \( \sqrt[3]{x y z} \leqslant \frac{x+y+z}{3}<1 \), which implies \( x y z<1 \). Therefore, \( x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,519 |
93. Let real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}=3$, prove the inequality $\frac{a^{2}}{2+b+c^{2}}+\frac{b^{2}}{2+c+a^{2}}$ $+\frac{c^{2}}{2+a+b^{2}} \geqslant \frac{(a+b+c)^{2}}{12}$, and determine when equality holds? (2008 Baltic Way Mathematical Olympiad) | 93. Since $a^{2}+b^{2}+c^{2}=3$, it follows that $-\sqrt{3} \leqslant a, b, c \leqslant \sqrt{3}$, hence $2+b+c^{2}$, $2+c+a^{2}$, $2+a+b^{2}$ are all positive. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{\left[\left(2+b+c^{2}\right)+\left(2+c+a^{2}\right)+\left(2+a+b^{2}\right)\right]} \\
\left(\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,520 |
94. Let $a, b, c, d$ be real numbers, prove the inequality: $(a+b+c+d)^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}+\right.$ $\left.d^{2}\right)+6 a b$. (1998 Polish Mathematical Olympiad) | 94. By the Cauchy-Schwarz inequality,
$$\begin{array}{c}
{[(a+b)+c+d]^{2} \leqslant\left(1^{2}+1^{2}+1^{2}\right)\left[(a+b)^{2}+c^{2}+d^{2}\right]} \\
(a+b+c+d)^{2} \leqslant 3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+6 a b
\end{array}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,521 |
95. Let $a, b, c, d$ be positive numbers, prove: $\frac{c}{a+2 b}+\frac{d}{b+2 c}+\frac{a}{c+2 d}+\frac{b}{d+2 a} \geqslant \frac{4}{3}$. (2005
Year Zhautykov Mathematical Olympiad Problem) | 95. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
{[c(a+2 b)+d(b+2 c)+a(c+2 d)+b(d+2 a)]} \\
\left(\frac{c}{a+2 b}+\frac{d}{b+2 c}+\frac{a}{c+2 d}+\frac{b}{d+2 a}\right) \geqslant(c+d+a+b)^{2}
\end{array}$$
Therefore,
$$\frac{c}{a+2 b}+\frac{d}{b+2 c}+\frac{a}{c+2 d}+\frac{b}{d+2 a} \geqslant \frac{(a+b+c+d)^{2}}{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,522 |
98. Let $x, y, z$ be positive numbers, and $x y + y z + z x = x + y + z$, prove:
$$\frac{1}{x^{2}+y+1}+\frac{1}{y^{2}+z+1}+\frac{1}{z^{2}+x+1} \leqslant 1$$ | 98. By Cauchy-Schwarz inequality,
$$\left(x^{2}+y+1\right)\left(1+y+z^{2}\right) \geqslant(x+y+z)^{2}$$
Therefore, $\frac{1}{x^{2}+y+1} \leqslant \frac{1+y+z^{2}}{(x+y+z)^{2}}$. Similarly,
$$\frac{1}{y^{2}+z+1} \leqslant \frac{1+z+x^{2}}{(x+y+z)^{2}}, \frac{1}{z^{2}+x+1} \leqslant \frac{1+x+y^{2}}{(x+y+z)^{2}}$$
Addi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,525 |
99. Let $a, b, c, d$ be positive numbers, prove: $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \geqslant 0$. (2009 Croatian National Training Team Problem) | 99. $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \geqslant 0 \Leftrightarrow \frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b} \geqslant 4$.
The proof is the same as for 97. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,526 |
101. Let $a, b, c$ be positive numbers, and $a b c \geqslant 1$, prove:
$$\frac{a+1}{a^{2}+a+1}+\frac{b+1}{b^{2}+b+1}+\frac{c+1}{c^{2}+c+1} \leqslant 2$$
(2008 Ukraine Mathematical Olympiad Problem) | 101. $\quad \frac{a+1}{a^{2}+a+1}+\frac{b+1}{b^{2}+b+1}+\frac{c+1}{c^{2}+c+1} \leqslant 2 \Leftrightarrow$
$$\frac{a^{2}}{a^{2}+a+1}+\frac{b^{2}}{b^{2}+b+1}+\frac{c^{2}}{c^{2}+c+1} \geqslant 1$$
Let $a=x^{3}, b=y^{3}, c=z^{3}$, then $x y z \geqslant 1$, then by Cauchy-Schwarz inequality we get
$$\begin{array}{l}
\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,528 |
102. Let $x, y, z$ be positive numbers, prove:
$$\frac{1+y z+z x}{(1+x+y)^{2}}+\frac{1+z x+x y}{(1+y+z)^{2}}+\frac{1+x y+y z}{(1+z+x)^{2}} \geqslant 1$$
(2010 Japan Mathematical Olympiad) | 102. By the Cauchy inequality, we have $(z(x+y)+1)\left(\frac{x+y}{z}+1\right) \geqslant(-x+y+\mathrm{F})^{2}$, so $\frac{1+y z+z x}{(1+x+y)^{2}} \geqslant \frac{z}{x+y \mp z}$, similarly $\frac{1+z x+x y}{(1+y+z)^{2}} \geqslant \frac{x}{x+y+z}, \frac{1+x y+y z}{(1+z+x)^{2}} \geqslant$ $\frac{y}{x+y+z}$, adding the thr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,530 |
105- Let $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive real numbers, satisfying: $\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\right.$ $\left.\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) \leqslant\left(n+\frac{1}{2}\right)^{2}$, prove: $\max \left\{a_{1} ; a_{2}, \cdots, a_{n} \right\} \l... | 105. By symmetry, without loss of generality, let $m=a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}=M$, we need to prove: $M \leqslant 4 m$. When $n=2$, the condition is $(m+M)\left(\frac{1}{m}+\frac{1}{M}\right) \leqslant \frac{25}{4}$. This is equivalent to $4(m+M)^{2} \leqslant 25 m M$, i.e., $(4 M-m)(M-4 m)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,533 |
107. Let $a, b, c$ be positive numbers, and $a+b+c=3$, prove: $\frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+$ $\frac{1}{2+c^{2}+a^{2}} \leqslant \frac{3}{4} \cdot$(2009 Iran National Training Team Problem) | 107. By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{\left[\left(2+a^{2}+b^{2}\right)+\left(2+b^{2}+c^{2}\right)+\left(2+c^{2}+a^{2}\right)\right]} \\
\left(\frac{a^{2}+b^{2}}{2+a^{2}+b^{2}}+\frac{b^{2}+c^{2}}{2+b^{2}+c^{2}}+\frac{c^{2}+a^{2}}{2+c^{2}+a^{2}}\right) \geqslant \\
\left(\sqrt{a^{2}+b^{2}}+\sqrt{... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,535 |
108. Let positive real numbers $a, b, c$ satisfy $a b + b c + c a = \frac{1}{3}$. Prove: $\frac{a}{a^{2} - b c + 1} + \frac{b}{b^{2} - c a + 1} + \frac{c}{c^{2} - a b + 1} \geqslant \frac{1}{a + b + c}$. (2009 Macedonia Mathematical Olympiad Problem) | 108. The denominator on the left side of the inequality is obviously positive.
By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\frac{a}{a^{2}-b c+1}+\frac{b}{b^{2}-c a+1}+\frac{c}{c^{2}-a b+1}= \\
\frac{a^{2}}{a^{3}-a b c+a}+\frac{b^{2}}{b^{3}-a b c+1}+\frac{c^{2}}{c^{3}-a b c+c} \geqslant \\
\frac{(a+b+c)^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,536 |
110. Let $a, b, c, d$ be positive real numbers, and satisfy $a b c d=1$, prove that: $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+$ $\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geqslant 1$. (2005 National Training Team for IMO Problem) | 110. Let $a=\frac{y z}{x^{2}}, b=\frac{z w}{y^{2}}, c=\frac{w x}{z^{2}}, d=\frac{x y}{w^{2}}$, then
$$\begin{array}{l}
\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geqslant 1 \Leftrightarrow \\
\frac{x^{4}}{\left(x^{2}+y z\right)^{2}}+\frac{y^{4}}{\left(y^{2}+z w\right)^{2}}+\frac{z^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,538 |
30. Let $x_{i}>0, i=1,2, \cdots, n . k \geqslant 1$, prove: $\sum_{i=1}^{n} \frac{1}{1+x_{i}} \sum_{i=1}^{n} x_{i} \leqslant \sum_{i=1}^{n} \frac{x_{i}^{k+1}}{1+x_{i}} \sum_{i=1}^{n} \frac{1}{x_{i}^{k}}$. | $$\begin{array}{l}
30. \text{Let } x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}, \text{ if } k \text{ is a positive integer, then} \\
\sum_{1 \leqslant i \neq j \leqslant n} \frac{x_{j}^{k+1}-x_{i}^{k+1}}{\left(1+x_{i}\right) x_{j}^{k}}= \\
\quad \sum_{i<j}^{n} \frac{x_{j}^{k+1}-x_{i}^{k+1}}{\left(1+x_{i}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,540 |
113. Given that $a, b, c$ are positive real numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove: $(a+b c+c)^{2}+$ $(b+c a+a)^{2}+(c+a b+b)^{2} \leqslant 27$. (2006 Poland Czechoslovakia Joint Competition Problem) | 113. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
(a+b c+c)^{2} \leqslant\left(a^{2}+b^{2}+c^{2}\right)\left(1^{2}+c^{2}+1^{2}\right)= \\
\left(2+c^{2}\right)\left(a^{2}+b^{2}+c^{2}\right)=3\left(-2+c^{2}\right)
\end{array}$$
Similarly, $(b+c a+a)^{2} \leqslant\left(b^{2}+c^{2}+a^{2}\right)\left(1^{2}+a^{2}+1^{2}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,542 |
115. Let $x, y, z$ be positive numbers, and $x+y+z=1$, prove: $\frac{x}{y^{2}+z}+\frac{y}{z^{2}+x}+\frac{z}{x^{2}+y} \geqslant \frac{9}{4}$. (2006 Serbia and Montenegro Mathematical Olympiad Problem) | 115. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
{\left[x\left(y^{2}+z\right)+y\left(z^{2}+x\right)+z\left(x^{2}+y\right)\right]\left(\frac{x}{y^{2}+z}+\frac{y}{z^{2}+x}+\frac{z}{x^{2}+y}\right) \geqslant} \\
(x+y+z)^{2}
\end{array}$$
It suffices to prove that
$$\begin{array}{l}
\quad 4(x+y+z)^{2}-9\left[x\left(y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,544 |
116. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, prove: $\frac{1}{\sum_{i=1}^{n} \frac{1}{1+a_{i}}}-\frac{1}{\sum_{i=1}^{n} \frac{1}{a_{i}}} \geqslant \frac{1}{n}$. (2001 Moldova Mathematical Olympiad Problem) | 116. Let $a=\sum_{i=1}^{n} \frac{1}{a_{i}}$, then $\sum_{i=1}^{n} \frac{1+a_{i}}{a_{i}}=\sum_{i=1}^{n}\left(1+\frac{1}{a_{i}}\right)=n+a$, by the Cauchy-Schwarz inequality we have
$$\sum_{i=1}^{n} \frac{a_{i}}{1+a_{i}} \sum_{i=1}^{n} \frac{1 \mp a_{i}}{a_{i}} \geqslant n^{2}$$
Therefore,
$$\sum_{i=1}^{n} \frac{a_{i}}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,545 |
118. Let positive real numbers $a, b, c$ satisfy $a^{3}+b^{3}+c^{3}=3$, prove: $\frac{1}{a^{2}+a+1}+\frac{1}{b^{2}+b+1}+$ $\frac{1}{c^{2}+c+1} \geqslant 1$. (2010 Chen Province Cup Mathematical Olympiad Problem) | 118. Since $(a-1)^{2}(a+1) \geqslant 0$, we have $a^{3}+2 \geqslant a^{2}+a+1$. Similarly, $b^{3}+2 \geqslant b^{2}+b+1$, $c^{3}+2 \geqslant c^{2}+c+1$, so
$$\frac{1}{a^{2}+a+1}+\frac{1}{b^{2}+b+1}+\frac{1}{c^{2}+c+1} \geqslant \frac{1}{a^{3}+2}+\frac{1}{b^{3}+2}+\frac{1}{c^{3}+2}$$
By the Cauchy-Schwarz inequality, w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,547 |
119. Let the equation $x^{4}+a x^{3}+2 x^{2}+b x+1=0$ have at least one real root, prove: $a^{2}+b^{2} \geqslant$ 8. (1993 International City Mathematics Competition Problem) | 119. By Cauchy-Schwarz inequality, we have $a^{2}+b^{2} \geqslant \frac{\left(x^{4}+2 x^{2}+1\right)^{2}}{x^{2}+x^{6}}$, it suffices to prove
$$\frac{\left(x^{4}+2 x^{2}+1\right)^{2}}{x^{2}+x^{6}} \geqslant 8 \Leftrightarrow\left(x^{2}-1\right)^{4} \geqslant 0$$ | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,548 |
120. Let real numbers $a, b, c \in[-1,1]$, and satisfy $1+2 a b c \geqslant a^{2}+b^{2}+c^{2}$, prove: $1+$ $2(a b c)^{n} \geqslant a^{2 n}+b^{2 n}+c^{2 n}$, where $n$ is any positive integer. (2010 IMC Problem) | 120. $1+2 a b c \geqslant a^{2}+b^{2}+c^{2}$ can be transformed into
$$(a-b c)^{2} \leqslant\left(1-b^{2}\right)\left(1-c^{2}\right)$$
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(a^{n-1}+a^{n-2} b c+\cdots+a b^{n-2} c^{n-2}+b^{n-1} c^{n-1}\right)^{2} \leqslant \\
\left(|a|^{n-1}+|a|^{n-2}|b c|+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,549 |
121. Let real numbers $x, y, z$ satisfy $x+y+z=0$, prove: $\frac{x(x+2)}{2 x^{2}+1}+\frac{y(y+2)}{2 y^{2}+1}+\frac{z(z+2)}{2 z^{2}+1} \geqslant$
0. (2011 Balkan Mathematical Olympiad Problem) | 121. $\frac{x(x+2)}{2 x^{2}+1}+\frac{1}{2}=\frac{(2 x+1)^{2}}{2\left(2 x^{2}+1\right)}, \frac{y(y+2)}{2 y^{2}+1}+\frac{1}{2}=\frac{(2 y+1)^{2}}{2\left(2 y^{2}+1\right)}, \frac{z(z+2)}{2 z^{2}+1}+$ $\frac{1}{2}=\frac{(2 z+1)^{2}}{2\left(2 z^{2}+1\right)}$, Therefore, we only need to prove $\frac{(2 x+1)^{2}}{2 x^{2}+1}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,550 |
31. Let $a_{i}$ be positive real numbers $(i=1,2, \cdots, n)$, and let $b_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, n)$, $C_{n}=\left(a_{1}-b_{1}\right)^{2}+\left(a_{2}-b_{2}\right)^{2}+\cdots+\left(a_{n}-b_{n}\right)^{2}$, $D_{n}=\left(a_{1}-b_{n}\right)^{2}+\left(a_{2}-b_{n}\right)^{2}+\cdots+\left(a_{n}-... | 31. Construct the sequences $x_{n}=2 C_{n}-D_{n}, y_{n}=D_{n}-C_{n}, n \in \mathbf{N}^{*}$, then
$$\begin{array}{l}
x_{n+1}-x_{n}=2\left(C_{n+1}-C_{n}\right)-\left(D_{n+1}-D_{n}\right)= \\
2\left(a_{n+1}-b_{n+1}\right)^{2}-\left(a_{n+1}-b_{n+1}\right)^{2}- \\
n\left(b_{n+1}^{2}-b_{n}^{2}\right)+2\left(b_{n+1}-b_{n}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,551 |
Example 2 Given that $x, y, z$ are positive real numbers, and satisfy $x^{4}+y^{4}+z^{4}=1$, find the minimum value of $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}$ + $\frac{z^{3}}{1-z^{8}}$. (1999 Jiangsu Province Mathematical Winter Camp Problem) | Solve by Cauchy-Schwarz inequality
$$\begin{array}{l}
\left(\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}}\right)\left[x^{5}\left(1-x^{8}\right)+y^{5}\left(1-y^{8}\right)+z^{5}\left(1-z^{8}\right)\right] \geqslant \\
\left(x^{4}+y^{4}+z^{4}\right)^{2}
\end{array}$$
By the AM-GM inequality, we get
S... | \frac{9}{8} \sqrt[4]{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,553 |
Example 3 Let $a, b, c, d$ be positive real numbers, and satisfy $a+b+c+d=1$. Prove that: $6\left(a^{3}+b^{3}+\right.$ $\left.c^{3}+d^{3}\right) \geqslant\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+\frac{1}{8}$. (8th Hong Kong Mathematical Olympiad Problem) | Prove that by the AM-GM inequality,
$$\begin{array}{l}
a^{3}+\left(\frac{a+b+c+d}{4}\right)^{3}+\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant 3 a\left(\frac{a+b+c+d}{4}\right)^{2} \\
b^{3}+\left(\frac{a+b+c+d}{4}\right)^{3}+\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant 3 b\left(\frac{a+b+c+d}{4}\right)^{2} \\
c^{3}+\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,554 |
Example 4 Let $\left\{a_{1}, a_{2}, a_{3}, \cdots\right\}$ be an infinite sequence of positive numbers. Prove the inequality $\sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 4 \sum_{n=1}^{N} a_{n}^{2}$ for any positive integer $N$. Here $\alpha_{n}$ is the average of $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$, i.e., $\alpha_{n} = \f... | Prove that $\alpha_{n}=\frac{a_{1}+a_{2}+a_{3}+\cdots+a_{n}}{n}$ satisfies
$$\begin{aligned}
\alpha_{n}^{2}-2 \alpha_{n} a_{n}= & \alpha_{n}^{2}-2 \alpha_{n}\left(n \alpha_{n}-(n-1) \alpha_{n-1}\right)= \\
& (1-2 n) \alpha_{n}^{2}+2(n-1) \alpha_{n} \alpha_{n-1} \leqslant \\
& (1-2 n) \alpha_{n}^{2}+(n-1)\left(\alpha_{n... | \sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 4 \sum_{n=1}^{N} a_{n}^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,555 |
Example 5 Given $a, b, c>0$, prove: $\sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(-a b^{2}+b c^{2}+c a^{2}\right)} \geqslant$ $a b c+\sqrt[3]{\left(a^{3}+a b c\right)\left(b^{3}+a b c\right)\left(c^{3}+a b c\right)}:(2001$ Korean Mathematical Olympiad $)$ | $$\begin{array}{l}
\sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)}= \\
\frac{1}{2} \sqrt{\left[b\left(a^{2}+b c\right)+c\left(b^{2}+c a\right)+a\left(c^{2}+a b\right)\right]:\left[c\left(a^{2} \pm b c\right)+a\left(b^{2}+c a\right)+b\left(c^{2}+a b\right)\right]} \geqslant \\
\frac{1}{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,556 |
Example 6 Let $a, b, c, \lambda > 0, a^{n-1} + b^{n-1} + c^{n-1} = 1 (n \geqslant 2)$, prove: $\frac{a^{n}}{b+\lambda c} +$ $\frac{b^{n}}{c+\lambda a} + \frac{c^{n}}{a+\lambda b} \geqslant \frac{1}{1+\lambda}$. (2006 China National Training Team Problem) | Prove that by Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(\frac{a^{n}}{b+\lambda c}+\frac{b^{n}}{c+\lambda a}+\frac{c^{n}}{a+\lambda b}\right)\left(a^{n-2}(b+\lambda c)+b^{n-2}(c+\lambda a)+c^{n-2}(a+\lambda b)\right) \geqslant \\
\left(a^{n-1}+b^{n-1}+c^{n-1}\right)^{2}=1
\end{array}$$
By the AM-GM inequality... | \frac{a^{n}}{b+\lambda c}+\frac{b^{n}}{c+\lambda a}+\frac{c^{n}}{a+\lambda b} \geqslant \frac{1}{1+\lambda} | Inequalities | proof | Yes | Yes | inequalities | false | 732,557 |
Example 7 Let $x, y, z$ be positive real numbers, and satisfy $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$, prove the inequality: $\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geqslant 1$. (2007 Asia Pacific Mathematical Olympiad Problem) | Prove that by Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(\frac{x^{2}}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}}{\sqrt{2 z^{2}(x+y)}}\right)(\sqrt{2(y+z)}+\sqrt{2(z+x)}+\sqrt{2(x+y)}) \geqslant \\
(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=1 \\
\left(\frac{y z}{\sqrt{2 x^{2}(y+z)}}+\frac{z x}{\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,558 |
1. Prove: The inequality $\frac{a^{2}}{(a+b)(a+c)}+\frac{b^{2}}{(b+c)(b+a)}+\frac{c^{2}}{(c+b)(c+a)} \geqslant \frac{3}{4}$ holds for all positive real numbers $a, b, c$. (2004 Croatian Mathematical Olympiad Problem) | $$\begin{array}{l}
\frac{a^{2}}{(a+b)(a+c)}+\frac{b^{2}}{(b-c)(b+a)}+\frac{c^{2}}{(c+b)(c+a)} \geqslant \\
\frac{(a+b+c)^{2}}{(a+b)(a+c)+(b+c)(b+a)+(c+b)(c+a)}= \\
\frac{(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)}= \\
\frac{(a+b+c)^{2}}{(a+b+c)^{2}+(a b+b c+c a)} \geqslant \\
\frac{(a+b+c)^{2}}{(a+b+c)^{2}+\frac{1}{... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,559 |
2. 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) | 2. 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
$$\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}} ... | \frac{3 \sqrt{3}}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,560 |
3- Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive numbers. Prove that: $\frac{x_{1}^{2}}{x_{2}^{2}}+\frac{x_{2}^{2}}{x_{3}^{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}^{2}}+\frac{x_{n}^{2}}{x_{1}^{2}} \geqslant \frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\cdots+$ $\frac{x_{n-1}}{x_{n}}+\frac{x_{n}}{x_{1}}$. (1981 German Nationa... | 3. By Cauchy-Schwarz inequality,
i.e.,
$$\begin{array}{l}
\left(\frac{x_{1}^{2}}{x_{2}^{2}}+\frac{x_{2}^{2}}{x_{3}^{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}^{2}}+\frac{x_{n}^{2}}{x_{1}^{2}}\right)\left(1^{2}+1^{2}+\cdots+1^{2}+1^{2}\right) \geqslant \\
\left(\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\cdots+\frac{x_{n-1}}{x_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,561 |
5. Let $a, b, c$ be positive numbers, and $a+b+c=1$, prove: $\frac{a}{1+a^{2}}+\frac{b}{1+b^{2}}+\frac{c}{1+c^{2}} \leqslant \frac{9}{10}$.
(1996 Polish Mathematical Olympiad Problem) | 5. By the AM-GM inequality, we have $a^{2}+\frac{1}{9} \geqslant \frac{2 a}{3}, b^{2}+\frac{1}{9} \geqslant \frac{2 b}{3}, c^{2}+\frac{1}{9} \geqslant \frac{2 c}{3}$, so
$$\begin{array}{l}
\frac{a}{1+a^{2}}+\frac{b}{1+b^{2}}+\frac{c}{1+c^{2}} \leqslant \frac{9}{10} \Leftrightarrow \\
\frac{a}{\frac{2 a}{3}+\frac{8}{9}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,564 |
6. Let $x, y, z$ be positive numbers, prove: $\sqrt{x^{2}+x y+y^{2}}+\sqrt{y^{2}+y z+z^{2}}+\sqrt{z^{2}+z x+x^{2}} \leqslant$ $\sqrt{3} x y z\left(\frac{1}{x^{2}} + \frac{1}{y^{2}}+\frac{1}{z^{2}}\right) \cdot($ (Mathematical Bulletin Problem 799) | 6. By the Cauchy-Schwarz inequality, we have $\sqrt{a}+\sqrt{b}+\sqrt{c} \leqslant \sqrt{3(a+b+c)}$, and applying $x y+y z+z x \leqslant x^{2}+y^{2}+z^{2}$, we get
$$\begin{array}{l}
\sqrt{x^{2}+x y+y^{2}}+\sqrt{y^{2}+y z+z^{2}}+\sqrt{z^{2}+z x+x^{2}} \leqslant \\
\sqrt{3\left(2 x^{2}+2 y^{2}+2 z^{2}+x y+y z+z x\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,565 |
7. Let $u, v, w$ be positive real numbers, satisfying the condition $u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=1$. Find the minimum value of $u+v+w$. (3rd China Girls Mathematical Olympiad Problem) | $$\text { 7. } \begin{aligned}
1= & u \sqrt{v w}+v \sqrt{w u}+w \sqrt{u v}=\sqrt{u v w}(\sqrt{u}+\sqrt{v}+\sqrt{w})= \\
& \sqrt{3 u v w} \sqrt{\frac{(\sqrt{u}+\sqrt{v}+\sqrt{w})^{2}}{3}} \leqslant \\
& \sqrt{3\left(\frac{u+v+w}{3}\right)^{3}} \sqrt{u+v+w}= \\
& \frac{1}{3}(u+v+w)^{2}
\end{aligned}$$
Therefore, the min... | \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,566 |
9. Let $a, b, c$ be positive numbers, prove: $\frac{1+\sqrt{3}}{3 \sqrt{3}}\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a+b+c+$ $\sqrt{a^{2}+b^{2}+c^{2}}$. (2002 Albanian Mathematical Olympiad Problem) | 9 .
$$\begin{array}{l}
\frac{1+\sqrt{3}}{3 \sqrt{3}}\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a+b+c+\sqrt{a^{2}+b^{2}+c^{2}} \Leftrightarrow \\
\frac{1}{3}\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{\sqrt{3}}{9}\left(a^{2}+b^{2}+c^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,568 |
11. Let $a, b, c$ be positive numbers, and $a+b+c=1$, prove: $\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geqslant 3\left(a^{2}+b^{2}+c^{2}\right)$. (2006 Romanian National Training Team Problem) | 11. Since $a, b, c$ are positive numbers, and $a+b+c=1$, the original inequality is equivalent to
$$\begin{array}{l}
(a+b+c)\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right) \geqslant 3\left(a^{2}+b^{2}+c^{2}\right) \Leftrightarrow \\
\frac{a^{2}(a+c)}{b}+\frac{b^{2}(a+b)}{c}+\frac{c^{2}(b+c)}{a} \geqslant 2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,570 |
12. Let $a, b, c$ be positive constants, and $x^{2}+y^{2}+z^{2}=1$, find
$$f(x, y, z)=\sqrt{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}}+\sqrt{a^{2} y^{2}+b^{2} z^{2}+c^{2} x^{2}}+\sqrt{a^{2} z^{2}+b^{2} x^{2}+c^{2} y^{2}}$$
the maximum and minimum values. (1999 China National Training Team Problem) | 12. By Cauchy-Schwarz inequality,
$$\left(a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \geqslant\left(a x^{2}+b y^{2}+c z^{2}\right)^{2}$$
Therefore, $\square$
$$\sqrt{a^{2} x^{2}+b^{2} y^{2}+c^{2} z^{2}} \geqslant a x^{2}+b y^{2}+c z^{2}$$
Similarly,
$$\begin{array}{l}
\sqrt{a^{2} y^{2}+b... | \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)} \text{ and } a+b+c | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,571 |
13. Let $0<a, b, c<1$, prove: $\sqrt{a b c}+\sqrt{(1-a)(1-b)(1-c)}<1$. (2002 Romanian Mathematical Olympiad) | 13. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\sqrt{a b c}+\sqrt{(1-a)(1-b)(1-c)}= \\
\sqrt{a} \cdot \sqrt{b c}+\sqrt{1-a} \cdot \sqrt{(1-b)(1-c)} \leqslant \\
\sqrt{a+(1-a)} \cdot \sqrt{b c+(1-b)(1-c)}= \\
\sqrt{b c+(1-b)(1-c)}=\sqrt{1-b-c+2 b c}
\end{array}$$
Since $0<b, c<1$, we have
$$1-b-c+2 b c<1-b^{2}-c^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,572 |
33. (1) Given real numbers $a, b, c$, satisfying $a+b+c=0$, prove: $a^{3}+b^{3}+c^{3}>0 \Leftrightarrow a^{5}+$ $b^{5}+c^{5}>0$
(2) Given real numbers $a, b, c, d$ satisfying $a+b+c+d=0$, prove: $a^{3}+b^{3}+c^{3}+d^{3}>0 \Leftrightarrow$ $a^{5}+b^{5}+c^{5}+d^{5}>0$. (2004 British National Training Team Selection Test) | 33. (1) $a+b+c=0$, then
$$\begin{array}{l}
c=-(a+b), a^{3}+b^{3}+c^{3}=a^{3}+b^{3}-(a+b)^{3}=-3 a b(a+b) \\
a^{5}+b^{5}+c^{5}=a^{5}+b^{5}-(a+b)^{5}=-5 a b(a+b)-\left(a^{2}+a b+b^{2}\right)
\end{array}$$
So $\square$
$$a^{3}+b^{3}+c^{3}>0 \Leftrightarrow a^{5}+b^{5}+c^{5}>0$$
(2) $a+b+c+d=0$, then $d=-(a+b+c)$, thus
$$... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,574 |
17. If $0<a, b, c<1$ satisfies $a b+b c+c a=1$, find the minimum value of $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$. (2004 Sichuan Province Mathematics Competition Problem) | 17. Since $(a+b+c)^{2} \geqslant 3(a b+b c+c a)=3$, therefore, $a+b+c \geqslant \sqrt{3}$. By the Cauchy-Schwarz inequality, we have
$$\left(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\right)[(1-a)+(1-b)+(1-c)] \geqslant 9$$
Thus,
$$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geqslant \frac{9}{3-(a+b+c)} \geqslant \frac{... | \frac{3(3+\sqrt{3})}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,579 |
18. Given that $x, y, z$ are positive numbers, prove: $x y z(x+2)(y+2)(z+2) \leqslant \left(1+ \frac{2(x y+y z+z x)}{3}\right)^{3} \cdot(2006$ Korean Mathematical Olympiad problem) | 18.
$$\begin{array}{l}
{\left[1+\frac{2(x y+y z+z x)}{3}\right]^{3}=} \\
\left\{\frac{1}{3}\left[x\left(\frac{1}{x}+y+z\right)+y\left(x+\frac{1}{y}+z\right)+z\left(x+y+\frac{1}{z}\right)\right]\right\}^{3} \geqslant \\
x y z\left(\frac{1}{x}+y+z\right)\left(x+\frac{1}{y}+z\right)\left(x+y+\frac{1}{z}\right)
\end{array}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,580 |
19. Find the maximum value of the expression $x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}$. (1990 Moscow Mathematical Olympiad problem) | 19. By Cauchy-Schwarz inequality, we have
$$\left|x \sqrt{1-y^{2}}+y \sqrt{1-x^{2}}\right|^{2} \leqslant\left(x^{2}+y^{2}\right)\left(2-x^{2}-y^{2}\right)$$
By the AM-GM inequality, we have
$$\left(x^{2}+y^{2}\right)\left(2-x^{2}-y^{2}\right) \leqslant\left[\frac{\left(x^{2}+y^{2}\right)+\left(2-x^{2}-y^{2}\right)}{2}... | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,581 |
20. Given real numbers $a, b, c, x, y, z$ satisfy $(a+b+c)(x+y+z)=3,\left(a^{2}+b^{2}+c^{2}\right)$. $\left(x^{2}+y^{2}+z^{2}\right)=4$, prove that $: a x+b y+c z \geqslant 0 .(2004$ China National Training Team Problem) | 20. Clearly $a^{2}+b^{2}+c^{2} \neq 0, x^{2}+y^{2}+z^{2} \neq 0$.
Let $\alpha=\sqrt[4]{\frac{a^{2}+b^{2}+c^{2}}{x^{2}+y^{2}+z^{2}}} \neq 0, a_{1}=\frac{a}{\alpha}, b_{1}=\frac{b}{\alpha}, c_{1}=\frac{c}{\alpha}, x_{1}=x a, y_{1}=y b, z_{1}=z c$,
then
$$\begin{array}{c}
\left(a_{1}+b_{1}+c_{1}\right)\left(x_{1}+y_{1}+z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,582 |
22. Let $a, b, c$ be positive numbers, and $abc=1$, prove: $\frac{1+ab^{2}}{c^{3}}+\frac{1+bc^{2}}{a^{3}}+\frac{1+ca^{2}}{b^{3}} \geqslant$ $\frac{18}{a^{3}+b^{3}+c^{3}} \cdot(2000$ Hong Kong Mathematical Olympiad Problem) | 22. Since $a b c=1$, we have
$$\frac{1+a b^{2}}{c^{3}}+\frac{1+b c^{2}}{a^{3}}+\frac{1+c a^{2}}{b^{3}}=\frac{a b c+a b^{2}}{c^{3}}+\frac{a b c+b c^{2}}{a^{3}}+\frac{a b c+c a^{2}}{b^{3}}$$
By the Cauchy-Schwarz inequality and the AM-GM inequality, we get
$$\begin{array}{l}
\left(c^{3}+a^{3}+b^{3}\right)\left(\frac{a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,584 |
34. Given that $a, b, c$ are positive real numbers, prove: $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{3}{2} \cdot \frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}$ | 34. From Example 4
$$\begin{array}{l}
\frac{a}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}-\frac{a+b+c}{2}= \\
\frac{(a+b+c)(a-b)^{2}}{2(b+c)(c+a)}+\frac{(a+b+c)(a-c)^{2}}{2(b+c)(a+b)}+\frac{(a+b+c)(b-c)^{2}}{2(c+a)(a+b)}
\end{array}$$
And
$$\begin{array}{l}
\frac{3}{2} \cdot \frac{a^{3}+b^{3}+c^{3}}{a^{2}+b^{2}+c^{2}}-\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,585 |
23. Let $x, y, z$ be positive real numbers, and $x y z=1$, prove:
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4}$$
(39th IMO Shortlist) | 23. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\frac{x^{4}}{x(1+y)(1+z)}+\frac{y^{4}}{y(1+z)(1+x)}+\frac{z^{4}}{z(1+x)(1+y)} \geqslant \\
\frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{x(1+y)(1+z)+y(1+z)(1+x)+z(1+x)(1+y)}
\end{array}$$
By Cauchy-Schwarz inequality, we have \(3\left(x^{2}+y^{2}+z^{2}\right) \geqslant(... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,586 |
24. Let real numbers $a, b, c$ satisfy $a^{2}+2 b^{2}+3 c^{2}=\frac{3}{2}$, prove: $3^{-a}+9^{-b}+27^{-c} \geqslant 1$. (First China Southeast Mathematical Olympiad) | 24. By the Cauchy-Schwarz inequality, $(a+2 b+3 c)^{2} \leqslant(1+2+3)\left(a^{2}+2 b^{2}+3 c^{2}\right)=9$, then $a+2 b+3 c \leqslant 3$.
Therefore,
$$3^{-a}+9^{-b}+27^{-c} \geqslant 3 \cdot \sqrt[3]{3^{-(a+2 b+3 c)}}=3 \cdot \sqrt[3]{3^{-3}}=1$$ | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 732,587 |
27. If $x, y, z > 0$, prove:
$$\frac{(x+1)(y+1)^{2}}{3 \sqrt[3]{z^{2} x^{2}}+1}+\frac{(y+1)(z+1)^{2}}{3 \sqrt[3]{x^{2} y^{2}}+1}+\frac{(z+1)(x+1)^{2}}{3 \sqrt[3]{y^{2} z^{2}}+1} \geq x+y+z+3$$
(2007 Bulgarian Mathematical Olympiad) | 27. By the AM-GM inequality, we have $x y+x+y \geqslant 3 \sqrt[3]{x^{2} y^{2}}$, i.e., $(x+1)(y+1) \geqslant 3 \sqrt[3]{x^{2} y^{2}}+1$, so
$$\begin{array}{l}
\frac{(x+1)(y+1)^{2}}{3 \sqrt[3]{z^{2} x^{2}}+1}+\frac{(y+1)(z+1)^{2}}{3 \sqrt[3]{x^{2} y^{2}}+1}+\frac{(z+1)(x+1)^{2}}{3 \sqrt[3]{y^{2} z^{2}}+1} \geqslant \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,590 |
28. Let $x, y, z$ be positive numbers, and $x+y+z \geqslant 1$, prove: $\frac{x \sqrt{x}}{y+z}+\frac{y \sqrt{y}}{z+x}+\frac{z \sqrt{z}}{x+y} \geqslant \frac{\sqrt{3}}{2}$. (2003 Moldova National Training Team Problem) | 28. By the AM-GM inequality, we have $x^{\frac{3}{2}}+y^{\frac{3}{2}}+y^{\frac{3}{2}} \geqslant 3 x^{\frac{1}{2}} y$, $x^{\frac{3}{2}}+z^{\frac{3}{2}}+z^{\frac{3}{2}} \geqslant 3 x^{\frac{1}{2}} z$. Adding these, we get
$$2\left(x^{\frac{3}{2}}+y^{\frac{3}{2}}+z^{\frac{3}{2}}\right) \geqslant 3 x^{\frac{1}{2}}(y+z)$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,591 |
29. Given that $a, b, c$ are non-negative real numbers, prove: $(a+b)^{3}+4 c^{3} \geqslant 4\left(\sqrt{a^{3} b^{3}}+\sqrt{b^{3} c^{3}}+\sqrt{c^{3} a^{3}}\right);$ (2008 Polish Mathematical Olympiad problem) | $$\begin{array}{l}
\text { 29. }(a+b)^{3}+4 c^{3}=a^{3}+b^{3}+3 a^{2} b+3 a b^{2}+4 c^{3}=\left(a^{3}+b^{3}+a^{2} b+a b^{2}\right)+ \\
2 a^{2} b+2 a b^{2}+4 c^{3}=2\left(a^{2} b+a b^{2}\right)+\left(a^{2}+b^{2}\right)(a+b)+4 c^{3} \geqslant \\
4 \sqrt{a^{3} b^{3}}+\left(a^{\frac{3}{2}}+b^{\frac{3}{2}}\right)^{2}+4 c^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,592 |
30. Let $x, y, z$ be positive real numbers, and $x+y+z=3$, prove: $\frac{x^{3}}{y^{3}+8}+\frac{y^{3}}{z^{3}+8}+\frac{z^{3}}{x^{3}+8} \geqslant$ $\frac{1}{9}+\frac{2}{27}(x y+y z+z x) \cdot(2008$ Iran Mathematical Olympiad problem $)$ | 30. By the mean inequality, we have
$$x^{2}+y^{2}+z^{2} \geqslant x y+y z+z x$$
Therefore,
$$\begin{array}{c}
(x+y+z)^{2} \geqslant 3(x y+y z+z x) \\
\frac{1}{9}+\frac{2}{27}(x y+y z+z x) \leqslant \frac{1}{3}
\end{array}$$
Thus, it suffices to prove that
$$\frac{x^{3}}{y^{3}+8}+\frac{y^{3}}{z^{3}+8}+\frac{z^{3}}{x^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,593 |
31. Given that $a, b, c$ are all positive real numbers, and $abc=1$, prove: $\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+$ $\frac{1}{a(c+a)} \geqslant \frac{3}{2}$. (2008 Tajikistan Mathematical Olympiad Problem) | 31. Let $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, then by the Cauchy-Schwarz inequality we have
$$\begin{aligned}
\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}= & \frac{x^{2}}{z^{2}+x y}+\frac{y^{2}}{x^{2}+y z}+\frac{z^{2}}{y^{2}+z x} \geqslant \\
& \frac{\left(x^{2}+y^{2}+z^{2}\right)^{2}}{x^{2} y^{2}+y^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,594 |
35. For the cubic equation $x^{3} + a x^{2} + b x + c = 0$ with three real roots, prove that $a^{2} - 3 b \geqslant 0$, and $\sqrt{a^{2} - 3 b}$ is not greater than the difference between the largest and smallest roots (1988 US Mathematical Olympiad problem) | 35. Let the three roots of the equation be $p, q, r$, and $p \leqslant q \leqslant r$. By the relationship between roots and coefficients, $a=$ - $(p+q+r), b=p q+q r+r p$. Thus,
$$\begin{aligned}
a^{2}-3 b= & (p+q+r)^{2}-3(p q+q r+r p)= \\
& p^{2}+q^{2}+r^{2}-(p q+q r+r p)= \\
& \frac{1}{2}\left[(p-q)^{2}+(q-r)^{2}+(r-... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,596 |
33. Let $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ be positive integers, prove: $\left(\sum_{i=1}^{n} a_{i}^{2} / \sum_{i=1}^{n} a_{i}\right)^{\frac{k n}{1}} \geqslant \prod_{i=1}^{n} a_{i}$. Here $k=$ max $\left\{a_{1}, a_{2}, a_{3}, \cdots, a_{n}\right\}, t=\min \left\{a_{1}, a_{2}, a_{3}, \cdots, a_{n}\right\}$, and indic... | 33. Using the Cauchy-Schwarz inequality, we get
$$\sum_{i=1}^{n} a_{i}^{2} \sum_{i=1}^{n} 1^{2} \geqslant\left(\sum_{i=1}^{n} a_{i}\right)^{2}$$, so
$$\sum_{i=1}^{n} a_{i}^{2} \geqslant \frac{\left(\sum_{i=1}^{n} a_{i}\right)^{2}}{n}$$
Thus, using the AM-GM inequality, we get
$$\frac{\sum_{i=1}^{n} a_{i}^{2}}{\sum_{i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,597 |
36. Given that $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ are positive numbers, and $a_{1} a_{2} a_{3} \cdots a_{n}=1$, prove: $\sqrt{a_{1}}+$ $\sqrt{a_{2}}+\cdots+\sqrt{a_{n}} \leqslant a_{1}+a_{2}+\cdots+a_{n}$. (1997 Moldova Mathematical Olympiad Problem) | 36. By Cauchy-Schwarz inequality and AM-GM inequality, we have
$$\begin{array}{l}
n\left(a_{1}+a_{2}+\cdots+a_{n}\right) \geqslant\left(\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}}\right)^{2}= \\
\left(\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}}\right)\left(\sqrt{a_{1}}+\sqrt{a_{2}}+\cdots+\sqrt{a_{n}}\right) \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,600 |
38. If $x, y, z \geqslant 0$, and satisfy $x^{2}+y^{2}+z^{2}=3$, prove: $\frac{x^{2009}-2008(x-1)}{y+z}+$ $\frac{y^{2009}-2008(y-1)}{z+x}+\frac{z^{2009}-2008(z-1)}{x+y} \geqslant \frac{1}{2}(x+y+z) \cdot$ (2009 Northern China Mathematical Invitational Tournament Problem) | 38. By the AM-GM inequality, we have $x^{2009}+2008=x^{2009}+1+1+\cdots+1 \geqslant 2009 x$, similarly, $y^{2009}+2008 \geqslant 2009 y, z^{2009}+2008 \geqslant 2009 z$. Therefore,
$$\begin{array}{l}
\frac{x^{2009}-2008(x-1)}{y+z}+\frac{y^{2009}-2008(y-1)}{z+x}+\frac{z^{2009}-2008(z-1)}{x+y} \geqslant \\
\frac{x}{y+z}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,602 |
40. Let $a, b, c$ be positive real numbers, and $a+b+c=1$, prove that: $(a b)^{\frac{5}{4}}+(b c)^{\frac{5}{4}}+(c a)^{\frac{5}{4}}< \frac{1}{4} \cdot$ (2004 China National Training Team Problem) | 40. Let $a=\max \{a, b, c\}$, by Cauchy-Schwarz inequality and AM-GM inequality, we have
$$\begin{array}{l}
{\left[(a b)^{\frac{5}{4}}+(b c)^{\frac{5}{4}}+(c a)^{\frac{5}{4}}\right]^{2} \leqslant} \\
{\left[(a b)^{2}+(b c)^{\frac{2}{2}}+(c a)^{2}\right]\left[(a b)^{\frac{1}{2}}+(b c)^{\frac{1}{2}}+(c a)^{\frac{1}{2}}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,604 |
41. Let $a, b, c$ be positive real numbers, prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{(a+b+c)^{2}} \geqslant \frac{7}{25}\left(\frac{1}{a}+\frac{1}{b}+\right.$ $\left.\frac{1}{c}+\frac{1}{a+b+c}\right)^{2} .(2010$ Iran Mathematical Olympiad Summer Camp Problem) | 41. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
{\left[1^{2}+1^{2}+1^{2}+\left(\frac{1}{3}\right)^{2}\right]\left[\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{(a+b+c)^{2}}\right] \geqslant} \\
{\left[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{3(a+b+c)}\right]^{2}}
\end{array}$$
That is,
$$\frac{1}{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,605 |
42. Let $a, b, c$ be positive real numbers, and $abc=\frac{9}{4}$, prove: $a^{3}+b^{3}+c^{3}>a \sqrt{b+c}+b \sqrt{c+a}+c \sqrt{a+b}$. (2002 JBMO Problem) | 42. By the AM-GM inequality, we have $a^{3}+b^{3}+c^{3} \geqslant 3 a b c>6$, so by the rearrangement inequality and the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
a^{3}+b^{3}+c^{3} \geqslant & \sqrt{6\left(a^{3}+b^{3}+c^{3}\right)} \geqslant \sqrt{2\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)}= \\
& \sqrt{\left(a^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,606 |
36. (1) Given that $x, y$ are positive numbers, and $x y \geqslant 1$, prove that $x^{3}+y^{3}+4 x y \geqslant x^{2}+y^{2}+x+y+2$.
(2) Given that $x, y$ are positive numbers, and $x y \geqslant 1$, prove: $2\left(x^{3}+y^{3}+x y+x+y\right) \geqslant 5\left(x^{2}+y^{2}\right)$. | 36. (1) Clearly, $x+y \geqslant 2$.
$$x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right) \geqslant 2\left(x^{2}-x y+y^{2}\right)$$
To prove $x^{3}+y^{3}+4 x y \geqslant x^{2}+y^{2}+x+y+2$, it suffices to prove
$$2\left(x^{2}+x y+y^{2}\right) \geqslant x^{2}+y^{2}+x+y+2 \Leftrightarrow x^{2}+y^{2}+2 x y \geqslant x+y+2$$
G... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,607 |
43. Let $a, b, c, d$ be positive real numbers, prove: $\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c d}+\frac{b^{2}+c^{2}+d^{2}}{b c+c d+d a}+\frac{c^{2}+d^{2}+a^{2}}{c d+d a+a b}+$ $\frac{d^{2}+a^{2}+b^{2}}{d a+a b+b c} \geqslant 4 .(2010$ Kazakhstan Mathematical Olympiad Problem) | 43. By the AM-GM inequality, we have
$$\begin{array}{l}
\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c d}+\frac{b^{2}+c^{2}+d^{2}}{b c+c d+d a}+\frac{c^{2}+d^{2}+a^{2}}{c d+d a+a b}+\frac{d^{2}+a^{2}+b^{2}}{d a+a b+b c} \geqslant \\
4 \sqrt[4]{\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c d}} \cdot \frac{b^{2}+c^{2}+d^{2}}{b c+c d+d a} \cdot... | 4 | Inequalities | proof | Yes | Yes | inequalities | false | 732,608 |
44. Let $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ be positive numbers, $S=\sum_{i=1}^{n} a_{i}$, prove: $\sum_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)^{2} \geqslant n\left(\frac{S}{n}+\right.$ $\left.\frac{n}{S}\right)^{2}$
(1980 Vietnam Mathematical Olympiad Problem) | 44. By the Cauchy-Schwarz inequality, we have $\sum_{i=1}^{n} a_{i} \sum_{i=1}^{n} \frac{1}{a_{i}} \geqslant n^{2}$, which implies $\sum_{i=1}^{n} \frac{1}{a_{i}} \geqslant \frac{n^{2}}{S}$. Again, by the Cauchy-Schwarz inequality, we get
$$\begin{array}{c}
n \sum_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)^{2} \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,609 |
45. Let $u_{1}, u_{2}, \cdots, u_{n}$ and $v_{1}, v_{2}, \cdots, v_{n}$ be real numbers, prove: $1+\sum_{i=1}^{n}\left(u_{i}+v_{i}\right)^{2} \leqslant$ $\frac{4}{3}\left(1+\sum_{i=1}^{n} a_{i}^{2}\right)\left(1+\sum_{i=1}^{n} b_{i}^{2}\right) \cdot(1970$ IMO Shortlist $)$ | 45.
$$\begin{array}{l}
1+\sum_{i=1}^{n}\left(u_{i}+v_{i}\right)^{2} \leqslant \frac{4}{3}\left(1+\sum_{i=1}^{n} a_{i}^{2}\right)\left(1+\sum_{i=1}^{n} b_{i}^{2}\right) \Leftrightarrow \\
3+\sum_{i=1}^{n} 3\left(u_{i}+v_{i}\right)^{2} \leqslant 4\left(1+\sum_{i=1}^{n} a_{i}^{2}\right)\left(1+\sum_{i=1}^{n} b_{i}^{2}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,610 |
50. Let $x, y, z$ be positive numbers, and $x+y=z=1$, prove that: $\sqrt{\frac{x y}{z+x y}}+\sqrt{\frac{y z}{x+y z}}+\sqrt{\frac{z x}{y+z x}} \leqslant \frac{3}{2} \cdot$(2010 Kyrgyzstan Mathematical Olympiad Problem) | 50. Since $x+y+z=1$, we have
$$\sqrt{\frac{x y}{z+x y}}=\sqrt{\frac{x y}{z(x+y+z)+x y}}=\sqrt{\frac{x y}{(x+z)(y+z)}}$$
Therefore, by the Cauchy-Schwarz inequality and using $(x+y)(y+z)(z+x) \geqslant 8 x y z$, we get
$$\begin{array}{l}
\sqrt{\frac{x y}{z+x y}}+\sqrt{\frac{y z}{x+y z}}+\sqrt{\frac{z x}{y+z x}}=\frac{\... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,615 |
Example 1 Let $a_{i}>0(i=1,2, \cdots, n), A_{n}=\frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right), G_{n}=$ $\sqrt[n]{a_{1} a_{2} \cdots a_{n}}$, prove the Popovic inequality:
$$\left(\frac{G_{n+1}}{A_{n+1}}\right)^{n+1} \geqslant\left(\frac{G_{n}}{A_{n}}\right)^{n}$$ | Proof that by the generalization of the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(n A_{n}+a_{n+1}\right)^{n+1}= \\
\left(A_{n}+A_{n}+\cdots+A_{n}+a_{n+1}\right)\left(A_{n}+A_{n}+\cdots+a_{n+1}+A_{n}\right) \cdots\left(a_{n+1}+A_{n}+\cdots+A_{n}\right) \geqslant \\
\left(\sqrt[n+1]{A_{n}^{n}} \cdot \sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,617 |
Example 2 Let $x_{i}>0, x_{i} y_{i}-z_{i}^{2}>0(i=1,2, \cdots, n)$, then
$$\frac{n^{3}}{\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} y_{i}-\left(\sum_{i=1}^{n} z_{i}\right)^{2}} \leqslant \sum_{i=1}^{n} \frac{1}{x_{i} y_{i}-z_{i}^{2}}$$
holds. Equality occurs if and only if $x_{1}=x_{2}=\cdots \cdots=x_{n} ; y_{1}=y_{2}=\cdots... | Prove that for $A_{i}=\sqrt{x_{i} y_{i}}+z_{i}, B_{i}=\sqrt{x_{i} y_{i}}-z_{i}(i=1,2, \cdots, n)$, using the Cauchy-Schwarz inequality and its generalization, we have
$$\begin{array}{l}
{\left[\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} y_{i}-\left(\sum_{i=1}^{n} z_{i}\right)^{2}\right] \sum_{i=1}^{n} \frac{1}{x_{i} y_{i}-z_{i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,619 |
Example 3 If $x, y, z$ are all positive real numbers, find the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)}$. (2003 Singapore Mathematical Olympiad) | Solve: By the generalization of Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
(1+5 x)(4 x+3 y)(5 y+6 z)(z+18) \geqslant \\
(\sqrt[4]{1 \cdot 4 x \cdot 5 y \cdot z}+\sqrt[4]{5 x \cdot 3 y \cdot 6 z \cdot 18})^{4}=5120 x y z
\end{array}$$
Therefore, the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+... | \frac{1}{5120} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,620 |
Example 4 Given $5 n$ real numbers $r_{i}, s_{i}, t_{i}, u_{i}, v_{i}(1 \leqslant i \leqslant n)$ all greater than 1, let $R=$ $\left(\frac{1}{n} \sum_{i=1}^{n} r_{i}\right), S=\left(\frac{1}{n} \sum_{i=1}^{n} s_{i}\right), T=\left(\frac{1}{n} \sum_{i=1}^{n} t_{i}\right), U=\left(\frac{1}{n} \sum_{i=1}^{n} u_{i}\right)... | Proof: Let $x_{1}, x_{2}, \cdots, x_{n} \in (1,+\infty)$, then by the generalized Cauchy inequality, we have
$$\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \geqslant\left(1+\sqrt[n]{x_{1} x_{2} \cdots x_{n}}\right)^{n}$$
i.e.,
$$\sqrt[n]{\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,621 |
Example 5 Let $a_{1}, a_{2}, \cdots, a_{n}$ be given non-zero real numbers. If the inequality
$$\begin{array}{l}
r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+r_{n}\left(x_{n}-a_{n}\right) \leqslant \\
\sqrt[m]{x_{1}^{m}+x_{2}^{m}+\cdots+x_{n}^{m}}-\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}
\en... | Substitute $x_{i}=0(i=1,2, \cdots, n)$ into the original inequality, we get
$$-\left(r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}\right) \leqslant-\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$
That is,
$$r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \geqslant \sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$
Substitute ... | r_{i}=\left(\frac{a_{i}}{\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}}\right)^{m-1}(i=1,2, \cdots, n) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,622 |
Example 6 Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, $p>0, q>0$, prove: $\frac{a_{1}^{p+q}}{a_{2}^{p}}+\frac{a_{2}^{p+q}}{a_{3}^{\prime \prime}}+\cdots+$ $\frac{a_{n-1}^{p+q}}{a_{n}^{q}}+\frac{a_{n}^{p+q}}{a_{1}^{p}} \geqslant a_{1}^{p}+a_{2}^{p}+\cdots+a_{n}^{p} \cdot$ (1984 National High Scho... | Prove that by Hölder's inequality,
$$\begin{array}{l}
\left(\frac{a_{1}^{p+q}}{a_{2}^{q}}+\frac{a_{2}^{p+q}}{a_{3}^{q}}+\cdots+\frac{a_{n-1}^{p+q}}{a_{n}^{q}}+\frac{a_{n}^{p+q}}{a_{1}^{q}}\right)^{\frac{p}{p+q}}\left(a_{2}^{p}+a_{3}^{p}+\cdots+a_{n}^{p}+a_{1}^{p}\right)^{\frac{1}{p+q}} \geqslant \\
a_{1}^{p}+a_{2}^{p}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,623 |
Example 7 Given that $a, b$ are positive constants, and $x$ is an acute angle, find the minimum value of the function $y=\frac{a}{\sin ^{n} x}+\frac{b}{\cos ^{n} x}$. | By Hölder's inequality, we have
$$\begin{array}{l}
\left(\frac{a}{\sin ^{n} x}+\frac{b}{\cos ^{n} x}\right)^{\frac{2}{n+2}}\left(\sin ^{2} x+\cos ^{2} x\right)^{\frac{n}{n+2}} \geqslant \\
\left(\frac{a}{\sin ^{n} x}\right)^{\frac{2}{n+2}}\left(\sin ^{2} x\right)^{\frac{n}{n+2}}+\left(\frac{b}{\cos ^{n} x}\right)^{\fra... | \left(a^{\frac{2}{n+2}}+b^{\frac{2}{n+2}}\right)^{\frac{n+2}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,624 |
Example 9 Let $a_{1}, a_{2}, \cdots, a_{m}$ be positive real numbers, and $\sum_{i=1}^{m} \frac{1}{a_{i}}=1$. Then for every $n \in \mathbf{N}^{*}$, we have
$$\left(\sum_{i=1}^{m} a_{i}\right)^{n}-\sum_{i=1}^{m} a_{i}^{n} \geqslant m^{2 n}-m^{n+1}$$ | Prove that because $\sum_{i=1}^{m} \frac{1}{a_{i}}=1 \geqslant m \sqrt[m]{\frac{1}{a_{1} a_{2} \cdots a_{m}}}$, so $a_{1} a_{2} \cdots a_{m} \geqslant m^{m \prime}$, according to the polynomial expansion theorem
$$\left(\sum_{i=1}^{m} a_{i}\right)^{n}=\sum \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} a_{1}^{n_{1}} a_{2}^{n_{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,626 |
Example 11 Let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, prove that: $\left(a_{1} a_{2}+ a_{2} a_{3}+\cdots+a_{n} a_{1}\right)\left(\frac{a^{2}}{a_{2}^{2}+a_{2}}+\frac{a_{2}}{a_{3}^{2}+a_{3}} \epsilon \cdots+\frac{a_{n}}{a_{1}^{2}+a_{1}}\right) \geqslant \frac{n}{n+1}$. (... | Prove that by the generalization of the Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)\left(\frac{a_{1}}{a_{2}^{2}+a_{2}}+\frac{a_{2}}{a_{3}^{2}+a_{3}}+\cdots+\frac{a_{n}}{a_{1}^{2}+a_{1}}\right) \cdot \\
{\left[a_{1}\left(a_{2}+1\right)+a_{2}\left(a_{3}+1\right)+\... | S \geqslant \frac{n}{n+1} | Inequalities | proof | Yes | Yes | inequalities | false | 732,628 |
38. Given $x, y, z \in [1,2]$, prove: $(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \geqslant 6\left(\frac{x}{y+z}+\right.$ $\left.\frac{y}{z+x}+\frac{z}{x+y}\right) \cdot(2006$ Vietnam Mathematical Olympiad Problem) | 38. Given $2 \geqslant x \geqslant y \geqslant z \geqslant 1$, we have
$$(x+y+z)\left(\frac{1}{x}+1+\frac{1}{z}\right)-9=\frac{(1-y)^{2}}{x}+\frac{(y-z)}{y}+\frac{(z-1)^{2}}{z}$$
Furthermore,
$$\begin{aligned}
\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}-\frac{3}{2}=\frac{1}{2}\left[\frac{(x-y)^{2}}{(y+z)(z+x)}+\frac{(y-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,629 |
Example 12 Given that $a, b, c$ are all positive numbers, and $ab + bc + ca = 1$, prove: $\sqrt{a^{3}+a} + \sqrt{b^{3}+b} + \sqrt{c^{3}+c} \geqslant 2 \sqrt{a+b+c}$. (2008 Iran National Training Team Problem) | Prove that from the generalization of the Cauchy-Schwarz inequality,
$$\left(\sqrt{a^{3}+a}+\sqrt{b^{3}+b}+\sqrt{c^{3}+c}\right)^{2}\left(\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}\right) \geqslant(a+b+c)^{3}$$
Therefore, it suffices to prove that \((a+b+c)^{2} \geqslant 4\left(\frac{a^{2}}{a^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,630 |
4. If $x_{1}, x_{2}, \cdots, x_{n}$ are all positive, then $x_{1}^{n+1}+x_{2}^{n+1}+\cdots+x_{n}^{n+1} \geqslant x_{1} x_{2} \cdots x_{n}\left(x_{1}+\right.$ $\left.x_{2}+\cdots+x_{n}\right)$. (College Mathematics Journal, Vol. 25, No. 4, 1994) | 4. By the generalization of the Cauchy-Schwarz inequality, we have
$$\left(x_{1}^{n+1}+x_{2}^{n+1}+\cdots+x_{n}^{n+1}\right)\left(x_{1}^{n+1}+x_{2}^{n+1}+\cdots+\right.$$
$$\begin{array}{l}
\left.x_{n}^{n+1}\right)\left(x_{2}^{n+1}+x_{3}^{n+1}+\cdots+x_{n}^{n+1}+x_{1}^{n+1}\right) \cdots\left(x_{n}^{n+1}+x_{1}^{n+1}+x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,634 |
6. Given that $\alpha, \beta$ are acute angles, prove: $\sin ^{3} \alpha+\cos ^{3} \alpha \cos ^{3} \beta+\cos ^{3} \alpha \sin ^{3} \beta \geqslant \frac{\sqrt{3}}{3}$. | 6. $\left(\sin ^{3} \alpha+\cos ^{3} \alpha \cos ^{3} \beta+\cos ^{3} \alpha \sin ^{3} \beta\right)\left(\sin ^{3} \alpha+\cos ^{3} \alpha \cos ^{3} \beta+\cos ^{3} \alpha \sin ^{3} \beta\right) \cdot$
$$(\Gamma+\Gamma+1) \geqslant\left(\sin ^{2} \alpha+\cos ^{2} \alpha \cos ^{2} \beta+\cos ^{2} \alpha \sin ^{2} \beta\... | \sin ^{3} \alpha+\cos ^{3} \alpha \cos ^{3} \beta+\cos ^{3} \alpha \sin ^{3} \beta \geqslant \frac{\sqrt{3}}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 732,636 |
7. Given that $\alpha, \beta$ are acute angles, prove: $\frac{1}{\sin ^{3} \alpha}+\frac{1}{\cos ^{3} \alpha \cos ^{3} \beta}+\frac{1}{\cos ^{3} \alpha \sin ^{3} \beta} \geqslant 9 \sqrt{3}$. | 7. By the generalization of Cauchy's inequality, we have
$$\begin{array}{l}
\left(\frac{1}{\sin ^{3} \alpha}+\frac{1}{\cos ^{3} \alpha \cos ^{3} \beta}+\frac{1}{\cos ^{3} \alpha \sin ^{3} \beta}\right)\left(\frac{1}{\sin ^{3} \alpha}+\frac{1}{\cos ^{3} \alpha \cos ^{3} \beta}+\frac{1}{\cos ^{3} \alpha \sin ^{3} \beta}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,637 |
9. Given $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ are two sets of positive real numbers, and $x_{1}, x_{2}, \cdots, x_{n}$ are positive variables, find the minimum value of the function $y=b_{1}^{k} x_{1}^{k}+b_{2}^{k} x_{2}^{k}+\cdots+b_{n}^{k} x_{n}^{k}$. | 9. By Hölder's inequality, we have
$$\begin{array}{l}
\left(b_{1}^{k} x_{1}^{k}+b_{2}^{k} x_{2}^{k}+\cdots+b_{n}^{k} x_{n}^{k}\right)^{\frac{1}{k}}\left[\left(\frac{a_{1}}{b_{1}}\right)^{\frac{k}{k-1}}+\left(\frac{a_{2}}{b_{2}}\right)^{\frac{k}{k-1}}+\cdots+\left(\frac{a_{n}}{b_{n}}\right)^{\frac{k}{k-1}}\right] \geqsl... | \frac{p^{k}}{\left[\left(\frac{a_{1}}{b_{1}}\right)^{\frac{k}{k-1}}+\left(\frac{a_{2}}{b_{2}}\right)^{\frac{k}{k-1}}+\cdots+\left(\frac{a_{n}}{b_{n}}\right)^{\frac{k}{k-1}}\right]^{k-1}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,639 |
10. (1) Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, $\alpha>\beta>0$, prove:
$$\sqrt{\frac{a_{1}^{\alpha}+a_{2}^{\alpha}+\cdots+a_{n}^{\alpha}}{n}}>\sqrt[\beta]{\frac{a_{1}^{\beta}+a_{2}^{\beta}+\cdots+a_{n}^{\beta}}{n}}$$ | 10. (1) By Hölder's inequality, we have
$$(1+1+\cdots+1)^{(\alpha-\beta) / \alpha}\left(a_{1}^{\alpha}+a_{2}^{\alpha}+\cdots+a_{n}^{\alpha}\right)^{\beta / \alpha} \geqslant a_{1}^{\beta}+a_{2}^{\beta}+\cdots+a_{n}^{\beta}$$
Raising both sides to the power of $\alpha$, we get
$$\left(a_{1}^{\alpha}+a_{2}^{\alpha}+\cdo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,641 |
(2) Given $a_{1}, a_{2}, \cdots, a_{n} ; p_{1}, p_{2}, \cdots, p_{n}$ are positive real numbers, $\alpha>\beta>0$, prove:
$$\sqrt[\alpha]{\frac{p_{1} a_{1}^{\alpha}+p_{2} a_{2}^{\alpha}+\cdots+p_{n} a_{n}^{\alpha}}{p_{1}+p_{2}+\cdots+p_{n}}}>\sqrt[\beta]{\frac{p_{1} a_{1}^{\beta}+p_{2} a_{2}^{\beta}+\cdots+p_{n} a_{n}^... | (2) By Hölder's inequality, we have
$$\begin{array}{l}
\left(p_{1}+p_{2}+\cdots+p_{n}\right)^{(\alpha-\beta) / \alpha}\left(p_{1} a_{1}^{\alpha}+p_{2} a_{2}^{\alpha}+\cdots+p_{n} a_{n}^{\alpha}\right)^{\beta / \alpha} \geqslant \\
p_{1} a_{1}^{\beta}+p_{2} a_{2}^{\beta}+\cdots+p_{n} a_{n}^{\beta}
\end{array}$$
Rearran... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,642 |
11. Given that $a, b$ are positive constants, and $x$ is an acute angle, find the minimum value of the function $y=a \sin ^{n} x+b \cos ^{n} x$. | 11.
$$\begin{array}{c}
\left(a \sin ^{n} x+b \cos ^{n} x\right)^{\frac{2}{n}}\left(a^{-\frac{2}{n-2}}+b^{-\frac{2}{n-2}}\right)^{\frac{n-2}{n}} \geqslant \\
a^{\frac{2}{n}} \sin ^{2} x\left(a^{-\frac{2}{n-2}}\right)^{\frac{n-2}{n}}+b^{\frac{2}{n}} \cos ^{2} x\left(b^{-\frac{2}{n-2}}\right)^{\frac{n-2}{n}}=1 \\
y^{2}\le... | \left(a^{-\frac{2}{n-2}}+b^{-\frac{2}{n-2}}\right)^{-\frac{n-2}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,643 |
12. Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$, prove:
(1) $\frac{1}{a_{1}^{k}}+\frac{1}{a_{2}^{k}}+\cdots+\frac{1}{a_{n}^{k}} \geqslant n^{k+1}$; | 12. (1) From the generalization of Cauchy's inequality, we have
$$\begin{array}{l}
\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(a_{1}+a_{2}+\cdots+a_{n}\right) \cdot \cdots \\
\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{\Gamma}{a_{1}^{k}}+\frac{1}{a_{2}^{k}}+\cdots+\frac{1}{a_{n}^{k}}\right) \geqslant \\
(1+1+\cdots... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,644 |
(3) $\left(a_{1}^{k}+\frac{1}{a_{1}^{k}}\right)\left(a_{2}^{k}+\frac{1}{a_{2}^{k}}\right) \cdots\left(a_{n}^{k}+\frac{1}{a_{n}^{k}}\right) \geqslant\left(n^{k}+\frac{1}{n^{k}}\right)^{n}$. | (3) By the generalization of Cauchy's inequality, we have
$$\left(a_{1}^{k}+\frac{1}{a_{1}^{k}}\right)\left(a_{2}^{k}+\frac{1}{a_{2}^{k}}\right) \cdots\left(a_{n}^{k}+\frac{1}{a_{n}^{k}}\right) \geqslant\left(\sqrt[n]{a_{1}^{k} a_{2}^{k} \cdots a_{n}^{k}}+\sqrt[n]{\frac{1}{a_{1}^{k} a_{2}^{k} \cdots a_{n}^{k}}}\right)^... | null | Inequalities | proof | Yes | Yes | inequalities | false | 732,646 |
13. Given $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1, k$ is a positive integer, find the minimum value of $\frac{1}{a_{1}^{k}\left(1+a_{1}\right)}+\frac{1}{a_{2}^{k}\left(1+a_{2}\right)}+\cdots+\frac{1}{a_{n}^{k}\left(1+a_{n}\right)}$. | 13. From $a_{1}+a_{2}+\cdots+a_{n}=1$, we get $1=a_{F}+a_{2}+\cdots+a_{n} \geqslant n \sqrt[n]{a_{1} a_{2} \cdots a_{n}}$, thus $\frac{1}{\sqrt[n]{a_{1} a_{2} \cdots a_{n}}} \geqslant n^{n}$. For $k$ being a positive integer, taking $k$ instances of $\left(a_{1}+a_{2}+\cdots+a_{n}\right)$, by the Cauchy-Schwarz inequal... | \frac{n^{k+2}}{n+1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,647 |
14. If $x_{i}>0(i=1,2, \cdots, n), n \geqslant 2$, and $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}=1$, prove that: $x_{1} x_{2} \cdots x_{n} \geqslant(n-1)^{n}$. | 14. The application of the generalization of Cauchy's inequality is as follows:
$$\begin{array}{l}
\prod_{i=1}^{n} \frac{x_{i}}{1+x_{i}}=\prod_{i=1}^{n}\left(1-\frac{1}{1+x_{i}}\right)= \\
\left(\frac{1}{1+x_{2}}+\frac{1}{1+x_{3}}+\cdots+\frac{1}{1+x_{n-1}}+\frac{1}{1+x_{n}}\right) . \\
\left(\frac{1}{1+x_{3}}+\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,648 |
15. Given that $x, y, z$ are positive numbers, and $x+y+z=1$, prove that: $\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+$
$$\frac{z^{4}}{x\left(1-x^{2}\right)} \geqslant \frac{1}{8}$$ | 15. By the generalization of Cauchy's inequality, we get
$$\begin{array}{l}
(y+x+z)[(1+y)+(1+z)+(1+x)][(1-y)+(1-z)+(1-x)] \cdot \\
{\left[\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+\frac{z^{4}}{x\left(1-x^{2}\right)}\right] \geqslant(x+y+z)^{4}}
\end{array}$$
Since $x+y+z=1$, we have
$$\fr... | \frac{1}{8} | Inequalities | proof | Yes | Yes | inequalities | false | 732,649 |
16. Prove: For positive numbers $a, b, c$, we have $\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1$. (42nd IMO Problem) | 16. By the generalization of Cauchy's inequality, we get
$$\text { LHS }=\sum \frac{a}{\sqrt{a^{2}+8 b c}}=\sum \frac{a^{\frac{3}{2}}}{\sqrt{a^{3}+8 a b c}} \geqslant \frac{\left(\sum a\right)^{\frac{3}{2}}}{\left[\sum\left(a^{3}+8 a b c\right)\right]^{\frac{1}{2}}}$$
To prove the original inequality, it suffices to p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,650 |
40. Let $a_{i} \in[-2,17], i=1,2, \cdots, 59$, and $a_{1}+a_{2}+\cdots+a_{59}=0$, prove: $a_{1}^{2}+$ $a_{2}^{2}+\cdots+a_{59}^{2} \leqslant 2006$. (2006 Boltic Way Mathematical Olympiad Problem) | 40. Since $a_{i} \in[-2,17], i=1,2, \cdots, 59$, we have $\left(-a_{i}+2\right)\left(a_{i}-17\right) \leqslant 0, i=1$, $2, \cdots, 59$, which means $a_{i}^{2}-15 a_{i}-34 \leqslant 0, i=1,2, \cdots, 59$. Noting that $a_{1} \pm a_{2}+\cdots+a_{59}=0$, adding the above inequalities yields $a_{1}^{2}+a_{2}^{2}+\cdots+a_{... | a_{1}^{2}+a_{2}^{2}+\cdots+a_{59}^{2} \leqslant 2006 | Inequalities | proof | Yes | Yes | inequalities | false | 732,651 |
17. Let $k \geqslant 1, a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, prove that: $\left(\frac{a_{1}}{a_{2}+a_{3}+\cdots+a_{n}}\right)^{k}+$ $\left(\frac{a_{2}}{a_{1}+a_{3}+\cdots+a_{n}}\right)^{k}+\cdots+\left(\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}}\right)^{k} \geqslant \frac{n}{(n-1)^{k}}$. (30th IMO Shortl... | 17. Let $s=a_{1}+a_{2}+a_{3}+\cdots+a_{n}$, when $k=1$
$$\begin{array}{l}
\frac{a_{1}}{a_{2}+a_{3}+\cdots+a_{n}}+\frac{a_{2}}{a_{1}+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}}= \\
\frac{s}{s-a_{1}}+\frac{s}{s-a_{2}}+\cdots+\frac{s}{s-a_{n}}-n
\end{array}$$
By the Cauchy-Schwarz inequality, we h... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,652 |
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