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56. Let $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ be non-negative real numbers, $c_{k}=\prod_{i=1}^{k} b_{i}^{\frac{1}{k}}, 1 \leqslant k \leqslant n$, prove: $n c_{n}+\sum_{k=1}^{n} k\left(a_{k}-1\right) c_{k} \leqslant \sum_{k=1}^{n} a_{k}^{k} b_{k} \cdot(2007$ China National Training Team Problem $... | 56. For $k=2,3, \cdots, n$ applying the AM-GM inequality we get
$$k a_{k} c_{k}=k a_{k} b_{k}^{\frac{1}{k}} c_{k-1}^{\frac{1}{k}} \cdots c_{k-1}^{\frac{1}{k}}\left(k-1 \text { factors of } c_{k-1}^{\frac{1}{k}}\right) \leqslant a_{k}^{k} b_{k}+(k-1) c_{k-1}$$
Summing up the above inequalities yields the result. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,362 |
14. Given $a->b>c$, prove: $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)>0$. (54th Moscow Mathematical Competition Question) | 14. From $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=-(a-b)(b-c)(c-a)$, we can get. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,363 |
58. Let $a, b, c, d$ be positive real numbers, prove:
$$\frac{a}{\sqrt[3]{a^{3}+63 b c d}}+\frac{b}{\sqrt[3]{b^{3}+63 c d a}}+\frac{c}{\sqrt[3]{c^{3}+63 d a b}}+\frac{d}{\sqrt[3]{d^{3}}+63 a b c} \geqslant 1$$
(2004 Polish Mathematical Olympiad Problem) | 58. Let $\frac{a}{\sqrt[3]{a^{3}+63 b c d}} \geqslant \frac{a^{p}}{a^{p}+b^{p}+c^{p}+d^{p}}, p$ is a real number.
$$\begin{array}{l}
\left(a^{p}+b^{p}+c^{p}+d^{p}\right)^{3}-\left(a^{p}\right)^{3}=\left(b^{p}+c^{p}+d^{p}\right)\left[\left(a^{p}\right)^{2}+\left(b^{p}\right)^{2}+\left(c^{p}\right)^{2}+\left(d^{p}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,365 |
59. Let $a, b$ be positive constants. $x_{1}, x_{2}, \cdots, x_{n}$ are positive real numbers. $n \geqslant 2$ is a positive integer. Find the maximum value of $y=$ $\frac{x_{1} x_{2} \cdots x_{n}}{\left(a+x_{1}\right)\left(x_{1}+x_{2}\right) \cdots\left(x_{n-1}+x_{n}\right)\left(x_{n}+b\right)}$. (1999 Polish Mathemat... | 59. Since
$$\begin{aligned}
a y & =\frac{a}{a+x_{1}} \cdot \frac{x_{1}}{x_{1}+x_{2}} \cdot \cdots \cdot \frac{x_{n-1}}{x_{n-1}+x_{n}} \cdot \frac{x_{n}}{x_{n}+b} \\
b y & =\frac{x_{1}}{a+x_{1}} \cdot \frac{x_{2}}{x_{1}+x_{2}} \cdot \cdots \cdot \frac{x_{n}}{x_{n-1}+x_{n}} \cdot \frac{b}{x_{n}+b}
\end{aligned}$$
By the... | \frac{1}{(\sqrt[n+1]{a}+\sqrt[n+1]{b})^{n+1}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,366 |
60. Let $a, b, c, d$ be positive real numbers, and $a^{2}+b^{2}+c^{2}+d^{2}=1$, prove: $a^{2} b^{2} c d+a b^{2} c^{2} d+$ $a b c^{2} d^{2}+a^{2} b c d^{2}+a^{2} b c^{2} d+a b^{2} c d^{2} \leqslant \frac{3}{32}$. (2008 Iran Mathematical Olympiad) | 60. Since $a, b, c, d$ are positive real numbers, $a^{2}+b^{2}+c^{2}+d^{2}=1 \geqslant 4 \sqrt[4]{a^{2} b^{2} c^{2} d^{2}}$, so $a b c d \leqslant$ $\frac{1}{16}$. Also,
$$\begin{array}{l}
a b+b c+c d+a d+a c+b d \leqslant \frac{a^{2}+b^{2}}{2}+\frac{b^{2}+c^{2}}{2}+\frac{c^{2}+d^{2}}{2}+\frac{a^{2}+a^{2}}{2}+ \\
\frac... | \frac{3}{32} | Inequalities | proof | Yes | Yes | inequalities | false | 732,367 |
66 Given $x_{1}, x_{2}, \cdots, x_{n}$ are positive real numbers, prove: $m=\min \left|x_{1}, x_{2}, \cdots, x_{n}\right|, M=\max \mid x_{1}$,
$$x_{2}, \cdots, x_{n} \in, A=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}, G=\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \text {, prove: } A-G \geqslant \frac{1}{n}(\sqrt{M}-$$
$\sqrt{m})^{2} \cd... | 66. $M=x_{1}, m=x_{2}$, without loss of generality
$$\begin{array}{l}
A-G \geqslant \frac{1}{n}(\sqrt{M}-\sqrt{m}) \Leftrightarrow n(A-G) \geqslant\left(\left(\sqrt{x_{1}}-\sqrt{x_{2}}\right)^{2} \Leftrightarrow\right. \\
x_{1}+x_{2}+\cdots+x_{n}-n \sqrt[n]{x_{1} x_{2} \cdots x_{a}} \geqslant x_{1}+x_{2}-2 \sqrt{x_{1} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,373 |
15. Let $a, b, c$ be non-negative real numbers. Prove that: $a^{4}+b^{4}+c^{4}-2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+$ $a^{2} b c+b^{2} c a+c^{2} a b \geqslant 0$. (45th Moscow Mathematical Olympiad Problem) | 15. Factorizing the left side of the inequality, we get
$$(a+b+c)[a b c-(-a+b+c)(a-b+c)(a+b-c)]$$
Since $a, b, c$ are non-negative real numbers, at most one of $a+b-c, a-b+c, -a+b+c$ is negative (otherwise, adding any two of these expressions or all three would lead to a contradiction). If two of these numbers are neg... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,374 |
68. Let $a, b, c$ be positive real numbers, try to prove $\frac{a}{\sqrt{a^{2}+9 b c}}+\frac{b}{\sqrt{b^{2}+9 c a}}+\frac{c}{\sqrt{c^{2}+9 a b}} \geqslant \frac{3}{\sqrt{10}}$ (2009 Taiwan Mathematical Olympiad Problem) | 68 . Let
$$\begin{array}{l}
\frac{a}{\sqrt{a^{2}+9 b c}} \geqslant \frac{a^{t}}{a^{t}+b^{t}+c^{t}} \cdot \frac{3}{\sqrt{10}} \Leftrightarrow 10\left(a^{t}+b^{t}+c^{t}\right)^{2} \geqslant \\
9 a^{2 t}+81 a^{2 t-2} b c \Leftrightarrow \\
a^{2 t}+10 b^{2 t}+10 c^{2 t}+20 a^{t} b^{t}+20 b^{t} c^{t}+20 c^{t} a^{t} \geqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,376 |
69. Let the real-coefficient polynomial $f(x)=x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n}$ have real roots $b_{1}, b_{2}, \cdots, b_{n}$, where $n \geqslant 2$. Prove that for $x>\max \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$, then
$$f(x+1) \geqslant \frac{1}{\frac{1}{x-b_{1}}+\frac{1}{x-b_{2}^{2}}+\cdots+\frac{1}... | 69. Since $n \geqslant 2$, we have $n^{2}-2 n(n-1) \leqslant 0$, so for any $t>0$,
$$\frac{n(n-1)}{2} t^{2}-n t+1 \geqslant 0$$
Therefore, we get
$$(1+t)^{n} \geqslant 1+n t+\frac{n(n-1)}{2} t^{2} \geqslant 2 n t$$
When $x>\max \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$, $f(x)$ is a polynomial with the leading coef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,377 |
70. Let $0<p<q, t_{1}, t_{2}, \cdots, t_{n} \in[p, q]$, and let $A$ and $B$ be the arithmetic means of the arrays $t_{1}, t_{2}, \cdots, t_{n}$ and $t_{1}^{2}, t_{2}^{2}, \cdots, t_{n}^{2}$, respectively. Prove: $\frac{A^{2}}{B} \geqslant \frac{4 p q}{(p+q)^{2}}$ (1981 Vietnam Mathematical Olympiad Problem) | 70. We prove a general conclusion, the famous Pölya-Szegö inequality: Let \(0 < m_{1} \leqslant a_{i} \leqslant M_{1}, 0 < m_{2} \leqslant b_{i} \leqslant M_{2}, i=1,2, \cdots, n\), then we have
$$\frac{\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}^{2}\right)}{\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,378 |
71 - Let $a, b, c$ be positive real numbers, prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant 3 \sqrt{\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}} \cdot(2007$ Mongolian Mathematical Olympiad Problem) | $$\begin{array}{l}
\text { 71. } \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant 3 \sqrt{\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}} \Leftrightarrow\left(a^{2} c+b^{2} a+c^{2} b\right)^{2}(a b+b c+c a) \geqslant \\
9 a^{2} b^{2} c^{2}\left(a^{2}+b^{2}+c^{2}\right) \Leftrightarrow \sum_{\text {oc }}\left(a^{5} b c^{2}+\sum_{\t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,379 |
72 Let $x_{1}, x_{2}, \cdots, x_{n} \in\left(0, \frac{\pi}{2}\right)$, and $\tan x_{1}+\tan x_{2}+\cdots+\tan x_{n} \leqslant n$, prove: $\sin x_{1} \sin x_{2} \cdots \sin x_{n} \leqslant \frac{1}{\sqrt{2^{n}}} \cdot$(2002 Bosnian Mathematical Olympiad Problem) | 72. Let $\tan x_{i}=a_{i}(i=1,2, \cdots, n)$, then $a_{1}+a_{2}+\cdots+a_{n} \leqslant n$, we only need to prove
$$\begin{array}{c}
\prod_{i=1}^{n} \frac{\sqrt{2}}{\sqrt{1+a_{i}^{2}}} \leqslant 1 \\
\prod_{i=1}^{n} \frac{\sqrt{2}}{\sqrt{1+a_{i}^{2}}} \leqslant \prod_{i=1}^{n} \sqrt{a_{i}}=\sqrt{\prod_{i=1}^{n} a_{i}} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,380 |
75. Let $n$ and $k$ be positive integers, and $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, such that $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove: $\sum_{k=1}^{n} \frac{1}{a_{n}^{k}} \geq n^{n+1}$.(1974 IMO Shortlist) | 75. By the mean inequality, we have $\left(\sum_{i=1}^{n} a_{i}\right)^{k} \sum_{i=1}^{n} \frac{1}{a_{i}^{k}} \geqslant\left(n \sqrt[n]{a_{1} a_{2} \cdots a_{n}}\right)^{k} n \sqrt[n]{\prod_{i=1}^{n} \frac{1}{a_{i}^{k}}}=n^{k+1}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,383 |
76. Let $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$, prove: $\prod_{k=1}^{n}\left(\frac{1}{a_{k}^{2}}-1\right) \geqslant\left(n^{2}-1\right)^{k}(2011$ Kazakhstan Mathematical Olympiad problem $)$ | 76. By the AM-GM inequality,
$$1-a_{1}=a_{2}+a_{3}+\cdots+a_{n} \geqslant(n-1) \cdot \sqrt[n-1]{a_{2} a_{3} \cdots a_{n}}=(n-1) \sqrt[n-1]{\frac{a_{1} a_{2} a_{3} \cdots a_{n}}{a_{1}}}$$
Similarly,
$$1-a_{k} \geqslant(n-1) \sqrt[n-1]{\frac{a_{1} a_{2} a_{3} \cdots a_{n}}{a_{k}}}, k=2,3, \cdots, n$$
Multiplying the $n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,384 |
16. Let $a, b, c, d$ be real numbers satisfying $a^{2}+b^{2}+c^{2}+d^{2} \leqslant 1$, then $(a+b)^{4}+(a+c)^{4}+$ $(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4} \leqslant 6$ 6. (28th IMO Shortlist Problem) | 16. Since
$$\begin{array}{l}
(a+b)^{4}=a^{4}+6 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \\
(a-b)^{4}=a^{4}-6 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}
\end{array}$$
Therefore,
$$\begin{array}{l}
(a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4}+(c+d)^{4}+ \\
(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+(b-c)^{4}+(b-d)^{4}+(c-d)^{4}= \\
6\l... | 6 | Inequalities | proof | Yes | Yes | inequalities | false | 732,385 |
77. Let $a, b, c$ be positive real numbers, and satisfy $a^{2}+b^{2}+c^{2}+(a+b+c)^{2} \leqslant 4$, prove: $\frac{a b+1}{(a+b)^{2}}+\frac{b c+1}{(b+c)^{2}}+\frac{c a+1}{(c+a)^{2}} \geqslant 3$ (201 + USA Mathematical Olympiad problem) | 77. Since $a^{2}+b^{2}+c^{2}+(a+b+c)^{2} \leqslant 4$, then
$$\begin{array}{l}
a^{2}+b^{2}+c^{2}+a b+b c+c a \leqslant 2 \\
\frac{2 a b+2}{(a+b)^{2}} \geqslant \frac{2 a b+a^{2}+b^{2}+c^{2}+a b+b c+c a}{(-a+b)^{2}}= \\
\frac{(a+b)^{2}+c^{2}+a b+b c+c a}{(a+b)^{2}}=\frac{(a+b)^{2}+(c+a)-(c+b)}{(a+b)^{2}}= \\
1+\frac{(c+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,386 |
Example 1 Let $x_{1}, x_{2}, \cdots, x_{n}$ be any real numbers, prove: $\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\cdots+$ $\frac{x_{n}}{1+x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}<\sqrt{n}$. (42nd IMO Shortlist Problem) | Prove that by the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{\left[\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}\right]^{2} \leqslant} \\
{\left[\left(\frac{x_{1}}{1+x_{1}^{2}}\right)^{2}+\left(\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,387 |
Example 2 Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, let $S=a_{1}+a_{2}+\cdots+a_{n}$, then $\sum_{k=1}^{n} \frac{a_{k}}{S-a_{k}} \geqslant$ $\frac{n}{n-1}$. (1976 British Mathematical Olympiad) | Proof Consider $(n-1) S=n S-S=n S-\left(a_{1}+a_{2}+\cdots+a_{n}\right)=\left(S-a_{1}\right)+$ $\left(S-a_{2}\right)+\cdots+\left(S-a_{n}\right)$ and $n^{2}=(1+1+\cdots+1)^{2}$. By Corollary 1, we have
$$\begin{array}{c}
\sum_{k=1}^{n}\left(S-a_{k}\right) \sum_{k=1}^{n} \frac{1}{S-a_{k}} \geqslant n^{2} \\
(n-1) S \cdo... | \sum_{k=1}^{n} \frac{a_{k}}{S-a_{k}} \geqslant \frac{n}{n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 732,388 |
Example 3 Let $a_{1}>a_{2}>\cdots>a_{n}>a_{n+1}$, prove that: $\frac{1}{a_{1}-a_{2}}+\frac{1}{a_{2}-a_{3}}+\cdots+$
$$\frac{1}{a_{n}-a_{n+1}}+\frac{1}{a_{n+1}-a_{1}}>0$$ | Proof. Change the structure, and rewrite the proof:
$$\left(a_{1}-a_{n+1}\right)\left[\frac{1}{a_{1}-a_{2}}+\frac{1}{a_{2}-a_{3}}+\cdots+\frac{1}{a_{n}-a_{n+1}}\right]>1$$
To apply the Cauchy-Schwarz inequality, write \(a_{1}-a_{n+1}\) in the following form \(\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\cdots+\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,389 |
Example 4 Let $a, b, c, d$ be positive real numbers satisfying $a b + c d = 1$, and let points $P_{i}\left(x_{i}, y_{i}\right)(i=1,2,3,4)$ be four points on the unit circle centered at the origin. Prove that: $\left(a y_{1} + b y_{2} + c y_{3} + \overline{d y_{4}}\right)^{2} + \left(\bar{a} x_{4} + b x_{3} + c x_{2} + ... | Let $\alpha=a y_{1}+b y_{2}+c y_{3}+d y_{4}, \beta=a x_{4}+b x_{3}+c x_{2}+d x_{1}$. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{\left[\left(\sqrt{a d} y_{1}\right)^{2}+\left(\sqrt{b c} y_{2}\right)^{2}+\left(\sqrt{b c} y_{3}\right)^{2}+\left(\sqrt{a d} y_{4}\right)^{2}\right]} \\
{\left[\left(\sqrt{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,390 |
Example 5 Let $a, b, x, y, k$ be positive numbers, and $k<2, a^{2}+b^{2}-k a b=x^{2}+y^{2}-k x y=1$, prove: $|a x-b y| \leqslant \frac{2}{\sqrt{4-k^{2}}},|a y+b x-k b y| \leqslant \frac{2}{\sqrt{4-k^{2}}}$. | Prove that because $a^{2}+b^{2}-k a b=1$, so $\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}=1$, similarly $\left(\frac{\sqrt{4-k^{2}}}{2} x\right)^{2}+\left(\frac{k x}{2}-y\right)^{2}=1$. Apply the Cauchy-Schwarz inequality:
$$\begin{array}{l}
\left(\left(a-\frac{k b}{2}\right)^{2}+\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,391 |
Example 8 Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, prove that
$$\frac{a_{1}}{1+a_{2}+a_{3}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}} \text { has a }$$
minimum value and calculate it. (1982 International ... | Prove
$$\frac{a_{1}}{1+a_{2}+a_{3}+\cdots+a_{n}}+1=\frac{1+\left(a_{1}+a_{2}+a_{3} \cdots+a_{n}\right)}{2-a_{1}}=\frac{2}{2-a_{1}}$$
Similarly,
$$\begin{array}{l}
\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\Gamma=\frac{2}{2-a_{2}}, \cdots, \\
\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}+1=\frac{2}{2-a_{n}}
\end{array}$$
... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 732,394 |
Example 9 Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$, prove: for every $n \in \mathbf{N}^{*}$, we have $(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1}$. (1988 National High School Mathematics League Question) | $$\begin{array}{l}
(a+b)^{n}-a^{n}-b^{n}+1=(a b)^{n}-a^{n}-b^{n}+1=\left(a^{n}-1\right)\left(b^{n}-1\right)= \\
(a-1)(b-1)\left(a^{n-1}+a^{n-2}+\cdots+a+1\right)\left(b^{n-1}+b^{n-2}+\cdots+b+1\right)= \\
\left(a^{n-1}+a^{n-2}+\cdots+a+1\right)\left(b^{n-1}+b^{n-2}+\cdots+b+1\right) \geqslant \\
{\left[(a b)^{\frac{n-1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,395 |
17. Given the sides of a triangle $a, b, c$ and its area $S$, prove that $a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S$, and specify the condition for equality. (Weitzenbock Inequality, 3rd IMO problem) | 17. $\left(a^{2}+b^{2}+c^{2}\right)^{2}-(4 \sqrt{3} S)^{2}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-3(a+b+c)(-a+b+$ c) $(a-b+c)(a+b-c)=2\left[\left(a^{2}-b^{2}\right)^{2}+\left(b^{2}-c^{2}\right)^{2}+\left(c^{2}-a^{2}\right)^{2}\right] \geqslant 0$. The equality holds when $a=b=c$, and $\triangle ABC$ is an equilateral trian... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,396 |
Example 10 Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}^{+}$, prove that: $\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}+\frac{x_{n}^{2}}{x_{1}} \geqslant x_{1}+$ $x_{2}+\cdots+x_{n-1}+x_{n}$. (1984 National High School Mathematics League Question) | Prove that by embedding the factor $x_{2}+x_{3}+\cdots+x_{n}+x_{1}$ on the left side of the inequality, it is equivalent to embedding the factor $x_{1}+x_{2}+\cdots+x_{n}$. Applying the Cauchy-Schwarz inequality, we have:
$$\begin{array}{l}
\left(\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n-1}^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,397 |
Example 11 For a positive integer $n \geqslant 3, x_{\mathrm{F}}, x_{2}, \cdots, x_{n}$ are positive real numbers, $x_{n+j}=x_{j}(1 \leqslant j \leqslant n-\mathrm{F})$, find the minimum value of $\sum_{i=1}^{n} \frac{x_{j}}{x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}}$. (1995 National Mathematical Olympiad Training Team ... | Let $a_{j}^{2}=\frac{x_{j}}{x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}}, b_{j}^{2}=x_{j}\left[x_{j+1}+2 x_{j+2}+\cdots+\right.$ $\left.(n-1) x_{j+n-1}\right]$, where $b_{j}$ is positive, then $a_{j} b_{j}=x_{j}$, by the Cauchy-Schwarz inequality
$$\sum_{j=1}^{n} a_{j}^{2} \cdot \sum_{j=1}^{n} b_{j}^{2} \geqslant\left(\su... | \frac{2}{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,398 |
Example 12 Let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, prove that: $\left(a_{1} a_{2}+ a_{2} a_{3}+\cdots+a_{n} a_{1}\right)\left(\frac{a_{1}}{a_{2}^{2}+a^{2}}+\frac{a_{2}}{a_{3}^{2}+a_{3}}+\cdots+\frac{a_{n}}{a_{1}^{2}+a_{1}}\right) \geqslant \frac{n}{n+1}$.
(2007 Nati... | Prove that by the Cauchy-Schwarz inequality,
$$\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)\left(\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n}}{a_{1}}\right) \geqslant\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=1$$
Therefore,
$$\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n}}{a_{1}} \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,399 |
Example 13 (1) Let three positive real numbers $a, b, c$ satisfy: $\left(a^{2} \mp b^{2}+c^{2}\right)^{2}>2\left(a^{4}+b^{4}+c^{4}\right)$. Prove: $a, b, c$ must be the lengths of the sides of some triangle.
(2) Let $n$ positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy:
$$\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n... | Prove by (1) and the Cauchy-Schwarz inequality:
$$\begin{array}{l}
(n-1)\left(a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4}\right)<\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)^{2}= \\
\left(\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+a_{4}^{2}+\cdots+a_{n}^{2}\right)^{2} \leqslant \\
(n-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,400 |
Example 14 Let $a_{1}, a_{2}, \cdots, a_{n}(n>1)$ be real numbers, and $A+\sum_{i=1}^{n} a_{i}^{2}<\frac{1}{n-1}\left(\sum_{i=1}^{n} a_{i}\right)^{2}$. Prove that for $1 \leqslant i<j \leqslant n$, we have $A<2 a_{i} a_{i}$. (38th American Putnam Mathematical Competition Problem) | Prove that by Cauchy-Schwarz inequality,
$$\begin{array}{l}
{\left[\left(a_{1}+a_{2}\right)+a_{3}+\cdots+a_{n}\right]^{2} \leqslant} \\
\left(1^{2}+1^{2}+\cdots+1^{2}\right)\left[\left(a_{1}+a_{2}\right)^{2}+a_{3}^{2}+\cdots+a_{n}^{2}\right]
\end{array}$$
Therefore,
$$\left(\sum_{-1}^{n} a_{1}\right)^{2} \leqslant(n-1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,401 |
Example 15 Let $\frac{3}{2} \leqslant x \leqslant 5$, prove: the inequality $2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x}<2 \sqrt{19}$. (2003 National High School Mathematics Competition Problem) | Prove that for $\lambda, \mu \in \mathbf{R}^{+}$, and satisfying $\lambda+2 \mu=3$, by the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
(2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x})^{2}= \\
\left(2 \frac{1}{\sqrt{\lambda}} \cdot \sqrt{\lambda(x+1)}+\frac{1}{\sqrt{\mu}} \sqrt{\mu(2 x-3)}+\sqrt{15-3 x}\right)^{2} \l... | 2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x}<2 \sqrt{19} | Inequalities | proof | Yes | Yes | inequalities | false | 732,402 |
Example 17 Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$ and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} \sqrt{\frac{i}{j}} x_{i} x_{j}=1$, find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$. (2001 National High School Mathematics League Additional Question)
The key is to find the maximum val... | Solve for the minimum value first, because
$$\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \geqslant \sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} \sqrt{\frac{i}{j}} x_{i} x_{j}=1 \Rightarrow \sum_{i=1}^{n} x_{i} \geqslant 1$$
Equality holds if and only if there exists $i$ s... | \left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,404 |
Example 18 Let $2 n$ real numbers $a_{1}, a_{2}, \cdots, a_{2 n}$ satisfy the condition $\sum_{i=1}^{2 n-1}\left(a_{i+1}-a_{i}\right)^{2}=1$, find the maximum value of $\left(a_{n+1}+\right.$ $\left.a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$. (2003 China Western Mathematical Olympiad) | First, when $n=1$, $\left(a_{2}-a_{1}\right)^{2}=1$, so $a_{2}-a_{1}= \pm 1$. It is easy to see that the maximum value desired at this point is 1.
When $n \geqslant 2$, let $x_{1}=a_{1}, x_{i+1}=a_{i+1}-a_{i}, i=1,2, \cdots, 2 n-1$, then $\sum_{i=2}^{2 n} x_{i}^{2}=1$, and $a_{k}=x_{1}+x_{2}+\cdots+x_{k}, k=1,2, \cdot... | \sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,405 |
Example 19 Positive real numbers $x, y, z$ satisfy $x y z \geqslant \mathrm{~F}$, prove: $\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+$ $\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$. (46th IMO problem) | Prove that the original inequality can be transformed into
$$\frac{x^{2}+y^{2}+z^{2}}{x^{5}+y^{2}+z^{2}}+\frac{x^{2}+y^{2}+z^{2}}{y^{5}+z^{2}+x^{2}}+\frac{x^{2}+y^{2}+z^{2}}{z^{5}+x^{2}+y^{2}} \leqslant 3$$
By the Cauchy-Schwarz inequality and the given condition \(xyz \geqslant 1\), we have
$$\left(x^{5}+y^{2}+z^{2}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,406 |
Example 20 Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ all be positive numbers, and $\sum_{i=1}^{n} x_{i}=1$, prove that: $\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant$ $\frac{\sum_{i=1}^{n} \sqrt{x_{i}}}{\sqrt{n-1}} \cdot($ 4th EMO Problem) | Let $y_{i}=1-x_{i}(i=1,2, \cdots, n)$, by the Cauchy-Schwarz inequality we have
$$\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2} \leqslant n \sum_{i=1}^{n} x_{i}=n$$
i.e., $\sum_{i=1}^{n} \sqrt{x_{i}} \leqslant \sqrt{n}$. Similarly, $\left(\sum_{i=k}^{n} \sqrt{y_{i}}\right)^{2} \leqslant n \sum_{i=1}^{n} x_{i}=n \sum_{i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,408 |
1. Let $P(x)$ be a polynomial with positive coefficients. If $P(x) P\left(\frac{1}{x}\right) \geqslant 1$ holds for $x=1$, then for all positive $x$, $P(x) P\left(\frac{1}{x}\right) \geqslant 1$. (2004 Boltic Way Competition Problem) | 1. Let the polynomial be $P(x)=\sum_{i=0}^{n} a_{i} x^{n-i}$, where $a_{i}>0, i=0,1,2, \cdots, n$, then by the Cauchy inequality we have
$$P(x) P\left(\frac{1}{x}\right)=\sum^{n} a_{i} x^{n-i} \sum_{0}^{n} a_{i}\left(\frac{1}{x}\right)^{n-i} \geqslant\left(\sum_{i}^{n} a_{i}\right)^{2}=[P(1)]^{2} \geqslant 1$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,409 |
2. (1) Let $x, y, z$ be positive numbers, and $k \geqslant 1, a=x+k y+k z, b=k x+y+k z, c=k x+k y+z$, prove: $\frac{x}{a}+\frac{y}{b}+\frac{z}{c} \geqslant \frac{3}{2 k+1}$. (1998 Greek National Training Team Problem) | 2. (1) Since $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=\frac{x^{2}}{x^{2}+k x y+k x z}+\frac{y^{2}}{k x y+y^{2}+k y z}+\frac{z^{2}}{k z x+k y z+z^{2}}$,
By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(\frac{x^{2}}{x^{2}+k x y+k x z}+\frac{y^{2}}{k x y+y^{2}+k y z}+\frac{z^{2}}{k z x+k y z+z^{2}}\right) \cdot... | \frac{x}{a}+\frac{y}{b}+\frac{z}{c} \geqslant \frac{3}{2 k+1} | Inequalities | proof | Yes | Yes | inequalities | false | 732,410 |
4. Let $x_{i}>0, x_{i} y_{i}-z_{i}^{2}>0(i=1,2, \cdots, n)$, then
$$\frac{n^{3}}{\sum_{i=1}^{n} x_{i} \cdot \sum_{i=1}^{n} y_{i}-\left(\sum_{i=1}^{n} z_{i}\right)^{2}} \leqslant \sum_{i=1}^{n} \frac{1}{x_{i} y_{i}-z_{i}^{2}}$$
holds, and determine the necessary and sufficient conditions for equality. (A generalization... | 4. Let $a_{i} b_{i}-c_{i}^{2}=d_{i}^{2}, d_{i}>0(i=1,2, \cdots, n)$, we first prove
$$\left(\sum_{i=1}^{n} c_{i}\right)^{2}+\left(\sum_{i=1}^{n} d_{i}\right)^{2} \leqslant\left(\sum_{i=1}^{n} a_{i}\right)\left(\sum_{i=1}^{n} b_{i}\right)$$
In fact, by the Cauchy-Schwarz inequality, the left side of the above equation ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,414 |
5. Let $a_{1}, a_{2}, \cdots, a_{n}$ be given non-zero real numbers, and $r_{1}, r_{2}, \cdots, r_{n}$ be real numbers. The inequality $\sum_{i=1}^{n} r_{i}\left(x_{i}-a_{i}\right) \leqslant \sqrt{\sum_{i=1}^{n} x_{i}^{2}}-\sqrt{\sum_{i=1}^{n} a_{i}^{2}}$ holds for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$. Find t... | 5. Let $x_{i}=0(i=1,2, \cdots, n)$, then $\sum_{i=1}^{n} r_{i} a_{i} \geqslant \sqrt{\sum_{i=1}^{n} a_{i}^{2}}$.
Let $x_{i}=2 a_{i}(i=1,2, \cdots, n)$, then $\sum_{i=1}^{n} r_{i} a_{i} \geqslant \sqrt{\sum_{i=1}^{n} a_{i}^{2}}$.
Therefore,
$$\sum_{i=1}^{n} r_{i} a_{i}=\sqrt{\sum_{i=1}^{n} a_{i}^{2}}$$
Let $x_{i}=r_{i... | r_{k}=\frac{a_{k}}{\sqrt{\sum_{i=1}^{n} a_{i}^{2}}}(k=1,2, \cdots, n) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,415 |
6. (1) Given that $a, b, c, d$ are positive real numbers, prove that $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \geqslant 2$. (1989 Sichuan Province Mathematics Competition Question) | 6. (1) Applying the Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
\left(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\right) \\
{[a(b+c)+b(c+d)+c(d+a)+d(a+b)] \geqslant(a+b+c+d)^{2}}
\end{array}$$
Since
$$\begin{array}{l}
(a+b+c+d)^{2}-2[a(b+c)+b(c+d)+c(d+a)+d(a+b)]= \\
a^{2}+b^{2}+c^{2}+d^{2}-2 a c-2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,416 |
7. Let $a_{1}, a_{2}, \cdots, a_{n}(n>1)$ be real numbers, and $A+\sum_{i=1}^{n} a_{i}^{2}<\frac{1}{n-1}\left(\sum_{i=1}^{n} a_{i}\right)^{2}$, prove: $A<2 a_{i} a_{j}(1 \leqslant i<j \leqslant n)$. (1981 Putnam Mathematical Competition) | 7. Considering the constant $n-1$ appearing in the problem, we apply the Cauchy-Schwarz inequality as follows:
$$\begin{aligned}
\left(\sum_{i=1}^{n} a_{i}\right)^{2}= & {\left[\left(a_{1}+a_{2}\right)+a_{3}+\cdots+a_{n}\right]^{2} \leqslant } \\
& \overbrace{(1+1+\cdots+1)}^{n-1 \uparrow}\left[\left(a_{1}+a_{2}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,419 |
8. Given that $a, b$ are positive constants, $0<x<\frac{\pi}{2}$, find the minimum value of $y=\frac{a}{\sin x}+\frac{b}{\cos x}$. | 8. By Cauchy's inequality
$$\begin{array}{c}
\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}\right)\left(\sin ^{2} x+\cos ^{2} x\right) \geqslant(\sqrt[3]{a} \sin x+\sqrt[3]{b} \cos x)^{2} \\
\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}\right)^{\frac{1}{2}}\left(\frac{a}{\sin x}+\frac{b}{\cos x}\right) \geqslant \\
(\sqrt[3]{a} \sin x+... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,420 |
9. In $\triangle A B C$, prove that:
$$\sin A+\sin B+5 \sin C \leqslant \frac{\sqrt{198+2 \sqrt{201}}(\sqrt{201}+3)}{40}$$ | 9. It is known that
$$\sin A+\sin B+5 \sin C \leqslant 2 \cos x(1+5 \sin x), x=\frac{C}{2}$$
Let $y=\cos x(1+5 \sin x)$ (where $x$ is an acute angle), and apply the Cauchy-Schwarz inequality:
$$\begin{array}{c}
y^{2}=\cos ^{2} x(1+5 \sin x)^{2}=25 \cos ^{2} x\left(\frac{1}{5}+\sin x\right)^{2}=25 \frac{\cos ^{2} x}{t^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,421 |
10. $P$ is a point inside $\triangle ABC$, and $D, E, F$ are the feet of the perpendiculars from $P$ to $BC, CA, AB$ respectively. Find all points $P$ that minimize $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$. (22nd IMO Problem) | 10. Let the three sides of $\triangle A B C$ be $A B=c, B C=a, C A=b$, the area be $S, P D=x, P E=y$, $P F=z$, then $a x+b y+c z=2 S$. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)(a x+b y+c z) \geqslant(a+b+c)^{2} \Rightarrow \\
\frac{a}{x}+\frac{b}{y}+\f... | P \text{ is the incenter of } \triangle ABC | Geometry | math-word-problem | Yes | Yes | inequalities | false | 732,422 |
11. Let $a, b, c, d$ all be positive numbers, prove the inequality:
$$\frac{a}{b+2 c+3 d}+\frac{b}{c+2 d+3 a}+\frac{c}{d+2 a+3 b}+\frac{d}{a+2 b+3 c} \geqslant \frac{2}{3}$$
(34th IMO Shortlist) | 11. When $x_{1}, x_{2}, x_{3}, x_{4}$ and $y_{1}, y_{2}, y_{3}, y_{4}$ are all positive real numbers, by the Cauchy-Schwarz inequality we have
$$\sum_{i=1}^{n} \frac{x_{i}}{y_{i}} \cdot \sum_{i=1}^{n} x_{i} y_{i} \geqslant\left(\sum_{i=1}^{n} x_{i}\right)^{2}$$
Let $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=(a, b, c, d)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,423 |
12. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, and let $S_{k}$ be the sum of the products of $a_{1}, a_{2}, \cdots, a_{n}$ taken $k$ at a time. Prove that: $S_{k} S_{n-k} \geqslant\left(\mathrm{C}_{n}^{k}\right)^{2} a_{1} a_{2} \cdots a_{n}(k=1,2, \cdots, n-1)$. (1990 Asia Pacific Mathematical Olympiad Prob... | 12. The number of permutations of taking $k$ elements from $a_{1}, a_{2}, \cdots, a_{n}$, denoted as $\mathrm{C}_{n}^{k}$, is equal to the number of permutations of taking $n-k$ elements from $a_{1}, a_{2}, \cdots, a_{n}$, denoted as $\mathrm{C}_{n}^{n-k}$. This can be seen as a one-to-one correspondence, i.e., taking ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,424 |
14. Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}$ all be positive numbers, prove that:
$$\frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{4}}+\frac{x_{3}}{x_{4}+x_{5}}+\frac{x_{4}}{x_{5}+x_{6}}+\frac{x_{5}}{x_{6}+x_{1}}+\frac{x_{6}}{x_{1}+x_{2}} \geqslant 3$$ | 14.
$$\begin{array}{l}
\left(\frac{x_{1}}{x_{2}+x_{3}}+\frac{x_{2}}{x_{3}+x_{4}}+\frac{x_{3}}{x_{4}+x_{5}}+\frac{x_{4}}{x_{5}+x_{6}}+\frac{x_{5}}{x_{6}+x_{1}}+\frac{x_{6}}{x_{1}+x_{2}}\right) \\
{\left[x_{1}\left(x_{2}+x_{3}\right)=+x_{2}\left(x_{3}+x_{4}\right)+x_{3}\left(x_{4}+x_{5}\right)+\right.} \\
\left.x_{4}\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,426 |
16. Given that $a_{1}, a_{2}, a_{3}, a_{4}$ are the four sides of a quadrilateral with perimeter $2 s$, prove:
$$\sum_{i=1}^{4} \frac{1}{a_{i}+s} \leqslant \frac{2}{9} \sum_{1 \leqslant i<j \leqslant 4} \frac{1}{\sqrt{\left(s-a_{i}\right)\left(s-a_{j}\right)}}$$
(2004 Romanian National Selection Exam Problem) | 16.
$$\frac{2}{9} \sum_{1 \leqslant i<j \leqslant 4} \frac{1}{\sqrt{\left(s-a_{i}\right)\left(s-a_{j}\right)}} \geqslant \frac{4}{9} \sum_{1 \leqslant i<j \leqslant 4} \frac{1}{\left(s-a_{i}\right)+\left(s-a_{j}\right)}$$
Therefore, it is sufficient to prove that $\sum_{i=1}^{4} \frac{1}{a_{i}+s} \leqslant \frac{4}{9}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,428 |
19. (1) Let $a, b, c$ be positive numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove:
$$\frac{1}{1+2 a b}+\frac{1}{1+2 b c}+\frac{1}{1+2 c a} \geqslant 1$$
(2004 Estonian Olympiad Problem) | 19. (1) By the Cauchy-Schwarz inequality, we have \(3\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2}\). Furthermore, by the arithmetic mean being no less than the harmonic mean, we have
$$\begin{array}{l}
\frac{1}{1+2 a b}+\frac{1}{1+2 b c}+\frac{1}{1+2 c a} \geqslant 3 \cdot \frac{3}{3+2 a b+2 b c+2 c a}= \\
\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,432 |
(2) Let $a, b, c$ be positive numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove:
$$\frac{1}{1+a b}+\frac{1}{1+b c}+\frac{1}{1+c a} \geqslant \frac{3}{2}$$
(2008 Croatian Olympiad Problem) | (2) $\frac{1}{1+a b}+\frac{1}{1+b c}+\frac{1}{1+c a} \geqslant \frac{9}{3+a b+b c+c a} \geqslant \frac{9}{3+a^{2}+b^{2}+c^{2}}=\frac{3}{2}$. | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,433 |
21. Let $a, b, c$ be positive numbers, prove: $a^{3}+b^{3}+c^{3} \geqslant a^{2} b+b^{2} c+c^{2} a$. (1983 British Mathematical Olympiad) | $2 F$
$$a^{3}+b^{3}+c^{3}=\frac{a^{2} b}{\frac{b}{a}}+\frac{b^{2} c}{\frac{c}{b}}+\frac{c^{2} a}{\frac{a}{c}} \geqslant \frac{\left(a^{2} b+b^{2} c + c^{2} a\right)^{2}}{a b^{2}+b c^{2}+c a^{2}}$$
So
$$\left(a^{3}+b^{3}+c^{3}\right)^{2} \geqslant \frac{\left(a^{2} b+b^{2} c+c^{2} a\right)^{4}}{\left(a b^{2}+b c^{2}+c ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,435 |
22. Let $a, b, c, d \in \mathbf{R}^{+}$, prove the inequality: $\frac{a}{b+2 c+d}+\frac{b}{c+2 d+a}+\frac{c}{d+2 a+b}+$ $\frac{d}{a+2 b+c} \geqslant 1$. (2005 Romanian Mathematical Olympiad) | 22. $\frac{a}{b+2 c+d}+\frac{b}{c+2 d+a}+\frac{c}{d+2 a+b}+\frac{d}{a+2 b+c}=$
$$\begin{array}{l}
\frac{a^{2}}{a(b+2 c+d)}+\frac{b^{2}}{b(c+2 d+a)}+\frac{c^{2}}{c(d+2 a+b)}+\frac{d^{2}}{d(a+2 b+c)} \geqslant \\
\frac{(a+b+c+d)^{2}}{a(b+2 c+d)+b(c+2 d+a)+c(d+2 a+b)+d(a+2 b+c)}= \\
\frac{(a+b+c+d)^{2}}{2 a b+4 a c+2 a d+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,436 |
23. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, $x_{n+1}=x_{1}+x_{2}+\cdots+x_{n}$, prove: $x_{n+1} \sum_{k=1}^{n}\left(x_{n+1}-\right.$ $\left.x_{k}\right) \geqslant\left[\sum_{k=1}^{n} \sqrt{x_{k}\left(x_{n+1}-x_{k}\right)}\right]^{2} .(1996$ Romanian National Training Team Exam Problem) | 23. Since $\sum_{k=1}^{n}\left(x_{n+1}-x_{\bar{k}}\right)=n x_{n+1}-\sum_{k=1}^{n} x_{k}=(n-1) x_{n+1}$, it suffices to prove
$$x_{n+1} \sqrt{n-1} \geqslant \sum_{k=1}^{n} \sqrt{x_{k}\left(x_{n+1}-x_{k}\right)}$$
which is equivalent to proving
$$\sum_{k=1}^{n} \sqrt{\frac{x_{k}}{x_{n+1}}\left(1-\frac{x_{k}}{x_{n+1}}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,437 |
24. Given $\sum_{i=1}^{n} x_{i}^{2}=\sum_{i=1}^{n} y_{i}^{2}=1$, prove the inequality: $\left(x_{1} y_{2}-x_{2} y_{1}\right)^{2} \leqslant 2\left|1-\sum_{i=1}^{n} x_{i} y_{i}\right|$. (2001 South Korea, 2004 2005 French Mathematical Olympiad Problem) | 24. By the Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left|\sum_{i=1}^{n} x_{i} y_{i}\right| \leqslant \sqrt{\sum_{i=1}^{n} x_{i}^{2}} \cdot \sqrt{\sum_{i=1}^{n} y_{i}^{2}}=1 \\
\left(x_{1} y_{2}-x_{2} y_{i}\right)^{2} \leqslant \sum_{1 \leqslant i<j \leqslant n}\left(x_{i} y_{j}-x_{j} y_{i}\right)^{2}=\left(\sum_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,438 |
25. If $x, y, z \geqslant 1$, and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, prove: $\sqrt{x+y+z} \geqslant \sqrt{x-1}+$ $\sqrt{y-1}+\sqrt{z-1} .(1998$ Iran Mathematical Olympiad Problem) | 25. Notice that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, by the Cauchy-Schwarz inequality we have
$$\sqrt{x+y+z} \sqrt{\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}} \geqslant \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$
and $\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=3-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1$, so the ine... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,439 |
26. Given that $x, y, z$ are positive numbers, and $x+y+z=1$, prove: $\sqrt{x y(1-z)}+\sqrt{y z(1-x)}+$ $\sqrt{z x(1-y)} \leqslant \sqrt{\frac{2}{3}}$. (2005 Srpska Mathematical Olympiad) | 26. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
(\sqrt{x y(1-z)}+\sqrt{y z(1-x)}+\sqrt{z x(1-y)})^{2} \leqslant \\
(x y+y z+z x)[(1-z)+(1-x)+(1-y)]
\end{array}$$
And
$$\begin{array}{l}
(1-z)+(1-x)+(1-y)=3-(x+y+z)=2 \\
3(x y+y z+z x)= x y+y z+z x+2(x y+y z+z x) \leqslant \\
x^{2}+y^{2}+z^{2}+2(x y+y z+z x)=(x+y+z)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,441 |
27. Given that $x, y, z$ are positive numbers, prove that $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \geqslant \sqrt{\frac{3}{2}(x+y+z)}$. (2005 Serbian Mathematical Olympiad Problem) | 27. Let $a=\frac{x}{x+y+z}, b=\frac{x}{x+y+z}, c=\frac{x}{x+y+z}$, then $a+b+c=1$, the original inequality
is transformed into proving
$$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \geqslant \sqrt{\frac{3}{2}}$$
By Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
\frac{a}{\sqrt{b+c}}+\frac{b}{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,442 |
28. (-1) Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, $\min \left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=a_{1}, \max \left\{a_{1}, a_{2}, \cdots\right.$, $\left.a_{n}\right\}=a_{n}$, prove the inequality:
$$a_{1}^{2} \leq a_{2}^{2}+\cdots+a_{n}^{2} \geq \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}+\frac... | 28. (1) Since $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, by the Cauchy-Schwarz inequality we have
$$\begin{array}{l}
n\left[\left(\frac{a_{1}+a_{n}}{2}\right)^{2}+\left(\frac{a_{1}+a_{n}}{2}\right)^{2}+a_{2}^{2}+a_{3}^{2}+\cdots+a_{n-1}^{2}\right]= \\
(1+1+1+\cdots+1)\left[\left(\frac{a_{1}+a_{n}}{2}\right)^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,443 |
(2) Let $n \geqslant 2, a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, determine the largest real number $C_{n}$ such that the inequality: $\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n} \geqslant\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{2}+C_{n}\left(a_{1}-a_{n}\right)^{2}$. (2010 Central European Mathe... | (2) When $a_{1}=a_{n}$, $C_{n}$ can take any real number. Below, let's assume $a_{1} \neq a_{n}$, and denote $a_{1}=a$, $a_{n}=c$, $a_{2}+\cdots+a_{n-1}=b$. Then, by the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
f\left(a_{1}, a_{2}, \cdots, a_{n}\right)=\frac{\frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}-\... | \frac{1}{2 n} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,444 |
$$\text { 30. Prove: The inequality } \frac{a^{2}}{(-a+b)(a+c)}+\frac{b^{2}}{(b+c)(b+a)}+\frac{c^{2}}{(c+b)(c+a)} \geqslant \frac{3}{4}$$
holds for all positive real numbers $a, b, c$. (2004 Croatian Mathematical Olympiad) | 30. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
{\left[\frac{a^{2}}{(a+b)(a+c)}+\frac{b^{2}}{(b+c)(b+a)}+\frac{c^{2}}{(c+b)(c+a)}\right]} \\
{[(a+b)(a+c)+(b+c)(b+a)+(c+b)(c+a)] \geqslant(a+b+c)^{2}}
\end{array}$$
And
$$\begin{array}{l}
(a+b)(a+c)+(b+c)(b+a)+(c+b)(c+a)= \\
a^{2}+b^{2}+c^{2}+3(a b+b c+c a)= \\
(a+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,446 |
31. If $x_{\mathrm{F}}, x_{2}, \cdots, x_{n}$ are positive real numbers, and $x_{1}+x_{2}+\cdots+x_{n} \leqslant n$, prove that
$$\frac{x_{1}}{1+(n-1) x_{1}}+\frac{x_{2}}{1+(n-1) x_{2}}+\cdots+\frac{x_{n}}{1+(n-1) x_{n}} \leqslant 1$$
(Strengthened version of a 2004 Singapore Mathematical Olympiad problem) | 31. Since
$$\begin{array}{c}
\frac{x_{i}}{1+(n-1) x_{i}}=\frac{1}{n-1} \cdot \frac{1+(n-1) x_{i}-1}{1+(n-1) x_{i}}= \\
\frac{1}{n-1}-\frac{1}{n-1} \cdot \frac{1}{1+(n-1) x_{i}}
\end{array}$$
Thus,
$$\begin{array}{l}
\frac{x_{1}}{1+(n-1) x_{1}}+\frac{x_{2}}{1+(n-1) x_{2}}+\cdots+\frac{x_{n}}{1+(n-1) x_{n}} \leqslant 1 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,447 |
33. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, satisfying the conditions:
$$\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}=m, \frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}{n}=1 \text {. }$$
Prove: For any $i(1 \leqslant i \leqslant n)$, if the condition ... | 33. Define $b_{k}=m-a_{k}$, obviously we have $b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}$, and from $a_{k}$ being positive, we get $b_{k} \leqslant m$. From the given conditions, $\sum_{k=1}^{n} b_{k}=0, \sum_{k=1}^{n} b_{k}^{2}=n\left(1-m^{2}\right)$, and $b_{i} \geqslant 0$, we first have $b_{1}+b_{2}+\c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,449 |
34. Let $a, b, c$ be positive numbers, prove: $\frac{a b}{3 a+4 b-5 c}+\frac{b c}{3 b+4 c+5 a}+\frac{c a}{3 c+4 a+5 b} \leqslant \frac{\bar{F}}{12}(a+b+c) .(2006$ Bulgarian National Training Team Problem) | 34. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
{[(a+b)+2(a+c)+3(b+c)]\left(\frac{1}{a+b}+\frac{2}{a+c}+\frac{3}{b+c}\right) \geqslant} \\
(1+2+3)^{2}=36
\end{array}$$
Therefore,
$$\begin{array}{l}
\frac{1}{3 a+4 b+5 c} \leqslant \frac{1}{36}\left(\frac{1}{a+b}+\frac{2}{a+c}+\frac{3}{b+c}\right) \\
\frac{a b}{3 a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,450 |
35. Prove that for any positive real numbers $a, b, c$, we have $1<\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}} \leqslant \frac{3 \sqrt{2}}{2} \cdot(2004$ China Western Mathematical Olympiad Problem) | 35. The left side of the inequality is easy to prove:
$$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{b^{2}+c^{2}}}+\frac{c}{\sqrt{c^{2}+a^{2}}}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1$$
Now we prove the right side.
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(\frac{a}{\sqrt{a^{2}+b^{2}}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,452 |
36. For all positive real numbers $a, b, c$, prove that $\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1$. (42nd IMO problem) | 36. By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(a \sqrt{a^{2}+8 b c}+b \sqrt{b^{2}+8 c a}+c \sqrt{c^{2}+8 a b}\right)\left(\frac{a}{\sqrt{a^{2}+8 b c}}+\right. \\
\left.\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}}\right) \geqslant(a+b+c)^{2}
\end{array}$$
Again by Cauchy-Schwarz inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,453 |
37. Let $x, y, z$ be positive numbers, and $\frac{F}{x}+\frac{1}{y}+\frac{1}{z}=1$, prove that: $\sqrt{x+y z}+\sqrt{y+z x}+$ $\sqrt{z+x y} \geqslant \sqrt{x y z}+\sqrt{x}+\sqrt{y}+\sqrt{z}=$ (2002 Asia Pacific Mathematical Olympiad Problem) | 37. Since $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, then
$$x y+y z+z x=x y z$$
$$\begin{array}{l}
\sum \sqrt{x+y z}=\sum \sqrt{x \frac{x y z}{x y+y z+z x}+y z}= \\
\sum \sqrt{\frac{y z(x+y)(x+z)}{x y+y z+z x}}=\sqrt{\frac{1}{x y+y z+z x}} \sum \sqrt{y z(x+y)(x+z)} \geqslant \\
\sqrt{\frac{1}{x y+y z+z x}} \sum \sqrt{y z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,454 |
38. (1) Let $x, y, z, a, b$ be positive numbers, prove: $\frac{x}{a y+b z}+\frac{y}{a z+b x}+\frac{z}{a x+b y} \geqslant \frac{3}{a+b}$. (2005 Romanian National Training Team Test) | 38. (1) By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(\frac{x}{a y+b z}+\frac{y}{a z+b x}+\frac{z}{a x+b y}\right)[x(a y+b z)+y(a z+b x)+z(a x+b y)] \geqslant \\
(x+y+z)^{2} \cdot(a+b)(x y+y z+z x)= \\
x(a y+b z)+y(a z+b x)+z(a x+b y) \\
\text { Also, }(x+y+z)^{2} \geqslant 3(x y+y z+z x), \text { so }
\end{ar... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,455 |
39. Let $x, y, z$ be real numbers, and $x^{2}+y^{2}+z^{2}=9$, prove: $2(x+y+z)-x y z \leqslant 10$. (2002 Vietnam Mathematical Olympiad) | 39. Let $x^{2} \geqslant y^{2} \geqslant z^{2}$, so $x^{2} \geqslant 3, 6 \geqslant y^{2}+z^{2} \geqslant 2 y z$. Using the Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
{[2(x+y+z)-x y z]^{2}=[2(y+z)+x(2-y z)]^{2} \leqslant} \\
{\left[(y+z)^{2}+x^{2}\right]\left[2^{2}+(2-y z)^{2}\right]=(2 y z+9)\left(y^{2} z^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,457 |
40. Given that $a, b, c$ are non-negative numbers, prove:
$$\begin{aligned}
a^{2}+b^{2}+c^{2} \leqslant & \sqrt{b^{2}-b c+c^{2}} \sqrt{c^{2}-c a+a^{2}}+\sqrt{c^{2}-c a+a^{2}} \sqrt{a^{2}-a b+b^{2}}+ \\
& \sqrt{a^{2}-a b+b^{2}} \sqrt{b^{2}-b c+c^{2}}
\end{aligned}$$
(2000 Vietnam Mathematical Olympiad Problem) | 40. By Cauchy-Schwarz inequality,
$$\begin{aligned}
\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right)= & {\left[\left(c-\frac{b}{2}\right)^{2}+\frac{3}{4} b^{2}\right]\left[\left(c-\frac{a}{2}\right)^{2}+\frac{3}{4} a^{2}\right] \geqslant } \\
& {\left[\left(c-\frac{b}{2}\right)\left(c-\frac{a}{2}\right)+\frac{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,458 |
41. Let positive real numbers $a, b, c$ satisfy $a b c \geqslant 2^{9}$, prove:
$$\frac{F}{\sqrt{1+a}}+\frac{F}{\sqrt{1+b}}+\frac{F}{\sqrt{1+c}} \geqslant \frac{3}{\sqrt{1+\sqrt[5]{a b c}}}$$
(2004 Taiwan Mathematical Olympiad Problem) | 41. Let \(a b c=\lambda^{3}, a=\lambda \frac{y z}{x^{2}}, b=\lambda \frac{z x}{y^{2}}, c=\lambda \frac{x y}{z^{2}}, x, y, z\) be positive numbers, then the inequality (1) is equivalent to
$$\frac{x}{\sqrt{x^{2}+\lambda y z}}+\frac{y}{\sqrt{y^{2}+\lambda z x}}+\frac{z}{\sqrt{z^{2}+\lambda x y}} \geqslant \frac{3}{\sqrt{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,459 |
43. Let $a, b, c$ be positive real numbers, prove:
$$\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}{a+b+c} \\
\text { ( VasileC... | 43. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(2 a^{2}+b^{2}\right)\left(2 a^{2}+c^{2}\right)=\left(a^{2}+a^{2}+b^{2}\right)\left(c^{2}+a^{2}+a^{2}\right) \geqslant \\
\left(a c+a^{2}+a b\right)^{2}=a^{2}(a+b+c)^{2}
\end{array}$$
Therefore,
$$\frac{a^{3}}{\left(2 a^{2}+b^{2}\right)\left(2 a^{2}+c^{2}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,461 |
45. Let $a, b, c$ be positive real numbers, and $a+b+c=1$, prove: $\frac{a}{1+b c}+\frac{b}{1+c a}+\frac{c}{1+a b} \geqslant \frac{9}{10}$.
(1997 Indian National Training Team Problem) | 45. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
{[a(1+b c)+b(1+c a)+c(1+a b)]\left(\frac{a}{1+b c}+\frac{b}{1+c a}+\frac{c}{1+a b}\right) \geqslant} \\
(a+b+c)^{2}
\end{array}$$
That is,
$$\begin{array}{c}
(1+3 a b c)\left(\frac{a}{1+b c}+\frac{b}{1+c a}+\frac{c}{1+a b}\right) \geqslant 1 \\
\frac{a}{1+b c}+\frac... | \frac{a}{1+b c}+\frac{b}{1+c a}+\frac{c}{1+a b} \geqslant \frac{9}{10} | Inequalities | proof | Yes | Yes | inequalities | false | 732,465 |
46. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, satisfying $\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i}^{2}=t$, prove: $\sum_{i \neq j} \frac{x_{i}}{x_{j}} \geqslant$ $\frac{(n-1)^{2} t}{t-1}$. (2006 Turkish Mathematical Olympiad Problem) | 46. By Cauchy-Schwarz inequality,
$$\sum_{i \neq j} \frac{x_{i}}{x_{j}}=\sum_{i \neq j} \frac{x_{i}^{2}}{x_{i} x_{j}} \geqslant \frac{(n-1)^{2}\left(\sum_{i=1}^{n} x_{i}\right)^{2}}{2 \sum_{i \neq j} x_{i} x_{j}}=\frac{(n-1)^{2} t^{2}}{t^{2}-t}=\frac{(n-1)^{2} t}{t-1}$$ | \frac{(n-1)^{2} t}{t-1} | Inequalities | proof | Yes | Yes | inequalities | false | 732,466 |
47. Given that $a, b, c$ are positive numbers, and $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}=2$, prove: $a b+b c+c a \leqslant \frac{3}{2} .(2005$ Iran Mathematical Olympiad problem) | 47. From $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}=2$, we get $\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}=1$. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
{\left[\left(a^{2}+1\right)+\left(b^{2}+1\right)+\left(c^{2}+1\right)\right] \cdot\left(\frac{a^{2}}{a^{2}+1}+\fr... | ab+bc+ca \leqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,467 |
$\begin{array}{l}\text { 48. Given } a, b, c, d, e \text { are positive numbers and } a b c d e=1 \text {, prove: } \frac{a+a b c}{1+a b+a b c d}+\frac{b+b c d}{1+b c+b c d e}+ \\ \frac{c+c d e}{1+c d+c d e a}+\frac{d+d e a}{1+d e+d e a b}+\frac{e+e a b}{1+e a+e a b c} \geqslant \frac{10}{3} \text {. (Crux Problem 2023... | 48. Let $a=\frac{y}{x}, b=\frac{z}{y}, c=\frac{u}{z}, d=\frac{v}{u}, e=\frac{x}{v}$, where $x, y, z, u, v$ are all positive numbers.
The original inequality is equivalent to
$$\frac{u+y}{x+z+v}+\frac{z+v}{x+y+u}+\frac{x+u}{y+z+v}+\frac{y+v}{x+z+u}+\frac{x+z}{y+u+v} \geqslant \frac{10}{3}$$
Adding 5 to both sides, and... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,468 |
49. Real numbers $x, y, z, t$ satisfy $x+y+z+t=0, x^{2}+y^{2}+z^{2}+t^{2}=1$, prove: $-1 \leqslant$ $xy + yz + zt + tx \leqslant 0$. (1996 Austrian-Polish Mathematical Olympiad Problem) | 49. Since $x+y+z+t=0$, then $y+t=-(x+z)$, and
$$x y+y z+z t+t x=(x+z)(y+t)=-(x+z)^{2} \leqslant 0$$
By the Cauchy-Schwarz inequality,
$$(x y+y z+z t+t x)^{2} \leqslant\left(x^{2}+y^{2}+z^{2}+t^{2}\right)\left(y^{2}+z^{2}+t^{2}+x^{2}\right)=1$$
Therefore, $-1 \leqslant x y+y z+z t+t x \leqslant 1$.
In summary, $-1 \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,469 |
50. Let $a, b, c$ be positive real numbers, and $a+b+c=1$, prove:
$$\frac{1}{a b+2 c^{2}+2 c}+\frac{1}{b c+2 a^{2}+2 a}+\frac{1}{c a+2 b^{2}+2 b} \geqslant \frac{1}{a b+b c+c a}$$
(2007 Turkish National Training Team Problem) | 50. We prove
$$I=\frac{a b+b c+c a}{a b+2 c^{2}+2 c}+\frac{a b+b c+c a}{b c+2 a^{2}+2 a}+\frac{a b+b c+c a}{c a+2 b^{2}+2 b} \geqslant 1$$
Since $a+b+c=1$, we have
$$\begin{aligned}
\frac{a b+b c+c a}{b c+2 a^{2}+2 a}= & \frac{a b+b c+c a}{b c+2 a^{2}+2 a(a+b+c)}=\frac{2(a b+b c+c a)}{2 b c+4 a^{2}+4 a(a+b+c)}= \\
& \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,470 |
51. Let $x, y, z$ be positive real numbers, and $x \geqslant y \geqslant z$, prove that: $\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y} \geqslant x^{2}+y^{2}+z^{2}$. (31st IMO Preliminary Problem) | 51. Let the left side of the inequality be $M$, and let $N=\frac{x^{2} z}{y}+\frac{y^{2} x}{z}+\frac{z^{2} y}{x}$. By the Cauchy-Schwarz inequality, we have
$$\left(\frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y}\right)\left(\frac{x^{2} z}{y}+\frac{y^{2} x}{z}+\frac{z^{2} y}{x}\right) \geqslant\left(x^{2}+y^{2}+z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,471 |
53. Let $n$ be a positive integer, and $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$ be real numbers. Prove:
$$\left(\sum_{i, j=1}^{n}\left|x_{i}-x_{j}\right|\right)^{2} \leqslant \frac{2\left(n^{2}-1\right)}{3} \sum_{i, j=1}^{n}\left(x_{i}-x_{j}\right)^{2}$$
(2) The necessary and sufficient condition for eq... | 53. (1) Without loss of generality, we can assume $\sum_{i=1}^{n} x_{i}=0$, then we have
$$\sum_{i, j=1}^{n}\left|x_{i}-x_{j}\right|=2 \sum_{i<j}\left(x_{j}-x_{i}\right)=2 \sum_{i=1}^{n}(2 i-n-1) x_{i}$$
By the Cauchy-Schwarz inequality, we get
$$\left(\sum_{i, j=1}^{n}\left|x_{i}-x_{j}\right|\right)^{2} \leqslant 4 \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,473 |
24. Given $a, b, c, d>0$, and $a+b+c+d=1$, prove: $a b+b c+c d+d a \leqslant \frac{1}{4}$. (1993 Indian Mathematical Olympiad Problem) | 24. $\begin{aligned} a b+b c+c d+d a= & \frac{1}{4}\left[(a+b+c+d)^{2}-(a-b+c-d)^{2}\right] \leqslant \\ & \frac{1}{4}(a+b+c+d)^{2}=\frac{1}{4}\end{aligned}$
24. $\begin{aligned} ab+bc+cd+da= & \frac{1}{4}\left[(a+b+c+d)^{2}-(a-b+c-d)^{2}\right] \leqslant \\ & \frac{1}{4}(a+b+c+d)^{2}=\frac{1}{4}\end{aligned}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,474 |
55. Let $a_{1}, a_{2}, \cdots, a_{n}$ be an infinite sequence of real numbers, such that for all positive integers $i$, there exists a real number $c$ such that $0 \leqslant a_{i} \leqslant c$, and $\left|a_{i}-a_{j}\right| \geqslant \frac{1}{i+j}$, for all positive integers $i, j(i \neq j)$. Prove: $c \geqslant 1$. (4... | For $n \geqslant 2$, let $\sigma(1), \sigma(2), \cdots, \sigma(n)$ be a permutation of $1,2, \cdots, n$, and satisfy $0 \leqslant a_{\sigma(1)}<a_{\sigma(2)}<\cdots<a_{\sigma(n)} \leqslant c$, then
$$\begin{array}{l}
c \geqslant a_{\sigma(n)}-a_{\sigma(1)} \geqslant\left(a_{\sigma(n)}-a_{\sigma(n-1)}\right)+ \\
\left(a... | c \geqslant 1 | Algebra | proof | Yes | Yes | inequalities | false | 732,476 |
56. Let real numbers $\alpha, \beta, \gamma$ satisfy $\beta \gamma \neq 0$, and $\frac{1-\gamma^{2}}{\beta \gamma} \geqslant 0$. Prove: $10\left(\alpha^{2}+\beta^{2}+\gamma^{2}-\right.$ $\left.\beta \gamma^{3}\right) \geqslant 2 \alpha \beta+5 \alpha \gamma$. (19th Greek Mathematical Olympiad Problem) | 56. Since $\frac{1-\gamma^{2}}{\beta \chi} \geqslant 0 \Leftrightarrow \beta \gamma\left(1-\gamma^{2}\right) \geqslant 0$-thus
$$10\left(\alpha^{2}+\beta^{2}+\gamma^{2}-\beta \gamma^{3}\right) \geqslant 10\left(\alpha^{2} \mp \beta^{2}+\gamma^{2}-\beta \gamma\right)$$
Therefore, it is sufficient to prove
$$10\left(\al... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,477 |
57. Given a positive integer $n(n \geqslant 2)$, let positive integers $a_{i}(i=1, 2, \cdots, n)$ satisfy $a_{1}<a_{2}<\cdots<a_{n}$ and $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$. Prove that for any real number $x$, we have
$$\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \frac{1}{2} \times \fra... | 57. When $x^{2} \geqslant a_{1}\left(a_{1}-1\right)$
$$\begin{aligned}
\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant & \left(\sum_{i=1}^{n} \frac{1}{2 a_{i} \mathrm{E} x}\right)^{2}= \\
& \frac{1}{4 x^{2}}\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} \leqslant \frac{1}{4 x^{2}} \leqslant \\
& \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,478 |
58-Let $n(n \geqslant 2)$ be a given positive integer, find all integer solutions $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ satisfying the conditions:
$$\begin{array}{l}
(-1) a_{1}+a_{2}+\cdots+a_{n} \geqslant n^{2} \\
(-2)=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \leqslant n^{3}+1
\end{array}$$
(-2002 Western Mathematica... | 58. Let $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ be an integer solution satisfying the conditions, then by the Cauchy-Schwarz inequality we have
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2} \geqslant \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \geqslant n^{3}$$
Combining this with $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n... | (n, n, \cdots, n) | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 732,479 |
59. Given non-negative real numbers $a, b, c$ satisfying $a+b+c=1$, prove:
$$\left.\left.2 \leqslant\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2} \leqslant(1+a)(1+b)(1+c)\right)\right.$$
and find the conditions under which equality holds. (2000 Austrian-Polish Mathematical Olympiad Problem... | 59. Let \( ab + bc + ca = M, abc = n \), then
\[
\begin{aligned}
(x-a)(x-b)(x-c) = & x^{3} - (a+b+c) x^{2} + (ab + bc + ca) x - abc = \\
& x^{3} - x^{2} + M x - n
\end{aligned}
\]
Let \( x = a \), then we have \( a^{3} = a^{2} - M a + n \), thus
\[
\begin{array}{l}
\sum a^{3} = \sum a^{2} - M \sum a + 3 n \\
\sum a^{4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,480 |
60. Let the integer $n \geqslant 4, a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers such that $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$. Prove:
$$\begin{array}{l}
\frac{a_{1}}{a_{2}^{2}+1}+\frac{a_{2}}{a_{3}^{2}+1}+\cdots+\frac{a_{n-1}}{a_{n}^{2}+1}+\frac{a_{n}}{a_{1}^{2}+1} \geqslant \\
\frac{4}{5}\left(a_{1} \sq... | 60. By Cauchy-Schwarz inequality, we have
$$\frac{a_{1}^{2}}{x_{1}}+\frac{a_{2}^{2}}{x_{2}}+\cdots+\frac{a_{n}^{2}}{x_{n}} \geqslant \frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}}{x_{1}+x_{2}+\cdots+x_{n}}$$
where $x_{1}, x_{2}, \cdots, x_{n}$ are positive real numbers. Therefore,
$$\begin{array}{l}
\frac{a_{1}}{a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,481 |
61. Let $a_{i} \in \mathbf{R}, i=1,2,3,4,5$, find
$$\frac{a_{1}}{a_{2}+3 a_{3}+5 a_{4}+7 a_{5}}+\frac{a_{2}}{a_{3}+3 a_{4}+5 a_{5}+7 a_{1}}+\cdots+\frac{a_{5}}{a_{1}+3 a_{2}+5 a_{3}+7 a_{4}}$$
the minimum value. (2004 Jilin Province Mathematics Competition Problem) | 61. Let the original expression be $A$, by Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
A\left(a_{1}\left(a_{2}+3 a_{3}+5 a_{4}+7 a_{5}\right)+a_{2}\left(a_{3}+3 a_{4}+5 a_{5}+7 a_{1}\right)+\cdots+\right. \\
\left.a_{5}\left(a_{1}+3 a_{2}+5 a_{3}+7 a_{4}\right)\right) \geqslant\left(a_{1}+a_{2}+a_{3}+a_{4}+a_... | \frac{5}{16} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,482 |
62. Given positive real numbers $a, b, c, d$, prove that
$$\frac{a^{3}+b^{3}+c^{3}}{a+b+c}+\frac{b^{3}+c^{3}+d^{3}}{b+c+d}+\frac{c^{3}+d^{3}+a^{3}}{c+d+a}+\frac{d^{3}+a^{3}+b^{3}}{d+a+b} \geqslant a^{2}+b^{2}+c^{2}+d^{2}$$
(US College Mathematics Competition Problem) | 62. By Cauchy-Schwarz inequality,
$$\begin{array}{c}
\left(a^{3}+b^{3}+c^{3}\right)(a+b+c) \geqslant\left(a^{2}+b^{2}+c^{2}\right)^{2} \\
(1+1+1)\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b+c)^{2}
\end{array}$$
Thus,
$$\begin{array}{l}
\left(a^{3}+b^{3}+c^{3}\right)(a+b+c) \geqslant\left(a^{2}+b^{2}+c^{2}\right)^{2} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,483 |
63. Let $P_{1}, P_{2}, \cdots, P_{n}(n \geqslant 2)$ be any permutation of $1,2, \cdots, n$. Prove that:
$$\frac{1}{P_{1}+P_{2}}+\frac{1}{P_{2}+P_{3}}+\cdots+\frac{1}{P_{n-2}+P_{n-1}}+\frac{1}{P_{n-1}+P_{n}}>\frac{n-1}{n-2}$$
(2002 Girls' Mathematical Olympiad Problem) | 63. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\left(\frac{1}{P_{1}+P_{2}}+\frac{1}{P_{2}+P_{3}}+\cdots+\frac{1}{P_{n-2}+P_{n-1}}+\frac{1}{P_{n-1}+P_{n}}\right) \\
{\left[\left(P_{1}+P_{2}\right)+\left(P_{2}+P_{3}\right)+\cdots+\left(P_{n-2}+P_{n-1}\right)+\left(P_{n-1}+P_{n}\right)\right] \geqslant} \\
(n-1)^{2}... | \frac{n-1}{n-2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,484 |
25. Let $n$ be a natural number greater than 2. Prove that the following inequality holds for any real numbers $a_{1}, a_{2}, \cdots, a_{n}$ if and only if $n=3$ or $n=5$: $\left(a_{1}-a_{2}\right)\left(a_{1}-a_{3}\right) \cdots\left(a_{1}-a_{n}\right)+\left(a_{2}-a_{1}\right) \cdot$ $\left(a_{2}-a_{3}\right) \cdots\le... | 25. When $n=3$
$$\begin{array}{l}
\left(a_{1}-a_{2}\right)\left(a_{1}-a_{3}\right)+\left(a_{2}-a_{1}\right)\left(a_{2}-a_{3}\right)+\left(a_{3}-a_{1}\right)\left(a_{3}-a_{2}\right)= \\
\frac{1}{2}\left\{\left[\left(a_{1}-a_{2}\right)\left(a_{1}-a_{3}\right)+\left(a_{2}-a_{1}\right)\left(a_{2}-a_{3}\right)\right]+\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,485 |
64. Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers, satisfying $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$. Prove that for any integer $k \geqslant 2$, there exist integers $a_{1}, a_{2}, \cdots, a_{n}$, not all zero, such that $\left|a_{i}\right| \leqslant k-1(i=1,2, \cdots, n)$ and $\left|a_{1} x_{1}+a_{2} x_{2}+\cdo... | 64. Without loss of generality, let $x_{i} \geqslant 0(i=1,2, \cdots, n)$, otherwise replace $x_{i}$ with $-x_{i}$. By the Cauchy-Schwarz inequality, we have
$$x_{1}+x_{2}+\cdots+x_{n} \leqslant \sqrt{n} \cdot \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}=\sqrt{n}$$
Therefore, for all $a_{i}=0,1,2, \cdots, k-1(i=1,2, \c... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,486 |
65. Let $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{n}$ be real numbers, and satisfy $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, $y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2}=1$. Prove: There exist not all zero numbers $a_{1}, a_{2}, \cdots$, taking values in $\{-1,0,1\}$, such that $| a_{1} x_{1} y_{1}+a_{2} x_{... | 65. By Cauchy-Schwarz inequality, we have
$$\begin{array}{c}
\left(\left|x_{1}\right| \cdot\left|y_{1}\right|+\left|x_{2}\right| \cdot\left|y_{2}\right|+\cdots+\left|x_{n}\right| \cdot\left|y_{n}\right|\right)^{2} \leqslant \\
\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)\left(y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2}\r... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,487 |
(2) Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers. Prove that: $\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}+\frac{x_{n}^{2}}{x_{1}} \geqslant x_{1}+$ $x_{2}+\cdots+x_{n}+\frac{4\left(x_{1}-x_{2}\right)^{2}}{x_{1}+x_{2}+\cdots+x_{n}} .(1984$ National High School Mathemat... | (2) Since $x^{2}+y^{2}-2xy=(x-y)^{2}$, we have $\frac{x^{2}}{y}=2x-y+\frac{(x-y)^{2}}{y}$. Therefore, $\frac{x_{k}^{2}}{x_{k+1}}=2x_{k}-x_{k+1}+\frac{\left(x_{k}-x_{k+1}\right)^{2}}{x_{k+1}}\left(k=1,2, \cdots, n, \text{ where } x_{n+1}=x_{1}\right)$.
Adding these $n$ equations, we get
$$\frac{x_{1}^{2}}{x_{2}}+\frac{x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,489 |
67. Given $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ are all positive numbers, and satisfy $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=$ $\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}\right)^{3}$. Prove:
$$\frac{b_{1}^{3}}{a_{1}}+\frac{b_{2}^{3}}{a_{2}}+\cdots+\frac{b_{n}^{3}}{a_{n}} \geqslant 1$$ | 67. By Cauchy's inequality,
$$\begin{array}{l}
\left(\frac{b_{1}^{3}}{a_{1}}+\frac{b_{2}^{3}}{a_{2}}+\cdots+\frac{b_{n}^{3}}{a_{n}}\right)\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right) \geqslant \\
\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}\right)^{2}= \\
\sqrt{\left(b_{1}^{2}+b_{2}^{2}++b_{n}^{2}\right)\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,491 |
68. Prove: For any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, the inequality
$$\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\cdots+\frac{n}{a_{1}+a_{2}+\cdots+a_{n}}<2\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)$$ | 68. By the Cauchy-Schwarz inequality,
$$\frac{k^{2}(k+1)^{2}}{4}=\left(\sum_{i=1}^{k} \frac{i}{\sqrt{a_{i}}} \cdot \sqrt{a_{i}}\right)^{2} \leqslant \sum_{i=1}^{k} \frac{i^{2}}{a_{i}} \sum_{i=1}^{n} a_{i}$$
Therefore,
$$\frac{k}{\sum_{i=1}^{k} a_{i}} \leqslant \frac{4}{k(k+1)^{2}} \sum_{i=1}^{k} \frac{i^{2}}{a_{i}}, \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,492 |
69. If real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=0$. Prove that:
$$\max _{1 \leqslant k \leqslant n} a_{k}^{2} \leqslant \frac{n}{3} \sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2}$$
(2006 China Mathematical Olympiad Problem) | 69. To prove that for any $1 \leqslant k \leqslant n$, $a_{k}^{2} \leqslant \frac{n}{3} \sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)^{2}$ is sufficient. Let $d_{k}=a_{k} - a_{k+1}, k=1,2, \cdots, n-1$. Then
$$\begin{array}{c}
a_{k}=a_{k} \\
a_{k+1}=a_{k}-d_{k}, a_{k+2}=a_{k}-d_{k}-d_{k+1}, \cdots, a_{n}=a_{k}-d_{k}-d_{k+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,493 |
70. Let $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n}$ be positive real numbers. Prove that: $\sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}} \leqslant \frac{A B}{A+B}$, where $A=\sum_{k=1}^{n} a_{k}, B=\sum_{k=1}^{n} b_{k}$. (1993 St. Petersburg City Mathematical Selection Test) | 70. By Cauchy's inequality,
$$\left(\sum_{i=1}^{n}\left(a_{i}-b_{i}\right)\right)^{2} \leqslant \sum_{i=1}^{n}\left(a_{i}+b_{i}\right)\left(-\sum_{i=1}^{n} \frac{\left(a_{i}-b_{i}\right)^{2}}{a_{i}+b_{i}}\right)$$
Therefore,
$$\begin{array}{l}
A B=\sum_{i=1}^{n} a_{i} \sum_{i=1}^{n} b_{i}=\frac{1}{4}\left[\left(\sum_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,494 |
26. Given $a, b>0$, prove: $\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+a+\frac{3}{4}\right) \geqslant\left(2 a+\frac{1}{2}\right)\left(2 b+\frac{1}{2}\right)$. (2005 Belarus Mathematical Olympiad Problem) | 26. Since
$$\begin{array}{c}
\left(a^{2}+b+\frac{3}{4}\right)\left(b^{2}+a+\frac{3}{4}\right)-\left(a^{2}+a+\frac{3}{4}\right)\left(b^{2}+b+\frac{3}{4}\right)= \\
\left(a^{2}-b^{2}\right)(a-b)=(a-b)^{2}(a+b) \geqslant 0 \\
a^{2}+a+\frac{3}{4}-\left(2 a+\frac{1}{2}\right)=\left(a-\frac{1}{2}\right)^{2} \geqslant 0 \\
b^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,496 |
75. Given that $x, y, z$ are positive numbers, prove that $\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}} \geqslant \sqrt{\frac{3}{2}(x+y+z)}$. (2005 Serbia Mathematical Olympiad Problem) | 75. By Cauchy-Schwarz inequality,
$$\left(\frac{x}{\sqrt{y+z}}+\frac{y}{\sqrt{z+x}}+\frac{z}{\sqrt{x+y}}\right)(x \sqrt{y+z}+y \sqrt{z+x}+z \sqrt{x+y}) \geqslant(x+y+z)^{2}$$
Again by Cauchy-Schwarz inequality,
$$\begin{array}{l}
x \sqrt{y+z}+y \sqrt{z+x}+z \sqrt{x+y}=\sqrt{x} \cdot \sqrt{x(y+z)}+ \\
\sqrt{y} \cdot \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,500 |
77. Given $a, b, c > 0$, prove: $\frac{a^{3}}{b+2 c}+\frac{b^{3}}{c+2 a}+\frac{c^{3}}{a+2 b} \geqslant \frac{a^{2}+b^{2}+c^{2}}{3}$. (1996 Ukrainian Mathematical Olympiad Problem) | $$\text { 77. } \begin{array}{l}
\frac{a^{3}}{b+2 c}+\frac{b^{3}}{c+2 a}+\frac{c^{3}}{a+2 b}=\frac{a^{4}}{a b+2 a c}+\frac{b^{4}}{b c+2 a b}+\frac{c^{4}}{a c+2 b c} \geqslant \\
\frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a b+2 a c+b c+2 a b+a c+2 b c}= \\
\left(a^{2}+b^{2}+c^{2}\right) \cdot \frac{a^{2}+b^{2}+c^{2}}{3(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,502 |
78. Given $a, b, c > 0$, and $a^{2} + b^{2} + c^{2} = 1$, prove: $\frac{a}{b^{2} + 1} + \frac{b}{c^{2} + 1} + \frac{c}{a^{2} + 1} \geqslant \frac{3}{4}(a \sqrt{a} + b \sqrt{b} + c \sqrt{c})^{2} \cdot(2002$ Mediterranean Mathematical Olympiad Problem) | 78. By Cauchy-Schwarz inequality,
$$\begin{array}{l}
\frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}=\frac{a^{2}}{a^{2} b^{2}+a^{2}}+\frac{b^{3}}{b^{2} c^{2}+b^{2}}+\frac{c}{c^{2} a^{2}+c^{2}} \geqslant \\
\frac{(a \sqrt{a}+b \sqrt{b}+c \sqrt{c})^{2}}{a^{2} b^{2}+a^{2}+b^{2} c^{2}+b^{2}+c^{2} a^{2}+c^{2}}= \\
\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,503 |
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