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1. Prove: For any triangle with side lengths $a, b, c$, the inequality
$$a(b-c)^{2}+b(c-a)^{2}+c(a-b)^{2}+4 a b c>a^{3}+b^{3}+c^{3}$$
holds.
(Hungarian Mathematical Olympiad problem) | 1. Since $a, b, c$ are the lengths of the three sides of a triangle, we have $a>0, b>0, c>0$, and $a+b-c>0, b+c-a>0, c+a-b>0$. Notice that
$$c(a-b)^{2}+4 a b c=c(a+b)^{2}$$
Taking the difference
$$\begin{aligned}
S= & a(b-c)^{2}+b(c-a)^{2}+c(a-b)^{2}+4 a b c-\left(a^{3}+b^{3}+c^{3}\right)= \\
& a\left[(b-c)^{2}-a^{2}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,218 |
86. Let $a, b, c$ be positive real numbers, prove that
$$\frac{1}{a+a b+a b c}+\frac{1}{b+b c+b c a}+\frac{1}{c+c a+c a b} \leqslant \frac{1}{3 \sqrt[3]{a b c}}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ | 86. By the AM-GM inequality,
$$a+a b+a b c \geqslant 3 \sqrt[3]{a \cdot a b \cdot a b c}=3 \sqrt[3]{a b c} \cdot \sqrt[3]{a^{2} b}$$
Similarly,
$$\begin{array}{l}
\quad b+b c+b c a \geqslant 3 \sqrt[3]{a b c} \cdot \sqrt[3]{b^{2} c}, c+c a+c a b \geqslant 3 \sqrt[3]{a b c} \cdot \sqrt[3]{c^{2} a} \\
\frac{1}{a+a b+a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,219 |
87. Let $a, b, c, d, e, f$ be positive real numbers, and $a+b+c+d+e+f=1, a c e+b d f \geqslant \frac{1}{108}$. Prove: $a b c+b c d+c d e+d e f+e f a+f a b \leqslant \frac{1}{2 c}$. (1998 Polish Mathematical Olympiad problem) | 87. Let $A=a c e+b d f, B=a b c+b c d+c d e+d e f+e f a+f a b$, then by the AM-GM inequality we have
$$\begin{array}{c}
A+B=(a+d)(b+e)(c+f) \leqslant\left[\frac{(a+d)+(b+e)+(c+f)}{3}\right]^{3}=\frac{1}{27} \\
B \leqslant \frac{1}{27}-A \leqslant \frac{1}{27}-\frac{1}{108}=\frac{1}{36}
\end{array}$$
Equality holds if ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,220 |
88. Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, and $n \geqslant 3$ be a positive integer. Prove that at least one of the following two inequalities holds:
where
$$\begin{array}{c}
\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}+x_{i+2}} \geqslant \frac{n}{2}, \sum_{i=1}^{n} \frac{x_{i}}{x_{i-1}+x_{i-2}} \geqslant... | 88. Since $x_{n+1}=x_{1}, x_{n+2}=x_{2}$, we have
$$\begin{array}{l}
\sum_{i=1}^{n} \frac{x_{i}}{x_{i-1}+x_{i-2}}=\sum_{i=1}^{n} \frac{x_{i+3}}{x_{i+1}+x_{i+2}} \\
\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}+x_{i+2}}+\sum_{i=1}^{n} \frac{x_{i}}{x_{i-1}+x_{i-2}}= \\
\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}+x_{i+2}}+\sum_{i=1}^{n} \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,221 |
90. Given positive real numbers $a, b, c$ satisfying $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$, prove that: $(a-1)(b-1)(c-1) \geqslant 8$. (2005 Croatian Mathematical Olympiad Problem) | 90. Multiply both sides of the known equation by $a b c$, we get $a b + b c + c a = a b c$, then $(a-1)(b-1)(c-\mathrm{F})=a b c-(a b + b c + c a)+(a+b+c)-1=(a+b+c)-1$
Moreover, by the AM-GM inequality, we have $a+b+c=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3+\left(\frac{a}{b}+\right.$ $\left.\frac{b}{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,223 |
92. Let $a, b, c$ be positive real numbers, and $a+b+c=1$, prove:
$$\frac{1}{a b+2 c^{2}+2 c}+\frac{1}{b c+2 a^{2}+2 a}+\frac{1}{c a+2 b^{2}+2 b} \geqslant \frac{1}{a b+b c+c a}$$ | 92. First, we will prove
$$\frac{a b+b c+c a}{a b+2 c^{2}+2 c} \geqslant \frac{a b}{a b+b c+c a}$$
In fact, the above inequality is equivalent to
$$a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c) \geqslant a^{2} b^{2}+2 a b c^{2}+2 a b c$$
Given that $2 a b c(a+b+c)=2 a b c$, by the AM-GM inequality we have $b^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,225 |
93. Let $a, b, c$ be real numbers, and $abc=1$, prove:
$$\begin{array}{l}
a^{2}+b^{2}+c^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+2\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant \\
6+2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)
\end{array}$$ | 93. Since $a b c=1$, we have
$$\begin{array}{l}
a^{2}+b^{2}+c^{2}+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=(a+b+c)^{2} \\
\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+2(a+b+c)=b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}+2 a b c(a+b+c)= \\
(a b+b c+c a)^{2} \\
2\left(\frac{a}{b}+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,226 |
94. Let $x, y, z$ be positive real numbers, and $x y z=1$, prove:
$$\begin{array}{l}
\text { (1) }(1+x+y)^{2}+(1+y+z)^{2}+(1+z+x)^{2} \geqslant 27 \\
\text { (2) }(1+x+y)^{2}+(1+y+z)^{2}+(1+z+x)^{2} \leqslant 3(x+y+z)^{2}
\end{array}$$
and the equality holds in both inequalities if and only if $x=y=z=1$. (2008 Iran Ma... | 94. (1) By Cauchy-Schwarz inequality and AM-GM inequality, we have
$$\begin{array}{l}
3\left[(1+x+y)^{2}+(1+y+z)^{2}+(1+z+x)^{2}\right] \geqslant \\
{[(1+x+y)+(1+y+z)+(1+z+x)]^{2}=} \\
{[3+2(x+y+z)]^{2} \geqslant\left(3+2 \times 3 \sqrt[3]{(x y z)^{2}}\right)^{2}=81}
\end{array}$$
Therefore, $(1+x+y)^{2}+(1+y+z)^{2}+(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,227 |
95. Find all positive integers $n$, such that if $a, b, c \geqslant 0$, and $a+b+c=3$, then $a b c\left(a^{n}+b^{n}+\right.$
$\left.c^{n}\right) \leqslant 3 .(2007$ Bulgaria Winter Mathematical Olympiad Problem) | 95. When $n \geqslant 3$, take $a=2, b=c=\frac{1}{2}$, then
$$a b c\left(a^{n}+b^{n}+c^{n}\right)=\frac{1}{2}\left(2^{n}+\frac{1}{2^{n}}+\frac{F}{2^{n}}\right)>3$$
When $n=1$, $a b c(a+b+c)=3 a b c \leqslant 3\left(\frac{a+b+c}{3}\right)^{3}=3$, the inequality holds.
When $n=2$, let $q=a b+b c+c a$, then $a^{2}+b^{2}+... | proof | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,228 |
2. Prove: $5+8 \cos \theta+4 \cos 2 \theta+\cos 3 \theta \geqslant 0$. (1978 National Mathematics Competition Problem) | 2.
$$\begin{array}{l}
5+8 \cos \theta+4 \cos 2 \theta+\cos 3 \theta= \\
5+8 \cos \theta+4\left(2 \cos ^{2} \theta-1\right)+\left(4 \cos ^{3} \theta-3 \cos \theta\right)= \\
1+5 \cos \theta+8 \cos ^{2} \theta+4 \cos ^{3} \theta= \\
(1+\cos \theta)\left(4 \cos ^{2} \theta+4 \cos \theta+1\right)= \\
(1+\cos \theta)(2 \cos... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,229 |
96. Let $x, y, z$ be positive real numbers, and $x+y+z=3$, prove:
$$\frac{x^{3}}{y^{3}+8}+\frac{y^{3}}{z^{3}+8}+\frac{z^{3}}{x^{3}+8} \geqslant \frac{1}{9}+\frac{2}{27}(x y+y z+z x)$$
(2008 Iran Mathematical Olympiad) | 96: By the AM-GM inequality, we have
$$\frac{x^{3}}{y^{3}+8}+\frac{y+2}{27}+\frac{y^{2}-2 y+4}{27} \geqslant 3 \sqrt[3]{\frac{x^{3}}{y^{3}+8} \cdot \frac{y+2}{27} \cdot \frac{y^{2}-2 y+4}{27}}=\frac{x}{3}$$
Similarly,
$$\begin{array}{l}
\frac{y^{3}}{z^{3}+8}+\frac{z+2}{27}+\frac{z^{2}-2 z+4}{27} \geqslant \frac{y}{3} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,230 |
98. Let positive numbers $a, b, c$ satisfy $a+b+c \leqslant \frac{3}{2}$, find the minimum value of $S=abc+\frac{1}{abc}$. (2008 Moldova Mathematical Olympiad Problem) | 98. By the AM-GM inequality, we have $\frac{3}{2} \geqslant a+b+c \geqslant 3 \sqrt[3]{a b c}$, thus $a b c \leqslant \frac{1}{8}$. By the AM-GM inequality, we get
$$\begin{aligned}
S= & a b c+\frac{1}{64 a b c}+\frac{63}{64 a b c} \geqslant 2 \sqrt{a b c \cdot \frac{1}{64 a b c}}+\frac{63}{64 a b c} \geqslant \\
& \fr... | \frac{65}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,232 |
101. Let $x, y, z$ be positive numbers, and $x+2 y+3 z=\frac{11}{12}$, prove: $6(3 x y+4 x z+2 y z+6 x$ $+3 y+4 z+72 x y z) \leqslant \frac{107}{18}$. (2006 JMMO Preliminary Problem) | $\begin{array}{l}\text { 101. Let } P=6(3 x y+4 x z+2 y z+6 x+3 y+4 z+72 x y z)=12\left(x+\frac{1}{2}\right)(2 y+ \\ \left.\frac{2}{3}\right)\left(3 z+\frac{3}{4}\right)-1 \leqslant 12\left(\frac{\left(x+\frac{1}{2}\right)+\left(2 y+\frac{2}{3}\right)+\left(3 z+\frac{3}{4}\right)}{3}\right)^{3}-1=12\left(\frac{5}{6}\ri... | \frac{107}{18} | Inequalities | proof | Yes | Yes | inequalities | false | 732,235 |
103. Let $a, b, c$ be positive numbers, prove the inequalities:
(1) $\frac{a}{3a^2 + b^2 + 2ac} + \frac{b}{3b^2 + c^2 + 2ab} + \frac{c}{3c^2 + a^2 + 2bc} \leq \frac{3}{2(a+b+c)}$;
(2) $\frac{a}{2a^2 + b^2 + c^2} + \frac{b}{2b^2 + c^2 + a^2} + \frac{c}{2c^2 + a^2 + b^2} \leq \frac{9}{4(a+b+c)}$.
(2009 Ukrainian City Mat... | 103. (1) By the AM-GM inequality, we have $a^{2}+b^{2} \geqslant 2 a b$, so
$$\begin{aligned}
3 a^{2}+b^{2}+2 a c= & 2 a^{2}+\left(a^{2}+b^{2}\right)+2 a c \geqslant \\
& 2 a^{2}+2 a b+2 a c=2 a(a+b+c)
\end{aligned}$$
Thus,
$$\frac{a}{3 a^{2}+b^{2}+2 a c} \leqslant \frac{a}{2 a(a+b+c)}=\frac{1}{2(a+b+c)}$$
Similarly,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,237 |
105. Let $a, b, c$ be positive numbers, prove the inequality: $\frac{a^{5}}{b c}+\frac{b^{5}}{c a}+\frac{c^{5}}{a b}+\frac{3}{2 a^{2} b^{2} c^{2}} \geqslant 2\left(a^{3}+b^{3}+c^{3}\right)+\frac{9}{2}-6 a b c$, and determine when equality holds (2008 Bulgarian Mathematical Olympiad problem) | 105. By the AM-GM inequality, $\frac{a^{5}}{b c}+\frac{b^{5}}{c a}+\frac{c^{5}}{a b}=\frac{a^{6}+b^{6}+c^{6}}{a b c} \geqslant \frac{\left(a^{3}+b^{3}+c^{3}\right)^{2}}{3 a b c}$, $\frac{\left(a^{3}+b^{3}+c^{3}\right)^{2}}{3 a b c}+3 a b c \geqslant 2\left(a^{3}+b^{3}+c^{3}\right), \frac{3}{2}\left(a b c+a b c+\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,239 |
Example 3 Let $0 \leqslant a, b, c \leqslant 1$, prove that: $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+$
(a) $(1-b)(1-c) \leqslant 1$. (9th US Mathematical Olympiad) | To prove that if we directly find a common denominator, the problem will become very complicated. Without loss of generality, we can assume \(0 \leqslant u \leqslant b \leqslant c \leqslant 1\), thus we have
$$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1} \leqslant \frac{a+b+c}{a+b+1}$$
Therefore, we can try to prov... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,240 |
3. Let $f(x)=1+\log _{x} 5, g(x)=\log _{x}^{2} 9+\log _{x}^{3} 8$, compare the size of $f(x)$ and $g(x)$.
(1985 Shanghai Mathematics Competition Problem) | 3. Since $f(x)=1+\log _{x} 5, g(x)=\log _{x}^{2} 9+\log _{x}^{3} 8=\log _{x} 3+\log _{x} 2=\log _{x} 6$, therefore $f(x)-g(x)=1-\log _{x} \frac{6}{5}$, thus, when $x=\frac{6}{5}$, $f(x)=g(x)$; when $1<x<\frac{6}{5}$, $f(x)>g(x)$; when $x>\frac{6}{5}$, $f(x)<g(x)$. | when\ x=\frac{6}{5},\ f(x)=g(x);\ when\ 1<x<\frac{6}{5},\ f(x)>g(x);\ when\ x>\frac{6}{5},\ f(x)<g(x) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,241 |
106. Let $a, b, c$ be positive real numbers, prove: $\frac{a^{2}}{b}+\frac{b^{3}}{c^{2}}+\frac{c^{4}}{a^{3}} \geqslant -a + 2b + 2c$. (2005 Ukrainian Mathematical Olympiad Problem) | 106. By the AM-GM inequality, $\frac{a^{2}}{b}+a+b \geqslant 3 a, \frac{b^{3}}{c^{2}}+c+c \geqslant 3 b, \frac{c^{4}}{a^{3}}+a+a+a \geqslant 4 c$, adding them up yields $\frac{a^{2}}{b}+\frac{b^{3}}{c^{2}}+\frac{c^{4}}{a^{3}} \geqslant -a+2 b+2 c$. | \frac{a^{2}}{b}+\frac{b^{3}}{c^{2}}+\frac{c^{4}}{a^{3}} \geqslant -a+2 b+2 c | Inequalities | proof | Yes | Yes | inequalities | false | 732,242 |
107. Let $a, b, c$ be positive numbers, and $a+b+c=1$, prove:
$$8\left(\frac{1}{2}-a b-b c-c a\right)\left(\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right) \geqslant 9$$
(2006 Macedonian Mathematical Olympiad Problem) | 107. Since
$$\begin{array}{l}
8\left(\frac{1}{2}-a b-b c-c a\right)=8\left[\frac{(a+b+c)^{2}}{2}-a b-b c-c a\right]= \\
4\left(a^{2}+b^{2}+c^{2}\right) \geqslant(a+b)^{2}+(b+c)^{2}+(c+a)^{2}
\end{array}$$
Therefore
$$\begin{array}{l}
\left.8\left(\frac{1}{2}-a b-b c-c a\right)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,243 |
108. Given that the heights of $\triangle ABC$ on sides $BC, CA, AB$ are $h_{1}, h_{2}, h_{3}$ respectively, and a point $M$ inside $\triangle ABC$ has distances to sides $BC, CA, AB$ as $u, v, w$ respectively, prove:
(1) $\frac{h_{1}}{u}+\frac{h_{2}}{v}+\frac{h_{3}}{w} \geqslant 9$;
(2) $h_{1} h_{2} h_{3} \geqslant 27... | 108. From the area relationship, we have $\frac{u}{h_{1}}+\frac{v}{h_{2}}+\frac{w}{h_{3}}=1$,
(1) By the AM-GM inequality, we get $\frac{h_{1}}{u}+\frac{h_{2}}{v}+\frac{h_{3}}{w}=\left(\frac{u}{h_{1}}+\frac{v}{h_{2}}+\frac{w}{h_{3}}\right)\left(\frac{h_{1}}{u}+\frac{h_{2}}{v}+\frac{h_{3}}{w}\right) \geqslant 9$;
(2) By... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,244 |
109. Find all real numbers $x, y, z, w$ such that
$$x+y+z=\frac{3}{2}, \sqrt{4 x-1}+\sqrt{4 y-1}+\sqrt{4 z-1} \geqslant 2+3 \sqrt{w-2}$$
(2007-2008 Hungarian Mathematical Olympiad) | 109. By the mean inequality,
$$\frac{\sqrt{4 x-1}+\sqrt{4 y-1}+\sqrt{4 z-1}}{3} \leqslant \sqrt{\frac{4 x-1+4 y-1+4 z-1}{3}}=1$$
Equality holds if and only if $x=y=z=\frac{1}{2}$. Therefore, $2+3 \sqrt{w-2} \leqslant \sqrt{4 x-1}+$ $\sqrt{4 y-1}+\sqrt{4 z-1} \leqslant 3$, hence, $\sqrt{w-2} \leqslant 0$. Thus, $w=2$, ... | x=y=z=\frac{1}{2}, w=2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,245 |
110. Let $a, b, c$ be real numbers greater than 1. Prove:
$$\log _{a} b c+\log _{b} c a+\log _{c} a b \geqslant 4\left(\log _{a b} c+\log _{b c} a+\log _{c a} b\right)$$
(2008 Croatian Mathematical Olympiad, 2010 Malaysian Mathematical Olympiad) | 110. Let $x=\lg a, y=\lg b, z=\lg c$, since $a, b, c$ are real numbers greater than 1, so $x, y, z >$ 0.
$$\begin{array}{l}
\log _{a} b c+\log _{b} c a+\log _{c} a b \geqslant 4\left(\log _{a b} c+\log _{b c} a+\log _{c a} b\right) \Leftrightarrow \\
\frac{x}{y}+\frac{x}{z}+\frac{y}{z}+\frac{y}{x}+\frac{z}{x}+\frac{z}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,246 |
114. Given that $x, y, z$ are positive numbers, and $x y z=1$, prove:
$$\frac{(x+y-1)^{2}}{z}+\frac{(y+z-1)^{2}}{x}+\frac{(z+x-1)^{2}}{y} \geqslant x+y+z$$ | 114. By the AM-GM inequality, we have $\frac{(x+y-1)^{2}}{z}+z \geqslant 2(x+y-1), \frac{(y+z-1)^{2}}{x}+x \geqslant$ $2(y+z-1), \frac{(z+x-1)^{2}}{y}+y \geqslant 2(z+x-1), x+y+z \geqslant 3 \sqrt[3]{x y z}=3$, i.e., $(x+$ $y+z)-3 \geqslant 0$.
Adding the first three inequalities, we get
$$\begin{array}{l}
\frac{(x+y-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,250 |
115. Given that $a, b, c$ are positive numbers, and $a+b+c=1$, prove: $\sqrt[3]{a+b}+\sqrt[3]{b+c}+$ $\sqrt[3]{c+a} \leqslant \sqrt[3]{18} .(2010$ Albanian Mathematical Olympiad Problem) | 115. By the AM-GM inequality, we have $(a+b)+\frac{2}{3}+\frac{2}{3} \geqslant 3 \sqrt[3]{\frac{4}{9}(a+b)}$, similarly, $(b+c)+$ $\frac{2}{3}+\frac{2}{3} \geqslant 3 \sqrt[3]{\frac{4}{9}(b+c)},(c+a)+\frac{2}{3}+\frac{2}{3} \geqslant 3 \sqrt[3]{\frac{4}{9}(c+a)}$, adding these three inequalities yields $3 \sqrt[3]{\fra... | \sqrt[3]{a+b}+\sqrt[3]{b+c}+\sqrt[3]{c+a} \leqslant \sqrt[3]{18} | Inequalities | proof | Yes | Yes | inequalities | false | 732,251 |
4. In $\triangle A B C$, prove: $a^{2}+b^{2}+c^{2} \geqslant 4 \sqrt{3} S$, where $S$ represents the area of $\triangle A B C$. (3rd IMO problem) | $$\begin{array}{l}
\text { 4. }\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) \\
\left(a^{2}+b^{2}+c^{2}\right)^{2}-3\left[2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)\right]= \\
{\left[\left(a^{4}+b^{4}+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,252 |
117. Given that $a, b$ are positive numbers, prove: $\sqrt{2}\left(\sqrt{a(a+b)^{3}}+b \sqrt{a^{2}+b^{2}}\right) \leqslant 3\left(a^{2}+b^{2}\right)$ (2004 Irish Mathematical Olympiad) | 117. By the mean inequality
$$\begin{array}{c}
\sqrt{2 a(a+b)^{3}}=\sqrt{2 a(a+b)(a+b)^{2}} \leqslant \frac{2 a(a+b)+(a+b)^{2}}{2} \\
\sqrt{2} b \sqrt{a^{2}+b^{2}}=\sqrt{2 b^{2}\left(a^{2}+b^{2}\right)} \leqslant \frac{2 b^{2}+\left(a^{2}+b^{2}\right)}{2}
\end{array}$$
Adding the two inequalities yields
$$\sqrt{2}\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,254 |
118. Given real numbers $a, b, c$ satisfy $a+b+c=0, a^{2}+b^{2}+c^{2}=1$, prove: $a^{2} b^{2} c^{2} \leqslant \frac{1}{54}$. And determine the condition for equality. (2009 Irish Mathematical Olympiad problem) | 118. Given real numbers $a, b$ satisfy $a+b+c=0, a^{2}+b^{2}+c^{2}=1$, prove: $a^{2} b^{2} c^{2} \leqslant \frac{1}{54}$. And determine the condition for equality. (2009 Irish Mathematical Olympiad problem) | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,255 |
119. Given positive real numbers $a, b, c$ satisfying $a+b+c \leqslant 3$, prove:
(1) $\frac{3}{2} \leqslant \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}<3$;
(2) $\frac{a+1}{a(a+2)}+\frac{b+1}{b(b+2)}+\frac{c+1}{c(c+2)} \geqslant 2$.
(2009 Fujian Province Mathematics Competition Problem) | 119. (1) Since $a, b, c$ are positive real numbers, we have $\frac{1}{a+1}<1, \frac{1}{b+1}<1, \frac{1}{c+1}<1$, so
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}<3$$
By the AM-GM inequality, we get
$$[(a+1)+(b+1)+(c+1)]\left[\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right] \geqslant 9$$
Since $a+b+c \leqslant 3$, we ha... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,256 |
120. Given positive real numbers $a, b, c$ satisfying $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant 16(a+b+c)$, prove:
$$\begin{array}{l}
\frac{1}{[a+b+\sqrt{2(a+c)}]^{3}}+\frac{1}{[b+c+\sqrt{2(b+a)}]^{3}}+ \\
\frac{1}{[c+a+\sqrt{2(c+b)}]^{3}} \leqslant \frac{8}{9}
\end{array}$$
(2010 Vietnam National Training Team Pr... | 120. By the AM-GM inequality, we have
$$\begin{aligned}
a+b+\sqrt{2(a+c)}= & a+b+\frac{\sqrt{2(a+c)}}{2}+\frac{\sqrt{2(a+c)}}{2} \geqslant \\
& 3 \sqrt[3]{(a+b) \frac{\sqrt{2(a+c)}}{2} \cdot \frac{\sqrt{2(a+c)}}{2}}
\end{aligned}$$
Thus,
$$[a+b+\sqrt{2(a+c)}]^{3} \geqslant \frac{27}{2}(a+b)(a+c)$$
Similarly,
$$\begin... | \frac{8}{9} | Inequalities | proof | Yes | Yes | inequalities | false | 732,257 |
121. Given positive real numbers $a, b, c$ satisfying $\frac{1}{a}+\frac{\Gamma}{b}+\frac{1}{c}=a+b+c$, prove:
$$\frac{1}{(2 a+b+c)^{2}}+\frac{1}{(a+2 b+c)^{2}}+\frac{1}{(2 a+b+2 c)^{2}} \leqslant \frac{3}{16}$$ | 121. By the mean inequality $(2 a+b+c)^{2}=[(a+b)+(a+c)]^{2} \geqslant 4(a+b)(a+c)$, hence
$$\begin{array}{l}
\frac{1}{(2 a+b+c)^{2}}+\frac{1}{(a+2 b+c)^{2}}+\frac{1}{(2 a+b+2 c)^{2}} \leqslant \\
\frac{1}{4(a+b)(a+c)}+\frac{1}{4(a+b)(b+c)}+\frac{1}{4(a+c)(b+c)}= \\
\frac{a+b+c}{2(a+b)(b+c)(c+a)}
\end{array}$$
Since $... | \frac{3}{16} | Inequalities | proof | Yes | Yes | inequalities | false | 732,258 |
122. Let $a, b, c$ be the lengths of the sides of a triangle, and let
$$\begin{array}{c}
A=\frac{a^{2}+b c}{b+c}+\frac{b^{2}+c a}{c+a}+\frac{c^{2}+a b}{a+b} \\
B=\frac{1}{\sqrt{(a+b-c)(b+c-a)}}+\frac{1}{\sqrt{(b+c-a)(c+a-b)}}+ \\
\frac{1}{\sqrt{(c+a-b)(a+b-c)}}
\end{array}$$
Prove: $A B \geqslant 9$ : (2009 China Math... | 122. Since
$$\begin{array}{l}
B=\frac{1}{\sqrt{(a+b-c)(b+c-a)}}+\frac{1}{\sqrt{(b+c-a)(c+a-b)}}+ \\
\frac{1}{\sqrt{(c+a-b)(a+b-c)}}=\frac{1}{\sqrt{b^{2}-(c-a)^{2}}}+ \\
\frac{1}{\sqrt{c^{2}-(a-b)^{2}}}+\frac{1}{\sqrt{a^{2}-(b-c)^{2}}} \geqslant \\
\frac{1}{b}+\frac{1}{c}+\frac{1}{a} \\
A-(a+b+c)=\frac{a^{2}+b c}{b+c}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,259 |
123. Given that $x, y, z$ are non-negative real numbers, and $x+y+z=2$, prove: $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+$
$xyz \leqslant 1$, and find the values of $x, y, z$ when the equality holds. (2009 Greek Mathematical Olympiad) | 123. Notice that
$$\begin{array}{l}
x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+x y z=\frac{1}{2}\left(2 x^{2} y^{2}+2 y^{2} z^{2}+2 z^{2} x^{2}+2 x y z\right)= \\
\frac{1}{2}(x y \cdot 2 x y+y z \cdot 2 y z+z x \cdot 2 z x+2 x y z) \leqslant \\
\frac{1}{2}\left[x y\left(x^{2}+y^{2}\right)+y z\left(y^{2}+z^{2}\right)+z x\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,260 |
124. For any three distinct positive real numbers $a, b, c$, prove:
$$\frac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}>8 a b c$$
(2009 Estonian National Team Selection Exam Problem) | 124. By factorization we get
$$\begin{aligned}
(a-b)^{3}+(b-c)^{3}+(c-a)^{3} & =3(a-b)(b-c)(c-a) \\
\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3} & =3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)
\end{aligned}$$
Therefore,
$$\frac{\left(a^{2}-b^{2}\ri... | (a+b)(b+c)(c+a)>8abc | Inequalities | proof | Yes | Yes | inequalities | false | 732,261 |
125. Let \(a, b, c\) be positive numbers, prove:
\[
\frac{a^{2}+2 b c}{(a+2 b)^{2}+(a+2 c)^{2}}+\frac{b^{2}+2 c a}{(b+2 c)^{2}+(b+2 a)^{2}}+\frac{c^{2}+2 a b}{(c+2 a)^{2}+(c+2 b)^{2}} \leqslant \frac{1}{2}
\]
(2004 USA Math Summer Camp Problem) | 125. Since $2\left[(a+2 b)^{2}+(a+2 c)^{2}\right] \geqslant[(a+2 b)+(a+2 c)]^{2}=4(a+$ $b+c)^{2}$, we have $(a+2 b)^{2}+(a+2 c)^{2} \geqslant 2(a+b+c)^{2}$. Similarly,
$$\begin{aligned}
(b+2 c)^{2}+ & (b+2 a)^{2} \geqslant 2(a+b+c)^{2},(c+2 a)^{2}+(c+2 b)^{2} \geqslant 2(a+b+c)^{2} \\
& \frac{a^{2}+2 b c}{(a+2 b)^{2}+(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,262 |
127. Let positive real numbers $a, b, c$ satisfy $(a+2 b)(b+2 c)=9$. Prove:
$$\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant 3$$
(2010 Northern China Mathematical Olympiad) | 127. Since $\sqrt{\frac{a^{2}+b^{2}}{2}} \geqslant \frac{a+b}{2} \cdot \sqrt{\frac{b^{3}+c^{3}}{2}} \geqslant \frac{b+c}{2}$, therefore
$$\begin{array}{c}
\sqrt{\frac{a^{2}+b^{2}}{2}}+2 \sqrt[3]{\frac{b^{3}+c^{3}}{2}} \geqslant \frac{a+b}{2}+(b+c)=\frac{(a+2 b)+(b+2 c)}{2} \geqslant \\
\sqrt{(a+2 b)(b+2 c)}=3
\end{arra... | 3 | Inequalities | proof | Yes | Yes | inequalities | false | 732,265 |
129. Let positive real numbers $x, y, z$ satisfy $xyz=1$. Prove that:
$$\frac{2}{(x+1)^{2}+2 y^{2}+1}+\frac{2}{(y+1)^{2}+2 z^{2}+1}+\frac{2}{(z+1)^{2}+2 x^{2}+1} \leqslant 1$$
(2010 Poland Czechoslovakia Joint Mathematical Olympiad) | 129. Let $x=\frac{a}{b}, y=\frac{c}{a}, z=\frac{b}{c}$, then by the mean value inequality we get
$$\begin{aligned}
(x+1)^{2}+2 y^{2}+1= & x^{2}+y^{2}+2 x+2 \geqslant 2 x y+2 x+2= \\
& \frac{2 c}{b}+\frac{2 a}{b}+2=\frac{2(a+b+c)}{b}
\end{aligned}$$
So
$$\begin{array}{l}
\frac{2}{(x+1)^{2}+2 y^{2}+1} \leqslant \frac{b}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,267 |
132. Let $a, b, c$ be real numbers, $A=\frac{a+b+c}{-3}, G=\sqrt[3]{a b c}, H=\frac{3}{1}+\frac{1}{a}+\frac{1}{b}$, prove: $\left(\frac{A}{G}\right)^{3} \geqslant \frac{1}{4}+\frac{3}{4} \cdot \frac{A}{H} \cdot(1992$ IMO Preliminary Problem $)$ | 132. $\left(\frac{A}{G}\right)^{3} \geqslant \frac{1}{4}+\frac{3}{4} \cdot \frac{4}{H} \Leftrightarrow(a+b+c)^{3} \geqslant$
$$\begin{aligned}
27 a b c\left(\frac{1}{4}+\frac{3}{4} \cdot \frac{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}{9}\right) \Leftrightarrow \\
4(a+b+c)^{3} \geqslant 27 a b c+9(a+b+c)(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,270 |
133. 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) | 133. Considering the inequality holds if and only if $x=y=z=\frac{1}{3}$, we have $\frac{x}{y^{2}+z}+\frac{81}{16} x\left(y^{2}+z\right) \geqslant \frac{9}{2} x$, which implies $\frac{x}{y^{2}+z} \geqslant \frac{9}{2} x-\frac{81}{16} x\left(y^{2}+z\right)$. Similarly, $\frac{y}{z^{2}+x} \geqslant \frac{9}{2} y-\frac{81... | \frac{9}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,271 |
134. Let $x, y$ be positive real numbers, prove: $x^{4}+y^{4}+\left(x^{2}+1\right)\left(y^{2}+1\right) \geqslant x^{3}(1+y)+$ $y^{3}(1+x)+x+y .(2000$ Czech and Slovak Mathematical Olympiad Problem) | 134. By the AM-GM inequality, we have $a^{4}+a^{2} b^{2} \geqslant 2 a^{3} b$. Taking $(a, b)=(x, y),(y, x)$, $(x, 1),(1, y),(1, x),(y, 1),(x, y)$, we get $x^{4}+x^{2} y^{2} \geqslant 2 x^{3} y, y^{4}+x^{2} y^{2} \geqslant 2 x y^{3}$, $x^{4}+x^{2} \geqslant 2 x^{3}, 1+x^{2} \geqslant 2 x, y^{4}+y^{2} \geqslant 2 y^{3},... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,272 |
135. Let $a, b, c, d$ be positive real numbers, prove: $\sqrt{a^{4}+c^{4}}+\sqrt{a^{4}+d^{4}}+\sqrt{b^{4} \pm c^{4}}+$ $\sqrt{b^{4}+d^{4}} \geqslant 2 \sqrt{2}(a d+b c)$. (2000 Turkish Mathematical Olympiad Problem) | 135. By the mean inequality,
$$\begin{array}{l}
\sqrt{a^{4}+c^{4}}+\sqrt{a^{4}+d^{4}}+\sqrt{b^{4}+c^{4}}+\sqrt{b^{4}+d^{4}} \geqslant \\
\sqrt{2}\left[\left(a^{2}+c^{2}\right)+\left(a^{2}+d^{2}\right)+\left(b^{2}+c^{2}\right)+\left(b^{2}+d^{2}\right)\right]= \\
2 \sqrt{2}\left[\left(a^{2}+d^{2}\right)+\left(b^{2}+c^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,273 |
6. Prove: For any real number $t$ the inequality $t^{4}-t+\frac{1}{2}>0$. (1995 Russian Mathematical Olympiad) | 6. Since $t^{4}-t+\frac{1}{2}=\left(t^{2}-\frac{1}{2}\right)^{2}+\left(t-\frac{1}{2}\right)^{2} \geqslant 0$. Because the equality cannot hold simultaneously, the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,274 |
137. Let $a, b, c$ be positive real numbers, prove: $\frac{a}{b c}+\frac{b^{5}}{c a}+\frac{c^{5}}{a b}+\frac{3}{2 a^{2} b^{2} c^{2}} \geqslant 2\left(a^{3}+b^{3}+c^{3}\right)+$ $\frac{9}{2}-6 a b c$. (2008 Bulgarian Mathematical Olympiad) | 137. By the AM-GM inequality, we have $\frac{a^{5}}{b c}+\frac{b^{5}}{c a}+\frac{c^{5}}{a b}=\frac{a^{6}+b^{6}+c^{6}}{a b c} \geqslant \frac{\left(a^{3}+b^{3}+c^{3}\right)^{2}}{3 a b c}$, $\frac{\left(a^{3}+b^{3}+c^{3}\right)^{2}}{3 a b c}+3 a b c \geqslant 2\left(a^{3}+b^{3}+c^{3}\right), \frac{3 a b c}{2}+\frac{3 a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,276 |
138. Let $a, b, c$ be positive real numbers, and $abc=1$, prove: $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a+b+c .(2003$ Czech and Slovak Mathematical Olympiad Problem) | 138. Since $a b c=1$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant a+b+c$ is equivalent to
$$\begin{array}{l}
a^{2} c+b^{2} a+c^{2} b \geqslant a+b+c \Leftrightarrow a^{2} c+b^{2} a+c^{2} b \geqslant \\
\left(a^{5} b^{2} c^{2}\right)^{\frac{3}{2}}+\left(a^{2} b^{5} c^{2}\right)^{\frac{3}{2}}+\left(a^{2} b^{2} c^{5}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,277 |
139-Let $x, y, z$ be positive numbers. And $x+y+z=3$, prove: $\frac{x^{2}}{x+y^{2}}+\frac{y^{2}}{y+z^{2}}+\frac{z^{2}}{z+x^{2}} \geqslant \frac{3}{2}$. (2010 Croatian Mathematical Olympiad problem) | 139. $\frac{x^{2}}{x+y^{2}}=\frac{x\left(x+y^{2}\right)-x y^{2}}{x+y^{2}}=x-\frac{x y^{2}}{x+y^{2}} \geqslant x-\frac{x y^{2}}{2 \sqrt{x y^{2}}}=x-\frac{y \sqrt{x}}{2}$
Similarly, we get
$$\begin{array}{l}
\frac{y^{2}}{y+z^{2}} \geqslant y-\frac{z \sqrt{y}}{2} \\
\frac{z^{2}}{z+x^{2}} \geqslant z-\frac{x \sqrt{z}}{2}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,278 |
Example 1 Let the length, width, and height of a rectangular prism be $a, b, c$, and its diagonal length be $l$. Try to prove: $l^{4}=$ a.) $\left(l^{4}-b^{4}\right)\left(l^{4}-c^{4}\right) \geqslant 512 a^{4} b^{4} c^{4}$. (2002 Hunan Mathematical Olympiad Problem) | Prove that the original inequality is equivalent to
$$\left(\frac{l^{4}}{a^{4}}-1\right)\left(\frac{l^{4}}{b^{4}}-1\right)\left(\frac{l^{4}}{c^{4}}-1\right) \geqslant 512$$
Let \( x=\frac{a^{2}}{l^{2}}, y=\frac{b^{2}}{l^{2}}, z=\frac{c^{2}}{l^{2}} \), then \( x+y+z=1 \), and the original inequality can be rewritten as... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,280 |
For example, $5 x, y, z$ are positive real numbers, and satisfy $x+y+z=1$. Find the minimum value of $\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+$ $\frac{z^{4}}{x\left(1-x^{2}\right)}$. (Mathematics Bulletin Problem) | Solving: According to symmetry, we estimate that equality holds when $x=y=z$. Applying the AM-GM inequality, we get
$$\begin{array}{l}
\frac{x^{4}}{y(1-y)(1+y)}+\frac{y}{8}+\frac{1-y}{16}+\frac{1+y}{32} \geqslant \\
4 \sqrt[4]{\frac{x^{4}}{y(1-y)(1+y)} \cdot \frac{y}{8} \cdot \frac{1-y}{16} \cdot \frac{1+y}{32}}=\frac{... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,284 |
7. Let $a, b, c$ be real numbers, prove: $\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \geqslant(a b+b c+c a-1)^{2}$.
(2007 Indonesia Training Team Problem) | 7. Because
$$\begin{array}{l}
\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)-(a b+b c+c a \\
a^{2} b^{2} c^{2}+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+a^{2}+b^{2}+c^{2}+1 \\
{\left[a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a b c(a+b+c)=2(a b+b c+c a)+1\right]=} \\
a^{2} b^{2} c^{2}+a^{2}+b^{2}+c^{2}+2(a b+b c+c a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,285 |
Example 7 Evaluate: For all positive real 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) | Prove that
\[
\left(a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}\right)^{2}-\left(a^{\frac{4}{3}}\right)^{2}=\left(b^{\frac{4}{3}}+c^{\frac{4}{3}}\right)\left(a^{\frac{4}{3}}+a^{\frac{4}{3}}+b^{\frac{1}{3}}+c^{\frac{4}{3}}\right) \geqslant
\]
\[
2 b^{\frac{2}{3}} c^{\frac{1}{3}} \cdot a^{\frac{2}{3}} b^{\frac{1}{3}}... | proof | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,287 |
Example 8 Let $x, y, z, w$ be real numbers, not all zero, find the maximum value of $\frac{x y+2 y z+z w}{x^{2}+y^{2}+z^{2}+w^{2}}$. (1985 Austrian-Polish Joint Competition Problem) | Consider only the case where $x \geqslant 0, y \geqslant 0, z \geqslant 0, w \geqslant 0$.
Introduce positive constants $\alpha, \beta, \gamma$, then we have $\alpha^{2} x^{2}+y^{2} \geqslant 2 \alpha x y, \beta^{2} y^{2}+z^{2} \geqslant 2 \beta y z, \gamma^{2} z^{2}+$ $w^{2} \geqslant 2 \gamma z w$, which means $\frac... | \frac{\sqrt{2}+1}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,288 |
Example 10 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and satisfy $a_{1}+a_{2}+\cdots+a_{n}<1$, prove that
$$\frac{a_{1} a_{2} \cdots a_{n}\left[1-\left(a_{1}+a_{2}+\cdots+a_{n}\right)\right]}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)} \leq... | Proof: Let $a_{n+1}=1-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$, obviously $a_{n+1}>0$, thus we obtain $n+1$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{n+1}$ whose sum is 1, and the inequality becomes
$$n^{n+1} a_{1} a_{2} \cdots a_{n} a_{n+1} \leqslant\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,290 |
Example 13 Let positive real numbers $a, b, c$ satisfy $abc=1$, prove that for integers $k \geqslant 2$, we have $\frac{a^{k}}{a+b}+$ $\frac{b^{k}}{b+c}+\frac{c^{k}}{c+a} \geqslant \frac{3}{2}$. (2007 China Southeast Mathematical Olympiad) | Prove that because
$$\frac{a^{k}}{a+b}+\frac{a+b}{4}+\frac{1}{2}+\frac{1}{2}+\cdots \frac{1}{2} \geqslant k \sqrt[k]{\frac{a^{k}}{2^{k}}}=\frac{k a}{2}$$
Therefore,
$$\frac{a^{k}}{a+b} \geqslant \frac{k a}{2}-\frac{a+b}{4}-\frac{k-2}{2}$$
Similarly, we can get
$$\frac{b^{b}}{b+c} \geqslant \frac{k b}{2}-\frac{b+c}{4}... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,293 |
1. Let $a, b, c$ be positive real numbers, then $\frac{(a+1)^{3}}{b}+\frac{(b+1)^{3}}{c}+\frac{(c+1)^{3}}{a} \geqslant \frac{81}{4}$. | 1. According to the mean inequality, we have $a+\frac{1}{2}+\frac{1}{2} \geqslant 3, \sqrt{a \cdot \frac{1}{2}} \cdot \frac{1}{2}=3 \sqrt{\frac{a}{4}}$, so $\frac{(a+1)^{3}}{b} \geqslant \frac{27 a}{b}$, similarly, $\frac{(b+1)^{3}}{c} \geqslant \frac{27 b}{c}, \quad \frac{(c+1)^{3}}{a} \geqslant \frac{27 c}{a}$. Addin... | \frac{81}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,294 |
8. Given natural numbers $a, b, n > 1$; $A_{n-1}$ and $A_{n}$ are $a$-ary numbers, $B_{n-1}$ and $B_{n}$ are $b$-ary numbers, $A_{n-1}, A_{n}, B_{n-1}, B_{n}$ are defined as
$A_{n}=x_{n} x_{n-1} \cdots x_{0}, A_{n-1}=x_{n-1} x_{n-2} \cdots x_{0}$ (written in $a$-ary) ; $B_{n}=x_{n} x_{n-1} \cdots x_{0}, B_{n-1}=x_{n-1}... | $$\begin{aligned}
& \text { 8. Since } A_{n}>0, B_{n}>0 \text {, it suffices to prove } A_{n} B_{n-1}-A_{n-1} B_{n}>0 . \\
& \text { And } A_{n}=x_{n} a^{n}+x_{n-1} a^{n-1}+\cdots+x_{0}, A_{n-1}=x_{n-1} a^{n-1}+x_{n-2} a^{n-2}+\cdots+x_{0} \text {, hence } \\
A_{n}= & x_{n} a^{n}+A_{n-1} .
\end{aligned}$$
Similarly, $... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 732,296 |
3. $x, y, z$ are positive real numbers, and satisfy $x^{4}+y^{4}+z^{4}=1$, find the minimum value of $\frac{x^{3}}{1-x^{8}}+\frac{y^{3}}{1-y^{8}}+\frac{z^{3}}{1-z^{8}}$. (2000 Jiangsu Province Mathematical Olympiad Problem) | 3. Applying the mean value theorem, we get
$-u^{8}\left(1-u^{8}\right)^{8}=\frac{1}{8} \cdot 8 u^{8}\left(1-u^{8}\right)\left(1-u^{8}\right) \cdots\left(1-u^{8}\right) \leqslant$
$$\frac{1}{8} \cdot\left(\frac{8 u^{8}+\left(1-u^{8}\right)+\cdots+\left(1-u^{8}\right)}{9}\right)^{8}=\frac{1}{8}\left(\frac{8}{9}\right)^{9... | \frac{9}{8} \cdot \sqrt[4]{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,297 |
(2) Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$, prove:
$$\prod_{k=1}^{n}\left(a_{k}+\frac{1}{a_{i}}\right) \geqslant\left(n+\frac{1}{n}\right)^{n}$$ | (2) Applying the AM-GM inequality, we get
So
$$1=\sum_{i \neq 1}^{n} a_{i} \geqslant n \cdot \sqrt[n]{\prod_{i=1}^{n} a_{i}}$$
$$\begin{array}{l}
\frac{1}{a_{1} a_{2} \cdots a_{n}} \geqslant n^{n} \\
a_{i}+\frac{1}{a_{i}}=a_{i}+\frac{1}{n^{2} a_{i}}+\cdots+\frac{1}{n^{2} a_{i}} \geqslant\left(n^{2}+1\right)\left(a_{i}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,300 |
7. If $x_{i}(i=1,2, \cdots, n)$ are positive numbers, and $\sum_{i=1}^{n} \frac{1}{1+x_{i}}=1$, prove that: $x_{1} x_{2} \cdots x_{n} \geqslant(n-1)^{n}$. (1992 Turkish National Training Team Problem) | 7. Let $1+x_{i}=a_{i}(i=1,2, \cdots, n) \Rightarrow \sum_{i=1} a_{i}=1$, for $i=1,2, \cdots, n$, we have
$$\begin{array}{l}
x_{i}=\frac{1-a_{i}}{a_{i}}=\frac{a_{1}+\cdots+a_{i-1}+a_{i+1}+\cdots+a_{n}}{-a_{i}} \geqslant \\
\quad \frac{(n-1) \cdot \sqrt[n-1]{a_{1} \cdots a_{i-1} a_{i+1} \cdots a_{n}}}{a_{i}}= \\
\frac{(n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,302 |
9. The sequence $\left.F_{n}\right\}$ is defined as follows: $F_{1}=1, F_{2}=2$, and for any $n \in \mathbf{N}$, $F_{n+2} \geqslant \boldsymbol{F}_{n+1}+F_{n}$. Prove that for any $n \in \mathbf{N}$, $\sqrt[n]{F_{n+1}} \geqslant 1+\frac{1}{\sqrt[n]{F_{n}}}$. (1992 St. Petersburg Mathematical Selection Exam) | 9. Let $F_{\theta}=F_{2}-F_{1}=1$, then $F_{k+1}=F_{k}+F_{k-1}, k=1,2, \cdots \cdots$, $n$, i.e., $1=\frac{F_{k}}{F_{k+1}}+$ $\frac{F_{k-1}}{F_{k+1}}, \vec{k}=1,2, \cdots, n$, thus $n=\sum_{k=1}^{n} \frac{F_{k}}{F_{k+1}}+\sum_{k=1}^{n} \frac{F_{k-1}}{F_{k+1}}$, and by the AM-GM inequality we get $\mathbf{L}=\frac{1}{n}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,304 |
10. (1) Let $n \geqslant 2, n \in \mathbf{N}$, then
$$n[(n+1)-1]<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}<n-(n-1) n^{-\frac{1}{n-1}}$$
(36th Putnam Mathematical Competition Problem)
(2) Let $n \geqslant 2, n \in \mathbf{N}^{*}, k \in \mathbf{N}^{*}$, then
$$\frac{1}{k n}+\frac{1}{k n+1}+\frac{1}{k n+2}+\cdots+\fra... | 10. (1) Apply the arithmetic mean-geometric mean inequality:
$$2+\frac{3}{2}+\frac{4}{3}+\cdots+\frac{n+1}{n}>n \sqrt[n]{2 \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n+1}{n}}=n(n+1)^{\frac{1}{n}}$$
Thus,
$$(1+1)+\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\cdots+\left(1+\frac{1}{n}\right)>n(n+1)^{\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,305 |
12. Given $5 n$ real numbers $r_{i}, s_{i}, t_{i}, u_{i}, v_{i}$ all greater than $1(1 \leqslant i \leqslant n)$, let $R=\left(\frac{1}{n} \sum_{i=1}^{n} r_{i}\right)$, $S=\left(\frac{1}{n} \sum_{i=1}^{n} s_{i}\right), T=\left(\frac{1}{n} \sum_{i=1}^{n} t_{i}\right), U=\left(\frac{1}{n} \sum_{i=1}^{n} u_{i}\right), V=\... | 12. When $x_{i}>1$, by the AM-GM inequality we have
$$\begin{array}{l}
\sqrt[n]{\prod_{i=1}^{n} \frac{x_{i}}{x_{i}+1}} \leqslant \frac{1}{n} \sum_{i=1}^{n} \frac{x_{i}}{x_{i}+1} \\
\sqrt[n]{\prod_{i=1}^{n} \frac{1}{x_{i}+1}} \leqslant \frac{1}{n} \sum_{i=1}^{n} \frac{1}{x_{i}+1}
\end{array}$$
Adding (1) and (2) gives $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,308 |
13. $(1)$ Given that $a, b, c$ are positive numbers, then $\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geqslant \frac{1}{1+a b c}$ (Aassila Inequality, 2000 China National Training Team Problem, 2006 Balkan Mathematical Olympiad Problem) | 13. (1)
$$\begin{array}{l}
(1+a b c)\left(\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\right)+3= \\
\left(\frac{1+a b c}{a+a b}+1\right)+\left(\frac{1+a b c}{b+b c}+1\right)+\left(\frac{1+a b c}{c+c a}+1\right)= \\
\frac{(a b c+a b)+(1+a)}{a+a b}+\frac{(a b c+b c)+(1+b)}{b+b c}+\frac{(a b c+c a)+(1+c)}{c+c a}= \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,309 |
(2) Given that $a, b, c$ are positive numbers, then $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \geqslant$ $\frac{9}{1+a b c} \cdot($ Crux Problem 2522) | (2) Using the arithmetic-geometric mean inequality twice, we get
$$\begin{array}{l}
\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \geqslant 3 \sqrt[3]{\frac{1}{1+a} \cdot \frac{1}{1+b} \cdot \frac{1}{1+c}}= \\
\frac{3}{\sqrt[3]{a b c}} \cdot \frac{1}{\sqrt[3]{\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\fra... | \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \geqslant \frac{9}{1+a b c} | Inequalities | proof | Yes | Yes | inequalities | false | 732,310 |
14. Let $x, y, z$ be positive numbers, then $\frac{x^{3}}{y z}+\frac{y^{3}}{z x}+\frac{z^{3}}{x y} \geqslant x+y+z$. (34th Canadian Mathematical Olympiad problem) | 14. Since $\frac{x^{3}}{y z}+y+z \geqslant 3 x, \frac{y^{3}}{z x}+z+x \geqslant 3 y, \frac{z^{3}}{x y}+x+y \geqslant 3 z$, adding the three inequalities yields $\frac{x^{3}}{y z}+$
$$\frac{y^{3}}{z x}+\frac{z^{3}}{x y} \geqslant x+y+z$$ | \frac{x^{3}}{y z}+\frac{y^{3}}{z x}+\frac{z^{3}}{x y} \geqslant x+y+z | Inequalities | proof | Yes | Yes | inequalities | false | 732,311 |
15. Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be positive numbers. Prove:
$$\sqrt[n]{a_{1} a_{2} \cdots a_{n}}+\sqrt[n]{b_{1} b_{2} \cdots b_{n}} \leqslant \sqrt[n]{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) \cdots\left(a_{n}+b_{n}\right)}$$
(1992 Irish Mathematical Olympiad, 64th Putnam... | 15. By the AM-GM inequality, we have
$$\begin{array}{l}
\frac{\sqrt[n]{a_{1} a_{2} \cdots a_{n}}}{\sqrt[n]{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right) \cdots\left(a_{n}+b_{n}\right)}} \leqslant \frac{1}{n}\left[\frac{a_{1}}{a_{1}+b_{1}}+\frac{a_{2}}{a_{2}+b_{2}}+\cdots+\frac{a_{n}}{a_{n}+b_{n}}\right] \\
\frac{\sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,312 |
16. Let $n$ be a positive integer, prove the inequality: $n^{n} \leqslant(n!)^{2} \leqslant\left[\frac{(n+1)(n+2)}{6}\right]^{n}$.
$(1995$ | 16. First prove the right inequality, i.e., prove $\sqrt[n]{(n!)^{2}} \leqslant \frac{(n+1)(n+2)}{6}$.
By the AM-GM inequality, we have
$$\begin{aligned}
\sqrt[n]{(n!)^{2}}= & \sqrt[n]{(1 \cdot n)(2 \cdot(n-1)) \cdots(k(n-k+1)) \cdots(n \cdot 1)^{2}} \leqslant \\
& \frac{\sum_{k=1}^{n} k(n-k+1)}{n}=\frac{\frac{n(n+1)(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,313 |
17. Let $a_{1}, a_{2}, \cdots, a_{n}$ be a permutation of $1,2, \cdots, n$, prove: $\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n} \leqslant$ $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}} \leqslant \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}$. (31st IMO Shortlist) | 17. Since $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$, we have
$$\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n-1}\right) \geqslant(1+1)(1+2) \cdots(1+n-1)=a_{1} a_{2} \cdots a_{n}$$
Thus,
$$\begin{aligned}
& \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,314 |
18. (1) Given $0 \leqslant x \leqslant 1, n$ is a positive integer, prove the inequality: $(1+x)^{n} \geqslant(1-x)^{n}+$ $2 n x \sqrt{\left(1-x^{2}\right)^{n-1}}$. (1988 Irish Mathematical Olympiad Problem) | $\begin{array}{l}\text { 18. }(1)(1+x)^{n}-(1-x)^{n}=[(1+x)-(1-x)] \cdot\left[(1+x)^{n-1}+(1+\right. \\ \left.x)^{n-2}(1-x)+\cdots+(1+x)(1-x)^{n-2}+(1-x)^{n-1}\right]=2 x\left[(1+x)^{n-1}+(1+\right. \\ \left.x)^{n-2}(1-x)+\cdots+(1+x)(1-x)^{n-2}+(1-x)^{n-1}\right] \geqslant 2 x \cdot n \cdot \\ \sqrt[n]{(1+x)^{n-1}(1+x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,315 |
(2) Given $a>0$, prove: $a^{n}+\frac{1}{a^{n}}-2 \geqslant n^{2}\left(a+\frac{1}{a}-2\right)$. (1992 Hungary, 2001 Belarus Mathematical Olympiad) | (2) $a^{n}+\frac{1}{a^{n}}-2 \geqslant n^{2}\left(a+\frac{1}{a}-2\right) \Leftrightarrow\left(a^{n}-1\right)^{2} \geqslant n^{2} a^{n-1}(a-1)^{2}$.
When $a=1$, the equality holds.
When $a \neq 1$, the inequality $\Leftrightarrow$
$$\left(a^{n-1}+a^{n-2}+\cdots+a+1\right)^{2} \geqslant n^{2} a^{n-1} \Leftrightarrow a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,316 |
19. (1) Given $x, y \in \mathbf{R}^{+}$, and $x+y=2$, prove: $x^{3} y^{3}\left(x^{3}+y^{3}\right) \leqslant 2$. (2002 Indian Mathematical Olympiad Problem) | 19. (1) Since $x, y \in \mathbf{R}^{+}$, we have $2=x+y \geqslant 2 \sqrt{x y}$, which means $x y \leqslant 1$.
$$\begin{array}{l}
x^{3} y^{3}\left(x^{3}+y^{3}\right)=(x y)^{3}\left[(x+y)^{3}-3 x y(x+y)\right]= \\
(x y)^{3}(8-6 x y)=2(x y)^{3}(4-3 x y)= \\
2[(x y)(x y)(x y)(4-3 x y)-] \leqslant 2\left[\frac{x y+x y+x y... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 732,317 |
10. Proposition: Let $a, b, c$ be non-negative real numbers. If $a^{4}+b^{4}+c^{4} \leqslant 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)$, then $a^{2}+b^{2}+c^{2} \leqslant 2(a b+b c+c a)$.
(1) Prove that the above proposition is correct;
(2) Write the converse of the above proposition, and determine whether the ... | 10. (-1) Let
$$\begin{aligned}
D= & 2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)= \\
& 4 a^{2} b^{2}-\left(a^{2}+b^{2}\right)^{2}+2 c^{2}\left(a^{2}+b^{2}\right)-c^{4}= \\
& (2 a b)^{2}-\left(a^{2}+b^{2}-c^{2}\right)^{2}
\end{aligned}$$
Since $D \geqslant 0$, then $(2 a b)^{2} \geqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,318 |
(2) Given $x, y \in \mathbf{R}^{+}$, and $x+y=2$, prove: $x^{2} y^{2}\left(x^{2}+y^{2}\right) \leqslant 2$. (2002 Irish Mathematical Olympiad problem) | (2) Since $x, y \in \mathbf{R}^{+}$, we have $2=x+y \geqslant 2 \sqrt{x y}$, which means $x y \leqslant 1$. Therefore,
$$\begin{array}{l}
x^{2} y^{2}\left(x^{2}+y^{2}\right)=(x y)(x y)\left[(x+y)^{2}-2 x y\right]= \\
2 x y(x y)(2-x y) \leqslant 2(x y)(2-x y) \leqslant \\
2\left(\frac{(x y)+(2-x y)}{2}\right)^{2}=2
\end... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,319 |
(2) Given that $a, b, c, d$ are positive real numbers, and $a b c d=1$, prove: $\frac{1}{a(b+1)}+\frac{1}{b(c+1)}$ $\frac{1}{c(d+1)}+\frac{1}{d(a+1)} \geqslant 2 .$(2004 Romanian Mathematical Olympiad) | (2) Let $a=\frac{x}{w}, b=\frac{y}{x}, c=\frac{z}{y}, d=\frac{w}{z}$. The inequality is transformed to proving $\frac{w}{x+y}+\frac{x}{y+z}+\frac{y}{z+w}+$ $\frac{z}{w+z} \geqslant 2$. This is the same as problem (1). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,322 |
23. Let $x_{i}(i=0,1,2, \cdots, n)$ be positive real numbers, $n \geqslant 2, n$ a natural number. Prove:
$$\left(\frac{x_{0}}{x_{1}}\right)^{n}+\left(\frac{x_{1}}{x_{2}}\right)^{n}+\cdots+\left(\frac{x_{n-1}}{x_{n}}\right)^{n}+\left(\frac{x_{n}}{x_{0}}\right)^{n} \geqslant \frac{x_{1}}{x_{0}}+\frac{x_{2}}{x_{1}}+\cdot... | 23. Selecting $n$ numbers in sequence and applying the AM-GM inequality, we get
$$\begin{array}{l}
\left(\frac{x_{0}}{x_{1}}\right)^{n}+\left(\frac{x_{1}}{x_{2}}\right)^{n}+\cdots+\left(\frac{x_{n-1}}{x_{n}}\right)^{n} \geqslant n \cdot \frac{x_{0}}{x_{1}} \frac{x_{1}}{x_{2}} \cdots \frac{x_{n-2}}{x_{n-1}} \frac{x_{n-1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,323 |
24. Let $x_{i}(i=0,1,2, \cdots, n)$ be positive real numbers, $n \geqslant 2, n$ a natural number, $x_{n+1}=x_{1}, x_{n+2}=x_{2}$, prove: $\sum_{i=1}^{n} \frac{x_{i}}{x_{i+1}+x_{i+2}}>(\sqrt{2}-1) n .(2005$ Moldova Mathematical Olympiad problem) | 24.
$$\begin{aligned}
\frac{x_{i}}{x_{i+1}+x_{i+2}}= & \frac{x_{i}+\frac{1}{2} x_{i+1}}{x_{i+1}+x_{i+2}}+\frac{\frac{1}{2} x_{i+1}+x_{i+2}}{x_{i+1}+x_{i+2}}-1 \geqslant \\
& 2 \sqrt{\frac{x_{i}+\frac{1}{2} x_{i+1}}{x_{i+1}+x_{i+2}} \cdot \frac{\frac{1}{2} x_{i+1}+x_{i+2}}{x_{i+1}+x_{i+2}}-1=} \\
& 2 \sqrt{\left[\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,324 |
25. Given that $x, y, z$ are positive numbers, prove: $x y z(x+2)(y+2)(z+2) \leqslant \left(1+ \frac{2(x y+y z+z x)}{3}\right)^{3} \cdot(2006$ Korean Mathematical Olympiad problem) | 25. Let $a=x y+y z+z x$, then by the AM-GM inequality we have $a \geqslant 3 \sqrt[3]{x y \cdot y z \cdot z x}=3 \sqrt[3]{(x y z)^{2}}$, i.e., $x y z \leqslant \sqrt{\left(\frac{a}{3}\right)^{3}}$. Also, by the inequality $(A+B+C)^{2} \geqslant 3(A B+B C+C A)$, we get
$$\begin{array}{l}
(x y+y z+z x)^{2} \geqslant 3(x ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,325 |
26. (1) Given that $a, b$ are positive numbers, prove: the inequality $2 \sqrt{a}+3 \sqrt[3]{b} \geqslant 5 \sqrt[5]{a b}$. (1996 Macedonia Mathematical Olympiad) | 26. (1) If $a, b$ are positive numbers, we prove the inequality $m \sqrt[m]{a}+n \sqrt[n]{b} \geqslant(m+n) \sqrt[m+n]{a b}$. Consider the arithmetic mean of $m$ numbers $\sqrt[m]{a}$ and $n$ numbers $\sqrt[n]{b}$ is not less than their geometric mean. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,326 |
(2) Let $x>y \geqslant 0$, prove: $x+\frac{4}{(x-y)(y+1)^{2}} \geqslant 3$. (1981 Yugoslav Mathematical Olympiad Problem) | $\begin{array}{c}\text { (2) By the AM-GM inequality, we have }(x-y)+\frac{y+1}{2}+\frac{y+1}{2}+\frac{4}{(x-y)(y+1)^{2}} \geqslant 4 \text {, thus } \\ x+\frac{4}{(x-y)(y+1)^{2}} \geqslant 3\end{array}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,327 |
(3) Non-negative real numbers $a, b, x, y$ are positive, and satisfy $a^{5}+b^{5} \leqslant 1$ and $x^{5}+y^{5} \leqslant 1$, prove: $a^{2} x^{3}+b^{2} y^{3} \leqslant 1$. (1983 Austrian-Polish Mathematical Olympiad Problem) | (3) By the mean inequality, we have
$$\begin{aligned}
a^{2} x^{3}+b^{2} y^{3} \leqslant & \frac{a^{5}+a^{5}+x^{5}+x^{5}+x^{5}}{5}+\frac{b^{5}+b^{5}+y^{5}+y^{5}+y^{5}}{5}= \\
& \frac{2\left(a^{5}+b^{5}\right)+3\left(x^{5}+y^{5}\right)}{5}=1
\end{aligned}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,328 |
27. Given that $x, y, z$ are positive numbers, prove: $\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant \frac{3}{4}$. | 27.
$$\begin{array}{l}
\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z}= \\
\frac{x}{(x+y)+(x+z)}+\frac{y}{(x+y)+(y+z)}+\frac{z}{(x+z)+(y+z)} \leqslant \\
\frac{x}{2 \sqrt{(x+y)(x+z)}}+\frac{y}{2 \sqrt{(x+y)(y+z)}}+\frac{z}{2 \sqrt{(x+z)(y+z)}}= \\
\frac{1}{4}\left(2 \sqrt{\frac{x}{(x+y)} \cdot \frac{x}{(x+z)}}+2 ... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 732,330 |
28. Given $a_{i}(i=1,2,3 \cdots, n, n \geqslant 3), \sum_{i=1}^{n} a_{i}^{n-1}=1$, then
$$\sum_{i=1}^{n} \frac{a_{i}^{n-2}}{1-a_{i}^{n-1}} \geqslant \frac{n}{n-1} \cdot \sqrt[n-1]{n}$$
(Generalization of a problem from the 30th IMO) | 28. By the AM-GM inequality,
$$\begin{array}{l}
(n-1) a_{i}^{n-1}\left[\left(1-a_{i}^{n-1}\right) \cdots\left(1-a_{i}^{n-1}\right)\right] \leqslant \\
{\left[\frac{(n-1) a_{i}^{n-1}+\left(1-a_{i}^{n-1}\right)+\cdots+\left(1-a_{i}^{n-1}\right)}{n}\right]^{n}=\left(\frac{n-1}{n}\right)^{n}}
\end{array}$$
Therefore,
$$\b... | \frac{n}{n-1} \cdot \sqrt[n-1]{n} | Inequalities | proof | Yes | Yes | inequalities | false | 732,331 |
30. Given that $x, y, z$ are positive numbers, and $x+y+z=1$, prove that: $\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+$
$$\frac{z^{4}}{x\left(1-x^{2}\right)} \geqslant \frac{1}{8}$$ | 30. Internal factor $\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y}{8}+\frac{1+y}{32}+\frac{1-y}{16} \geqslant 4$ $\sqrt[4]{\frac{x^{4}}{y\left(1-y^{2}\right)} \cdot \frac{y}{8} \cdot \frac{1+y}{32} \cdot \frac{1-y}{16}}=\frac{x}{2}$, similarly $\frac{y^{4}}{z\left(1-z^{2}\right)}+\frac{z}{8}+\frac{1+z}{32}+\frac{1-z}{16... | \frac{1}{8} | Inequalities | proof | Yes | Yes | inequalities | false | 732,333 |
33. Real numbers $t_{1}, t_{2}, \cdots, t_{n}$ are $n$ real numbers, satisfying $0<t_{1} \leqslant t_{2} \leqslant \cdots \leqslant t_{n}<1$, prove:
$$\left(1-t_{n}\right)^{2}\left[\frac{t_{1}}{\left(1-t_{1}^{2}\right)^{2}}+\frac{t_{2}^{2}}{\left(1-t_{2}^{3}\right)^{2}}+\cdots+\frac{t_{n}^{n}}{\left(1-t_{n}^{n+1}\right... | 33. Let $a_{k}=\frac{\left(1-t_{n}\right)^{2} t_{k}^{k}}{\left(1-t_{k}^{t+1}\right)^{2}} \leqslant \frac{\left(1-t_{k}\right)^{2} t_{k}^{k}}{\left(1-t_{h}^{t+1}\right)^{2}}=\frac{t_{k}^{k}}{\left(1+t+t_{s}^{2}+\cdots+t_{k}^{k}\right)^{2}}$.
By the AM-GM inequality, and noting that $0 < \sqrt[k+1]{1 \cdot t \cdot t_{k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,336 |
34. Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ non-negative real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=n$, prove:
$$\frac{a_{1}^{2}}{1+a_{1}^{4}}+\frac{a_{2}^{2}}{1+a_{2}^{4}}+\cdots+\frac{a_{n}^{2}}{1+a_{n}^{4}} \leqslant \frac{1}{1+a_{1}}+\frac{1}{1+a_{2}}+\cdots+\frac{1}{1+a_{n}}$$
(1994 Hefei High School Mathematics... | 34. Since for any real number $a$ we have $1+a^{4} \geqslant 2 a^{2}$, the left side $\leqslant \frac{n}{2}$, on the other hand, by the arithmetic harmonic mean inequality we know the right side $\geqslant \frac{n^{2}}{\left(1+a_{1}\right)+\left(1+a_{2}\right)+\cdots+\left(1+a_{n}\right)}=\frac{n}{2}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,337 |
35. Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n$ positive real numbers, and $a_{1} a_{2} \cdots a_{n}=1$. Prove that:
$$\sum_{i=1}^{n} \frac{a_{i}^{n}\left(1+a_{i}\right)}{A} \geqslant \frac{n}{2^{n-1}}$$
where $A=\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)$. (A generalization of a problem from the... | 35. By the AM-GM inequality, we have
$$\begin{array}{l}
\frac{a_{1}^{n}}{\left(1+a_{2}\right)\left(1+a_{3}\right) \cdots\left(1+a_{n-1}\right)\left(1+a_{n}\right)}+\frac{1+a_{2}}{2^{n}}+\frac{1+a_{3}}{2^{n}}+\cdots+ \\
\frac{1+a_{n-1}}{2^{n}}+\frac{1+a_{n}}{2^{n}} \geqslant \frac{a_{1}}{2^{n-1}}
\end{array}$$
$$\begin{... | \frac{n}{2^{n-1}} | Inequalities | proof | Yes | Yes | inequalities | false | 732,338 |
36. Let $a, x_{1}, x_{2}, \cdots, x_{n} (n \geqslant 2)$ be positive real numbers. Prove that: $\frac{a^{x_{1}-x_{2}}}{x_{1}+x_{2}}+\frac{a^{x_{2}-x_{3}}}{x_{2}+x_{3}}+\cdots+$ $\frac{a^{x_{n}-x_{1}}}{x_{n}+x_{1}} \geqslant \frac{n^{2}}{2\left(x_{1}+x_{2}+\cdots+x_{n}\right)}$, and determine the necessary and sufficien... | 36. Since $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ are all positive real numbers, we have
$$\begin{array}{l}
2\left(x_{1}+x_{2}+\cdots+x_{n}\right)=\left(x_{1}+x_{2}\right)+\left(x_{2}+x_{3}\right)+\cdots+\left(x_{n-1}+x_{n}\right)+ \\
\left(x_{n}+x_{1}\right) \geqslant n \sqrt[n]{\left(x_{1}+x_{2}\right)\left(x_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,339 |
38. Not allowed to use calculus. Prove:
(1) If $a, b, c, d$ are real numbers, then $a^{6}+b^{6}+c^{6}+d^{6}-6 a b c d \geqslant-2$ holds, and indicate when equality occurs. | 38. (1) The given inequality is transformed into $a^{6}+b^{6}+c^{6}+d^{6}+1+1 \geqslant 6 a b c d$.
According to the arithmetic-geometric mean inequality, we have
$$\frac{a^{6}+b^{6}+c^{6}+d^{6}+1+1}{6} \geqslant \sqrt[6]{a^{6} b^{6} c^{6} d^{6} \cdot 1 \cdot 1}=|a b c d| \geqslant a b c d$$
The equality in the arith... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,341 |
(2) For which positive integers $k$, does the inequality $a^{k}+b^{k}+c^{k}+d^{k}-k a b c d \geqslant M_{k}$ hold for all real numbers $a, b, c, d$? Find the maximum possible value of $M_{k}$, and indicate when equality holds. (2004 Australian Mathematical Olympiad problem) | (2) Notice that, when $k$ is odd, because the selection of sufficiently large negative values for $a, b, c, d$ results in a sufficiently large negative value for $a^{k}+b^{k}+c^{k}+d^{k}-k a b c d$, such a number $M_{k}$ does not exist.
When $k=2$, taking $a=b=d=r$, we get $a^{2}+b^{2}+c^{2}+d^{2}-2 a b c d=4 r^{2}-2 ... | 4-k | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,342 |
39. Positive numbers $a, b$ satisfy $a+b=1$, prove: $\left(\frac{1}{a^{3}}-a^{2}\right)\left(\frac{1}{b^{3}}-b^{2}\right) \geqslant\left(\frac{31}{4}\right)^{2}$. (Mathematical Bulletin Problem 1808) | 39. After factorization and applying the Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
\left(\frac{1}{a^{3}}-a^{2}\right)\left(\frac{1}{b^{3}}-b^{2}\right)=\frac{\left(1-a^{5}\right)\left(1-b^{5}\right)}{a^{3} b^{3}}= \\
\frac{(1-a)(1-b)\left(1+a+a^{2}+a^{3}+a^{4}\right)\left(1+b+b^{2}+b^{3}+\right.}{a^{3} b^{3}... | \left(\frac{31}{4}\right)^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,343 |
40. Given a positive integer $n \geqslant 2$, find all $m \in \mathbf{N}^{+}$, such that for $a_{i} \in \mathbf{R}^{+}, i=1,2,3 \cdots, n$, satisfying $a_{1} a_{2} \cdots a_{n}=1$, then $a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}$. (1999 Poland Austria Mathemat... | 40. Let $x=a_{1}=a_{2}=\cdots=a_{n-1}>0, a_{n}=\frac{1}{x^{n-1}}$, then
$$(n-1) x^{m}+\frac{1}{x^{(n-1) m}} \geqslant \frac{n-1}{x}+x^{n-1}$$
From this, we get $m \geqslant n-1$. Now, assuming $m \geqslant n-1$, then
$$\begin{array}{l}
(n-1)\left(a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}\right)+n(m-n+E)= \\
{\left[\left(a_... | m \geqslant n-1 | Inequalities | proof | Yes | Yes | inequalities | false | 732,344 |
41. For each positive integer $n>1$, prove the inequality:
$$n\left[(n+1)^{\frac{2}{n}}-1\right]<\sum_{j=1}^{n} \frac{2 j+1}{j^{2}}<n\left(1-n^{-\frac{2}{n-1}}\right)+4$$
(1994 British Mathematical Olympiad) | 41. Clearly, $2 j+1=(j+1)^{2}-j^{2}$, thus, by the AM-GM inequality, we have
$$\begin{array}{l}
\sum_{j=1}^{n} \frac{2 j+1}{j^{2}}=\sum_{j=1}^{n}\left[\frac{(j+1)^{2}}{j^{2}}-1\right]=\sum_{j=1}^{n} \frac{(j+1)^{2}}{j^{2}}-n> \\
\sqrt[n]{\frac{2^{2}}{1^{2}} \cdot \frac{3^{2}}{2^{2}} \cdots \cdot \frac{(n+1)^{2}}{n^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,345 |
42. Let the set $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=\{1,2, \cdots, n\}$, prove: $\frac{1}{2}+\frac{2}{3}+\cdots+\frac{n-1}{n} \leqslant$ $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}$. (31st IMO Preliminary Problem) | 42. Since $a_{1}, a_{2}, \cdots, a_{n}$ is a permutation of $1,2, \cdots, n$, we have
$$\left(a_{1}+1\right)\left(a_{2}+1\right) \cdots\left(a_{n-1}+1\right) \geqslant(1+1)(2+1) \cdots(n-1+1)=a_{1} a_{2} \cdots a_{n}$$
Thus,
$$\begin{array}{l}
\quad \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,346 |
43. Let $x, y, z$ be positive real numbers, and satisfy $x y + y z + z x = 1$. Prove: $\frac{27}{4}(x+y)(y+z)(z+x) \geqslant (\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x})^{2} \geqslant 6 \sqrt{3}$. (2006 Turkey National Training Team Problem) | 43. By the AM-GM inequality, we have
$$\begin{array}{l}
(x+y)(y+z)(z+x)=(x+y+z)(x y+y z+z x)-x y z \geqslant \\
(x+y+z)(x y+y z+z x)-\frac{1}{9}(x+y+z)(x y+y z+z x)= \\
\frac{8}{9}(x+y+z)(x y+y z+z x)
\end{array}$$
Since $x y+y z+z x=1$, we have
$$\begin{array}{l}
\quad(x+y)(y+z)(z+x) \geqslant \frac{8}{9}(x+y+z) \\
\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,347 |
44. Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ be positive real numbers. Prove that: $x_{1}+2 x_{2}+3 x_{3}+\cdots+n x_{n} \leqslant$ $\frac{n(n-1)}{2}+x_{1}+x_{2}^{2}+x_{3}^{3}+\cdots+x_{n}^{n}$. (2001 Polish Mathematical Olympiad Problem) | 44. Since
$$\frac{n(n-1)}{2}=\sum_{k=1}^{n}(k-1)$$
By the AM-GM inequality, we have
$$\begin{array}{l}
\frac{n(n-1)}{2}+x_{1}+x_{2}^{2}+x_{3}^{3}+\cdots+x_{n}^{n}=\sum_{k=1}^{n}\left[(k-1)+x_{k}^{k}\right]= \\
\sum_{k=1}^{n}\left(1+1+\cdots+1+x_{k}^{k}\right) \geqslant \sum_{k=1}^{n} k \sqrt[k]{1^{k-1} x_{k}^{k}}=\sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,348 |
45. Let $n$ be a positive integer, prove that: $\left(2 n^{2}+3 n+1\right)^{n} \geqslant 6^{n}(n!)^{2}$. (1992 Balkan Mathematical Olympiad) | 45. By the AM-GM inequality, $\frac{1^{2}+2^{2}+\cdots+n^{2}}{n} \geqslant \sqrt[n]{1^{2} \cdot 2^{2} \cdots \cdots \cdot n^{2}}$, $\frac{(n+1)(2 n+1)}{6} \geqslant \sqrt[n]{(n!)^{2}}$, which means $\left(2 n^{2}+3 n+1\right)^{n} \geqslant 6^{n}(n!)^{2}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,349 |
46. Let $a_{1}, a_{2}, \cdots, a_{n}$ be $n(n \geqslant 2)$ positive numbers that are not all equal, and satisfy $\sum_{k=1}^{n} a_{k}^{-2 n}=1$. Prove that: $\sum_{k=1}^{n} a_{k}^{2 n}-n^{2} \sum_{1 \leqslant i<j \leqslant n} a_{i}^{-2} a_{j}^{-2} > n^{2} .(2003$ National Training Team Exam Question) | 46. The original inequality is equivalent to
$$\sum_{k=1}^{n} a_{k} \sum_{k=1}^{n} \frac{1}{a_{k}}-n^{2} \sum_{-1 \leqslant in^{2}$$
We now prove (1).
Notice that for any $x>0$, by the arithmetic-geometric mean inequality, we have $\sum_{k=0}^{n-1} x^{n-1-2 k} \geqslant n$, with equality if and only if $x=1$. And $\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,350 |
13. $a_{1}, a_{2}, \cdots, a_{n}$ are real numbers. If the sum of any two of them is non-negative, then for any non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, the inequality $a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n} \geqslant a_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\cdots+a_{... | 13. From the given conditions $x_{i} \geqslant 0, a_{i}+a_{j} \geqslant 0(i, j=1,2, \cdots, n)$, we have
$$\begin{array}{l}
a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}=\left(a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}\right) \cdot 1= \\
\left(a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}\right)\left(x_{1}+x_{2}+\cdots+x_{n}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,352 |
47. Let $x_{1}, x_{2}, x_{3}, x_{4}$ be positive real numbers, and $x_{1} x_{2} x_{3} x_{4}=1$. Prove that: $\sum_{k=1}^{4} x_{k}^{3} \geqslant \max \left\{\sum_{k=1}^{4} x_{k}\right.$, $\left.\sum_{k=1}^{4} \frac{1}{x_{k}}\right\} \cdot$ (1997 Iran Mathematical Olympiad) | 47. Since $x_{1} x_{2} x_{3} x_{4}=1$, we have
$$\begin{array}{l}
\sum_{k=1}^{4} x_{k}^{3}+8=\sum_{k=1}^{4}\left(x_{k}^{3}+1+1\right) \geqslant \sum_{k=1}^{4} 3 x_{k}=\sum_{k=1}^{4} x_{k}+2 \sum_{k=1}^{4} x_{k} \geqslant \\
\sum_{k=1}^{4} x_{k}+2 \times 4 \sqrt[4]{x_{1} x_{2} x_{3} x_{4}}=\sum_{k=1}^{4} x_{k}+8
\end{ar... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,353 |
48. Given $x_{1} x_{2} \cdots x_{n}=Q^{n}$, where $x_{i} \geqslant 1, i=1,2, \cdots, n$, and let $x_{n+1}=x_{1}$, prove: when $n$ is an even number greater than 2, $\sum_{i=1}^{n} x_{i} x_{i+1}-\sum_{i=1}^{n} x_{i} \geqslant \frac{n}{2}\left(Q^{2}-1\right)$. Does this hold when $n$ is odd?
(1989 National Training Team ... | 48. Since $x_{i} \geqslant 1, i=1,2, \cdots, n$, therefore, $\left(x_{i}-1\right)\left(x_{i+1}-1\right) \geqslant 0$, which means $2 x_{i} x_{i+1}-$ $x_{i}-x_{i+1}+1 \geqslant x_{i} x_{i+1}$. Adding these inequalities, we get $2 \sum_{i=1}^{n} x_{i} x_{i+1}-2 \sum_{i=1}^{n} x_{i}+n \geqslant \sum_{i=1}^{n} x_{i} x_{i+1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,354 |
52. Let positive real numbers $x, y, z$ satisfy $x+y+z=x y z$, find the minimum value of $x^{7}(y z-1)+y^{7}(z x-1)+$ $z^{7}(x y-1)$. (2003 China National Training Team Problem) | 52. Since $x, y, z > 0$, and $x + y + z = xyz$, we have $z(xy - 1) = x + y$, similarly $y(zx - 1) = z + x$, and $x(yz - 1) = y + z$. By the AM-GM inequality, we have $xyz = x + y + z \geqslant 3 \cdot \sqrt[3]{xyz} \Rightarrow xyz \geqslant 3 \sqrt{3}$, with equality holding if and only if $x = y = z = \sqrt{3}$. There... | 162 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,358 |
53. Let $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}$ be positive real numbers, and satisfy the condition $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+$ $\frac{1}{1+x_{n}}+\frac{1}{1+x_{n+1}}=1$, prove that: $x_{1} x_{2} \cdots x_{n} x_{n+1} \geqslant n^{n+1}$. (1992 Turkish Training Team Problem) | 53. Let $\frac{1}{1+x_{i}}=t_{i}(i=1,2, \cdots, n, n+1)$, then
$$\begin{array}{l}
t_{1}+t_{2}+\cdots+t_{n}+t_{n+1}=1, x_{i}=\frac{1-t_{i}}{t_{i}}, i=1,2, \cdots, n, n+1 \\
x_{i}=\frac{1-t_{i}}{t_{i}}=\frac{t_{1}+t_{2}+\cdots+t_{i-1}+t_{i+1}+t_{i+2}+\cdots+t_{n}+t_{n+1}}{t_{i}} \geqslant \\
\frac{n \sqrt[n]{t_{1} t_{2} ... | x_{1} x_{2} \cdots x_{n} x_{n+1} \geqslant n^{n+1} | Inequalities | proof | Yes | Yes | inequalities | false | 732,359 |
54. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, $n>1$, and let $g_{n}$ denote their geometric mean. Let $A_{1}$, $A_{2}, \cdots, A_{n}$ denote the arithmetic means $A_{k}=\frac{a_{1}+a_{2}+\cdots+a_{k}}{k}(k=1,2, \cdots, n)$, and let $G_{n}$ denote the geometric mean of $A_{1}, A_{2}, \cdots, A_{n}$. Pr... | 54. For convenience, let $A_{0}=0$, then for $k=1,2, \cdots, n$, we have
$$\frac{a_{k}}{A_{k}}=\frac{k A_{k}-(k-1) A_{k-1}}{A_{k}}=k-(k-1) \frac{A_{k-1}}{A_{k}}$$
Let $x_{1}=1, x_{k}=\frac{A_{k-1}}{A_{k}}, k=2,3, \cdots, n$, then we have
$$\sqrt[n]{\frac{G_{n}}{A_{n}}}=\sqrt[n^{2}]{\frac{A_{1} A_{2} \cdots A_{n}}{A_{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,360 |
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