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Theorem 2 (Chebyshev's Inequality) Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \cdots$ $\leqslant b_{n}$, then
$$\sum_{k=1}^{n} a_{k} \sum_{k=1}^{n} b_{k} \leqslant n \sum_{k=1}^{n} a_{k} b_{k}$$
Let $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \geqsla... | $\begin{array}{l}\text { Prove } \quad n \sum_{k=1}^{n} a_{k} b_{k}-\sum_{k=1}^{n} a_{k} \sum_{k=1}^{n} b_{k}=\sum_{k=1}^{n} \sum_{j=1}^{n}\left(a_{k} b_{k}-a_{k} b_{j}\right) \\ =\sum_{k=1}^{n} \sum_{j=1}^{n}\left(a_{j} b_{j}-a_{j} b_{k}\right)=\frac{1}{2} \sum_{k=1}^{n} \sum_{j=1}^{n}\left(a_{k} b_{k}+a_{j} b_{j}-a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,773 |
Example 2 Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Try to prove: $\frac{1}{a^{3}(b+c)}+$ $\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. (36th IMO) | Prove that starting from the condition $abc=1$, the left side of the original inequality can be transformed. Let the left side be $S$, then
$$\begin{aligned}
S & =\frac{(abc)^{2}}{a^{3}(b+c)}+\frac{(abc)^{2}}{b^{3}(c+a)}+\frac{(abc)^{2}}{c^{3}(a+b)} \\
& =\frac{bc}{a(b+c)} \cdot bc+\frac{ac}{b(c+a)} \cdot ac+\frac{ab}{... | S \geqslant \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 733,775 |
$\square$ Example 4 Let $x, y, z$ be positive real numbers, and $x y z=1$, prove:
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4} .$$
(39th IMO Shortlist) | Prove that if $x \leqslant y \leqslant z$, then
$$\frac{1}{(1+y)(1+z)} \leqslant \frac{1}{(1+z)(1+x)} \leqslant \frac{1}{(1+x)(1+y)} .$$
By Chebyshev's inequality, we have
$$\begin{aligned}
& \frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \\
\geqslant & \frac{1}{3}\left(x^{3}+y^{3}+z^{3}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,777 |
Example 6 Let $\frac{1}{2} \leqslant p \leqslant 1, a_{i} \geqslant 0,0 \leqslant b_{i} \leqslant p(i=1,2, \cdots, n, n \geqslant$ 2), if $\sum_{i=1}^{n} a_{i}=\sum_{i=1}^{n} b_{i}=1$, prove: $\sum_{i=1}^{n} b_{i} \prod_{\substack{1<j \leqslant i \\ j \neq i}} a_{j} \leqslant \frac{p}{(n-1)^{n-1}}$. (32nd IMO Shortlist... | Let $A_{i}=a_{1} a_{2} \cdots a_{i-1} a_{i+1} \cdots a_{n}$, by the rearrangement inequality, without loss of generality, assume $b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n}, A_{1} \geqslant A_{2} \geqslant \cdots \geqslant A_{n}$. Since $0 \leqslant b_{i} \leqslant p$, and $\sum_{i=1}^{n} b_{i}=1, \frac{1}{... | \sum_{i=1}^{n} b_{i} A_{i} \leqslant p\left(\frac{1}{n-1}\right)^{n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 733,779 |
$\square$ Example 14 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 Mathematics Selection Test) | Let
$$\begin{aligned}
D & =\left(\sum_{k=1}^{n} a_{k}\right)\left(\sum_{k=1}^{n} b_{k}\right)-\left[\sum_{k=1}^{n}\left(a_{k}+b_{k}\right)\right]\left[\sum_{k=1}^{n} \frac{a_{k} b_{k}}{a_{k}+b_{k}}\right] \\
& =\left(\sum_{k=1}^{n} a_{k}\right)\left(\sum_{j=1}^{n} b_{j}\right)-\left[\sum_{k=1}^{n}\left(a_{k}+b_{k}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,780 |
$\square$ Example 7 Let $x, y, z \in \mathbf{R}^{+}$, and $x+y+z=1$, prove that: $\frac{x y}{\sqrt{x y+y z}}+$ $\frac{y z}{\sqrt{y z+x z}}+\frac{x z}{\sqrt{x z+x y}} \leqslant \frac{\sqrt{2}}{2}$. (2006 China National Training Team Test) | To prove that since $(x+y)(y+z)(z+x) \leqslant\left[\frac{2(x+y+z)}{3}\right]^{3}=\frac{8}{27}$. Therefore, we only need to prove a slightly stronger conclusion:
$$\begin{aligned}
& \frac{x y}{\sqrt{x y+y z}}+\frac{y z}{\sqrt{y z+x z}}+\frac{x z}{\sqrt{x z+x y}} \\
\leqslant & \frac{3 \sqrt{3}}{4} \sqrt{(x+y)(y+z)(z+x)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,781 |
$\square$ Example 8 Let $a_{1}, a_{2}, \cdots, a_{N}$ be positive real numbers, and $2 S=a_{1}+a_{2}+\cdots+a_{N}$, $n \geqslant m \geqslant 1$, then $\sum_{k=1}^{N} \frac{a_{k}^{n}}{\left(2 S-a_{k}\right)^{m}} \geqslant \frac{(2 S)^{n-m}}{N^{n-m-1}(N-1)^{m}}$. | Assume without loss of generality that $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{N}$, then
$$\frac{1}{2 S-a_{1}} \geqslant \frac{1}{2 S-a_{2}} \geqslant \cdots \geqslant \frac{1}{2 S-a_{N}}$$
By Chebyshev's inequality, we have
$$\sum_{k=1}^{N} \frac{a_{k}^{n}}{\left(2 S-a_{k}\right)^{m}} \geqslant \frac{1}{... | \sum_{k=1}^{N} \frac{a_{k}^{n}}{\left(2 S-a_{k}\right)^{m}} \geqslant \frac{(2 S)^{n-m}}{N^{n-m-1}(N-1)^{m}} | Inequalities | proof | Yes | Yes | inequalities | false | 733,782 |
$\square$ Example 2 Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Prove:
$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1$ (41st IMO Problem) | Proof divided into two cases:
(1) $a-1+\frac{1}{b}, b-1+\frac{1}{c}, c-1+\frac{1}{a}$ are not all positive.
Without loss of generality, assume $a-1+\frac{1}{b} \leqslant 0$, then $a \leqslant 1, b \geqslant 1$, so
$$b-1+\frac{1}{c} \geqslant \frac{1}{c}>0, c-1+\frac{1}{a} \geqslant c>0 \text {, }$$
Thus $\quad\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,784 |
$\square$ Example 3 Given that $\alpha, \beta$ are two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, and the domain of the function $f(x)-\frac{2 x-t}{x^{2}+1}$ is $[\alpha, \beta]$.
(1) Find $g(t)=\max f(x)-\min f(x)$;
(2) Prove that for $u_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2,3)$, i... | (1) Let $\alpha \leqslant x_{1}t\left(x_{1}+x_{2}\right)-2 x_{1} x_{2}+\frac{1}{2}>0$, so $f\left(x_{2}\right)-f\left(x_{1}\right)>0$, hence $f(x)$ is an increasing function on the interval $[\alpha, \beta]$.
Since $\alpha+\beta=t, \alpha \beta=-\frac{1}{4}, \beta-\alpha=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}=\sqrt{... | \frac{3}{4} \sqrt{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,785 |
$\square$ Example 4 In the Cartesian coordinate system $x O y$, the sequence of points $\left\{A_{n}\right\}$ on the positive $y$-axis and the sequence of points $\left\{B_{n}\right\}$ on the curve $y=\sqrt{2 x}(x \geqslant 0)$ satisfy $\left|O A_{n}\right|=\left|O B_{n}\right|=\frac{1}{n}$. The intercept of the line $... | Prove (1) It is easy to know $A_{n}\left(0, \frac{1}{n}\right), B_{n}\left(b_{n}, \sqrt{2 b_{n}}\right)\left(b_{n}>0\right)$.
From $\left|O B_{n}\right|=\frac{1}{n}$ we get $b_{n}^{2}+2 b_{n}=\left(\frac{1}{n}\right)^{2}$, hence $b_{n}=\sqrt{\left(\frac{1}{n}\right)^{2}+1}-1, n \in \mathbf{N}^{\cdot}$.
Next, the interc... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,786 |
$\square$ Example 5 Let $a_{1}, a_{2}, \cdots, a_{n} ; b_{1}, b_{2}, \cdots, b_{n} \in[1,2]$, and $\sum_{i=1}^{n} a_{i}^{2}=$ $\sum_{i=1}^{n} b_{i}^{2}$, prove that: $\sum_{i=1}^{n} \frac{a_{i}^{3}}{b_{i}} \leqslant \frac{17}{10} \sum_{i=1}^{n} a_{i}^{2}$. Also, determine the necessary and sufficient conditions for equ... | Prove that since $a_{i}, b_{i} \in [1,2], i=1; 2, \cdots, n$, therefore
$$\frac{1}{2} \leqslant \frac{\sqrt{\frac{a_{i}^{3}}{b_{i}}}}{\sqrt{a_{i} b_{i}}}=\frac{a_{i}}{b_{i}} \leqslant 2$$
Thus, $\left(\frac{1}{2} \sqrt{a_{i} b_{i}}-\sqrt{\frac{a_{i}^{3}}{b_{i}}}\right)\left(2 \sqrt{a_{i} b_{i}}-\sqrt{\frac{a_{i}^{3}}{... | \sum_{i=1}^{n} \frac{a_{i}^{3}}{b_{i}} \leqslant \frac{17}{10} \sum_{i=1}^{n} a_{i}^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 733,787 |
$\square$ Example 6 Let $n \geqslant 3, n \in \mathbf{N}^{*}$, and let $a_{1}, a_{2}, \cdots, a_{n}$ be a sequence of real numbers, where $2 \leqslant a_{i} \leqslant 3, i=1,2,3, \cdots, n$. If we take $S=a_{1}+a_{2}+\cdots+a_{n}$, prove: $\frac{a_{1}^{2}+a_{2}^{2}-a_{3}^{2}}{a_{1}+a_{2}-a_{3}}+\frac{a_{2}^{2}+a_{3}^{2... | Let \( A_{i}=\frac{a_{i}^{2}+a_{i+1}^{2}-a_{i+2}^{2}}{a_{i}+a_{i+1}-a_{i+2}} \)
\[ =a_{i}+a_{i+1}+a_{i+2}-\frac{2 a_{i} a_{i+1}}{a_{i}+a_{i+1}-a_{i+2}} \]
From \(\left(a_{i}-2\right)\left(a_{i+1}-2\right) \geqslant 0\), we get
\[ -2 a_{i} a_{i+1} \leqslant -4\left(a_{i}+a_{i+1}-2\right). \]
Noting that \(1=2+2-3 \leq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,788 |
$\square$ side 7 Let $0 \leqslant a_{1}, a_{2}, \cdots, a_{n} \leqslant 1$, then $\frac{a_{1}}{a_{2}+a_{3}+\cdots+a_{n}+1}+$
$$\frac{a_{2}}{a_{1}+a_{3}+\cdots+a_{n}+1}+\cdots+\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}+1}+\left(1-a_{1}\right)(1-$$
$\left.a_{2}\right) \cdots\left(1-a_{n}\right) \leqslant 1$. (Generalizatio... | Suppose $0 \leqslant a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n} \leqslant 1$, then
$$\begin{aligned}
& \frac{a_{1}}{a_{2}+a_{3}+\cdots+a_{n}+1}+\frac{a_{2}}{a_{1}+a_{3}+\cdots+a_{n}+1}+\cdots+ \\
& \frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n-1}+1}+\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{n}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,789 |
Example 8 Let $n \in \mathbf{N}^{\cdot}, x_{0}=0, x_{i}>0, i=1,2,3, \cdots, n$, and $\sum_{i=1}^{n} x_{i}=1$, prove that: $1 \leqslant \sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{0}+x_{1}+x_{2}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+\cdots+x_{n}}}<$ $\frac{\pi}{2}$. (1996 China Mathematical Olympiad Problem) | Prove that from $\sum_{i=1}^{n} x_{i}=1$ and the AM-GM inequality, we have
$$\begin{array}{l}
\sqrt{1+x_{0}+x_{1}+x_{2}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+\cdots+x_{n}} \\
\leqslant \frac{1+x_{0}+x_{1}+x_{2}+\cdots+x_{i-1}+x_{i}+\cdots+x_{n}}{2}=1 .
\end{array}$$
Thus, the left inequality holds.
Since $0 \leqslant x_{0... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,790 |
Example 15 Let $x_{1}, x_{2}, \cdots, x_{n} \in(0,1)$, and $x_{1}+x_{2}+\cdots+x_{n}=1$, prove that: $(n-1)\left(\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+\cdots+\frac{1}{1-x_{n}}\right) \geqslant(n+1)\left(\frac{1}{1+x_{1}}\right.$ $\left.+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}\right) \cdot(2004$ Romanian Mathematical O... | Prove
\[
(n-1)\left(\frac{1}{1-x_{1}}+\frac{1}{1-x_{2}}+\cdots+\frac{1}{1-x_{n}}\right)-(n+1).
\]
\[
\begin{aligned}
& \left(\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}\right) \\
= & \sum_{k=1}^{n}\left[(n-1) \frac{1}{1-x_{k}}-(n+1) \frac{1}{1+x_{k}}\right] \\
= & 2 \sum_{k=1}^{n} \frac{n x_{k}-1}{1-x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,791 |
$\square$ Example 9 Let $a, b, c$ be the lengths of the three sides of a triangle with a perimeter not exceeding $2 \pi$. Prove that $\sin a, \sin b, \sin c$ can form the lengths of the three sides of a triangle. (2004 China National Training Team Selection Test Problem) | Proof: From the given, we know $0c$, i.e., $\frac{a+b}{2}>\frac{c}{2}$. Since $0\sin \frac{c}{2}>0$$
If $\pi0$$
From the given, we know $|a-b|\cos \frac{c}{2}>0$$
Multiplying both sides of (1) and (2) respectively, we get
$$2 \sin \frac{a+b}{2} \cos \frac{a-b}{2}>2 \sin \frac{c}{2} \cos \frac{c}{2}$$
That is,
$$\si... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,792 |
$\square$ Example 10 Let $a, b, c$ be positive numbers, their sum equals 1, prove: $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geqslant \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$. (29th Russian Mathematical Olympiad Problem) | Without loss of generality, let $a \geqslant b \geqslant c$, then $1-c^{2} \geqslant 1-b^{2} \geqslant 1-a^{2}$, thus
$$\frac{1}{1-a^{2}} \geqslant \frac{1}{1-b^{2}} \geqslant \frac{1}{1-c^{2}} \text {. }$$
Notice that $\frac{1}{1-a}-\frac{2}{1+a}=\frac{3 a-1}{1-a^{2}}$, so it suffices to prove
$$\frac{3 a-1}{1-a^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,793 |
Example 11 Let $n$ be a given natural number, $n \geqslant 3$, and for $n$ given real numbers $a_{1}, a_{2}, \cdots$, $a_{n}$, denote the minimum value of $\left|a_{i}-a_{j}\right|(1 \leqslant i, j \leqslant n)$ as $m$. Find the maximum value of the above $m$ under the condition $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$... | Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$. Then $a_{2}-a_{1} \geqslant m, a_{3}-a_{2} \geqslant m$,
$$\begin{aligned}
a_{3}-a_{1} & \geqslant 2 m, \cdots, a_{n}-a_{n-1} \geqslant m, a_{j}-a_{i} \geqslant(j-i) m(1 \leqslant i<j \leqslant n) \\
& \sum_{1 \leqslant i<j \leqslant n}\left(a_{i}-a... | \sqrt{\frac{12}{n\left(n^{2}-1\right)}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,794 |
Example 13 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 Question) | $$\begin{array}{l}
\text { Prove } \quad \frac{x^{2} y}{z}+\frac{y^{2} z}{x}+\frac{z^{2} x}{y}-\left(x^{2}+y^{2}+z^{2}\right) \\
=\frac{x^{2}}{z}(y-z)+\frac{y^{2} z}{x}+\frac{z^{2} x}{y}-\left(y^{2}+z^{2}\right) \\
\geqslant \frac{y^{2}}{z}(y-z)+2 \sqrt{y z} \cdot z-\left(y^{2}+z^{2}\right) \\
=\frac{y^{2}}{z}(y-z)-(y-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,796 |
$\square$ Example 14 Let $n(n>3)$ be a natural number, and $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ positive numbers, with $x_{1} x_{2} \cdots x_{n}=1$. Prove:
$$\frac{1}{1+x_{1}+x_{1} x_{2}}+\frac{1}{1+x_{2}+x_{2} x_{3}}+\cdots+\frac{1}{1+x_{n}+x_{n} x_{1}}>1$$
(From the Russian Mathematical Olympiad) | Prove that if $x_{n+k}=x_{k}(k=1,2, \cdots, n)$, then
$$\begin{aligned}
& \sum_{k=1}^{n} \frac{1}{1+x_{k}+x_{k} x_{k+1}} \\
> & \sum_{k=1}^{n} \frac{1}{1+\sum_{i=0}^{n-2} \prod_{j=k}^{k+i} x_{j}} \\
= & \sum_{k=1}^{n} \frac{x_{n} \prod_{j=1}^{k-1} x_{j}}{x_{n} \prod_{j=1}^{k-1} x_{j}\left(1+\sum_{i=1}^{n-2} \prod_{j=k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,797 |
Example $15\{a_{n}\}$ is defined as follows: $a_{1}=\frac{1}{2}$, and $a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-a_{n}+1}, n=1$, $2, \cdots$. Prove that for each positive integer $n$ we have $\sum_{i=1}^{n} a_{i}<1$. (2004 China National Training Team Problem, 2003 Romania National Training Team Problem) | Prove that taking the reciprocal of both sides of the original recurrence relation yields
$$\frac{1}{a_{n+1}}=\frac{1}{a_{n}^{2}}-\frac{1}{a_{n}}+1, \quad \frac{1}{a_{n+1}}-1=\frac{1}{a_{n}}\left(\frac{1}{a_{n}}-1\right)$$
Let \( x_{n}=\frac{1}{a_{n}}-1 \), then \(\frac{1}{x_{n+1}}=\frac{1}{x_{n}}-\frac{1}{\frac{1}{a_... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,798 |
Example 16 Let $a, b, c$ be the lengths of the three sides of a triangle, and $a+b+c$ $=1$. If the positive integer $n \geqslant 2$, prove: $\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<1+$ $\frac{\sqrt[n]{2}}{2} \cdot$ (2003 Asia Pacific Mathematical Olympiad Problem) | Proof: Let $a \geqslant b \geqslant c > 0$, and since $a + b + c = 1$, then
$$\begin{aligned}
\left(b+\frac{c}{2}\right)^{n} & =b^{n}+\mathrm{C}_{n}^{1} b^{n-1} \frac{c}{2}+\mathrm{C}_{n}^{2} b^{n-2}\left(\frac{c}{2}\right)^{2}+\cdots+\mathrm{C}_{n}^{n}\left(\frac{c}{2}\right)^{n} \\
& \geqslant b^{n}+\left[\mathrm{C}_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,799 |
$\square$ Example 17 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1, p, q$ be positive constants, $m$ be a positive integer, and $m \geqslant 2$, prove: $\sqrt[m]{p+q}+(n-1) \sqrt[m]{q}<$
$$\sum_{i=1}^{n} \sqrt[m]{p a_{i}+q} \leqslant \sqrt[m]{p n^{m-1}+q n^{m}} .$$ | Prove (1) First, prove the lower bound is obvious $0(\sqrt[m]{p+q}-\sqrt[m]{q}) a_{i}+\sqrt[m]{q}$.
Since $a_{1}+a_{2}+\cdots+a_{n}=1$, summing over $i$ gives
$$\sum_{i=1}^{n} \sqrt[m]{p a_{i}+q}>\sum_{i=1}^{n}(\sqrt[m]{p+q}-\sqrt[m]{q}) a_{i}+\sqrt[m]{q}=\sqrt[m]{p+q}+(n-1) \sqrt[m]{q}$$
(2) Next, prove the upper boun... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,800 |
$\square$ Example 18 Prove that if two given positive numbers $p<q$, then for any $a, b, c, d, e$ $\in[p, q]$, we have
$$(a+b+c+d+e)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right) \leqslant 25+6\left(\sqrt{\frac{p}{q}}-\sqrt{\frac{q}{p}}\right)^{2},$$
and determine the necessary and sufficient... | Prove from Lagrange's identity:
$$\begin{aligned}
& \sum_{i=1}^{n} a_{i}^{2} \sum_{i=1}^{n} b_{i}^{2}=\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2}+\sum_{1 \leqslant i<j<n}\left(a_{i} b_{j}-a_{i} b_{i}\right)^{2} \\
\text { we get } \quad & (a+b+c+d+e)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,801 |
Example 16 Let $0<\theta_{i} \leqslant \frac{\pi}{4}, i=1,2,3,4$, prove: $\tan \theta_{1} \tan \theta_{2} \tan \theta_{3} \tan \theta_{4}$
$\leqslant \sqrt{\frac{\sin ^{8} \theta_{1}+\sin ^{8} \theta_{2}+\sin ^{8} \theta_{3}+\sin ^{8} \theta_{4}}{\cos ^{8} \theta_{1}+\cos ^{8} \theta_{2}+\cos ^{8} \theta_{3}+\cos ^{8} ... | Prove the original inequality is to prove
$$\begin{array}{l}
\frac{\sin ^{8} \theta_{1}+\sin ^{8} \theta_{2}+\sin ^{8} \theta_{3}+\sin ^{8} \theta_{4}}{\sin ^{2} \theta_{1} \sin ^{2} \theta_{2} \sin ^{2} \theta_{3} \sin ^{2} \theta_{4}} \geqslant \frac{\cos ^{8} \theta_{1}+\cos ^{8} \theta_{2}+\cos ^{8} \theta_{3}+\cos... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,802 |
$$\square \square$$ Example 19 Given that $a, b, c, d$ are positive real numbers, prove:
$$\begin{aligned}
& \sqrt{(a+c)^{2}+(b+d)^{2}} \leqslant \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \\
\leqslant & \sqrt{(a+c)^{2}+(b+d)^{2}}+\frac{2|a d-b c|}{\sqrt{(a+c)^{2}+(b+d)^{2}}}
\end{aligned}$$ | Proof: Let $P(a, b), Q(-c,-d)$ be in the first and third quadrants of the Cartesian coordinate system $x O y$, respectively, then
$$\begin{array}{c}
|O P|=\sqrt{a^{2}+b^{2}},|O Q|=\sqrt{c^{2}+d^{2}} \\
|P Q|=\sqrt{(a+c)^{2}+(b+d)^{2}}
\end{array}$$
In $\triangle O P Q$, by $|O P|+|O Q| \geqslant|P Q|$ we get
Therefor... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,803 |
$\square$
Example 20 Let $a \geqslant b \geqslant c \geqslant 0$, and $a+b+c=3$, prove: $a b^{2}+b c^{2}+c a^{2} \leqslant \frac{27}{8}$. And determine the condition for equality. (2002 Hong Kong Mathematical Olympiad Problem) | Proof: Let $f(a, b, c) = a b^2 + b c^2 + c a^2$, then
$$f(a, c, b) = a c^2 + c b^2 + b a^2$$
Since $f(a, b, c) - f(a, c, b)$
$$\begin{array}{l}
= a b^2 + b c^2 + c a^2 - (a c^2 + c b^2 + b a^2) \\
= -[a b(a - b) + c^2(a - b) + c(b^2 - a^2)] \\
= -(a - b)(a b + c^2 - a c - b c) \\
= -(a - b)(a - c)(b - c) \leqslant 0
\... | a = b = \frac{3}{2}, c = 0 | Inequalities | proof | Yes | Yes | inequalities | false | 733,804 |
Example 21 Positive numbers $a, b, c$ satisfy $a+b+c=1$, prove that: $\frac{1+a}{1-a}+\frac{1+b}{1-b}+$ $\frac{1+c}{1-c} \leqslant 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)$. (2004 Japan Mathematical Olympiad Problem) | For positive real numbers $\alpha, \beta, \gamma, x, y, z$, set $\alpha \beta \gamma = x y z = 1, x, y, z \in [\alpha, \gamma]$.
Since $(\alpha + \beta + \gamma) - (x + y + z)$
$$=\left\{\begin{array}{l}
\beta \gamma(x - \alpha)(z - \alpha) + x z(\gamma - y)(y - \beta) \geqslant 0, \quad y \geqslant \beta, \\
\alpha \b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,805 |
$\square$ Example 22 Let $a, b, c, d$ be positive real numbers, satisfying $a b+c d=1$, and let points $P_{i}\left(x_{i}, y_{i}\right)$ $(i=1,2,3,4)$ be four points on the unit circle centered at the origin. Prove:
$$\left(a y_{1}+b y_{2}+c y_{3}+d y_{4}\right)^{2}+\left(a x_{4}+b x_{3}+c x_{2}+d x_{1}\right)^{2} \leqs... | Prove that if $u=a y_{1}+b y_{2}, v=c y_{3}+d y_{4}, u_{1}=a y_{4}+b y_{3}, v_{1}=c y_{2} +d y_{1}$, then
$$u^{2} \leqslant\left(a y_{1}+b y_{2}\right)^{2}+\left(a x_{1}-b x_{2}\right)^{2}=a^{2}+b^{2}+2 a b\left(y_{1} y_{2}-x_{1} x_{2}\right),$$
i.e., $\square$
$$y_{1} y_{2}-x_{1} x_{2} \leqslant \frac{a^{2}+b^{2}-u^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,806 |
Example 23 Let $n (n \geqslant 3)$ be a positive integer. Prove: For positive real numbers $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant$ $x_{n}$, the inequality
$\frac{x_{1} x_{2}}{x_{3}}+\frac{x_{2} x_{3}}{x_{4}}+\cdots+\frac{x_{n-1} x_{n}}{x_{1}}+\frac{x_{n} x_{1}}{x_{2}} \geqslant x_{1}+x_{2}+\cdots+x_{n}$ hold... | First, we prove a lemma: If \(0 < x \leqslant y, 0 < a \leqslant 1\), then
\[ x + y \leqslant a x + \frac{y}{x} \]
In fact, from \(a x \leqslant x \leqslant y\) we get \((1-a)(y - a x) \geqslant 0\), i.e., \(a^2 x + y \geqslant a x + a y\). Therefore, \(x + y \leqslant a x + \frac{y}{x}\).
Now, let \((x, y, a) = \lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,807 |
$\square$ Example 1 For a natural number $n \geqslant 2$ and $x_{1}, x_{2}, \cdots, x_{n} \in[0,1]$, prove that: $\sum_{k=1}^{n} x_{k}-$ $\sum_{1 \leqslant k \leq i \leqslant n} x_{k} x_{j} \leqslant 1$. (1994 Romanian Mathematical Olympiad) | Prove for positive integer $n(n \geqslant 2)$ using mathematical induction.
When $n=2$, since $x_{1}, x_{2} \in[0,1]$, then
$$x_{1}+x_{2}-x_{1} x_{2}=1-\left(1-x_{1}\right)\left(1-x_{2}\right) \leqslant 1 .$$
Assume when $n=m$ (where $m \geqslant 2$), for $x_{1}, x_{2}, \cdots, x_{m} \in[0,1]$, we have
$$\sum_{k=1}^{m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,808 |
Example 2 Let $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 3)$ be non-negative real numbers, and $x_{1}+x_{2}+\cdots+$ $x_{n}=1$, prove: $x_{1}^{2} x_{2}+x_{2}^{2} x_{3}+\cdots+x_{n-1}^{2} x_{n}+x_{n}^{2} x_{1} \leqslant \frac{4}{27}$. (1992 Taipei Mathematical Olympiad Problem) | Prove by mathematical induction on $n$. When $n \geqslant 3$, since the inequality is cyclic symmetric about $x_{1}$, $x_{2}$, $x_{3}$, we may assume $x_{1} \geqslant x_{2}, x_{1} \geqslant x_{3}$. If $x_{2}<x_{3}$, then
$$\begin{array}{l}
x_{1}^{2} x_{2}+x_{2}^{2} x_{3}+x_{3}^{2} x_{1}-\left(x_{1}^{2} x_{3}+x_{3}^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,809 |
Example 3 Let $r_{1}, r_{2}, \cdots, r_{n}$ be real numbers greater than or equal to 1, prove that $\frac{1}{r_{1}+1}+$ $\frac{1}{r_{2}+1}+\cdots+\frac{1}{r_{n}+1} \geqslant \frac{n}{\sqrt[n]{r_{1} r_{2} \cdots r_{n}}+1}$. (39th IMO Shortlist Problem) | Prove that when $n=1$, the inequality obviously holds. Below, we use mathematical induction to prove that the inequality holds for $n=2^{k} (k=1,2, \cdots)$.
When $k=1$, we have
$$\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}-\frac{2}{\sqrt{r_{1} r_{2}}+1}=\frac{\left(\sqrt{r_{1} r_{2}}-1\right)\left(\sqrt{r_{1}}-\sqrt{r_{2}}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,810 |
$\square$ Example 4 Given $x_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, satisfying $\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}$ $=0$, prove: $\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n}$. (1989 National High School Mathematics League Question) | We prove the strengthened proposition using mathematical induction: For $n(n \geqslant 2)$ real numbers $x_{1}$, $x_{2}, \cdots, x_{n}$, if $\sum_{i=1}^{n}\left|x_{i}\right| \leqslant 1$ and $\sum_{i=1}^{n} x_{i}=0$, then $\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n}$.
(1) When $n=2$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,811 |
$\qquad$
Example 5 Let $\theta \in\left(0, \frac{\pi}{2}\right), n$ be a positive integer greater than 1, prove:
$$\left(\frac{1}{\sin ^{n} \theta}-1\right)\left(\frac{1}{\cos ^{n} \theta}-1\right) \geqslant 2^{n}-2 \sqrt{2^{n}}+1$$ | Prove that when $n=2$, the left and right sides of the inequality are equal, so the proposition holds for $n=2$. Assume the proposition holds for $n=k (k \geqslant 2)$, i.e.,
$$\left(\frac{1}{\sin ^{k} \theta}-1\right)\left(\frac{1}{\cos ^{2} \theta}-1\right) \geqslant 2^{k} -2 \sqrt{2^{k}}+1,$$
then for $n=k+1$, we ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,812 |
$\square$
Example 17 In $\triangle A B C$, prove: $a^{2}\left(\frac{b}{c}-1\right)+b^{2}\left(\frac{c}{a}-1\right)+$ $c^{2}\left(\frac{a}{b}-1\right) \geqslant 0 .(2006$ Moldova Mathematical Olympiad Problem) | Prove the inequality by multiplying both sides by $2abc$, transforming it into proving the inequality
$$2 a^{3} b(b-c)+2 b^{3} c(c-a)+2 c^{3} a(a-b) \geqslant 0 .$$
Since $2 a^{3} b(b-c)+2 b^{3} c(c-a)+2 c^{3} a(a-b)$ can be transformed using the Sum of Squares (SOS) method:
$$\begin{aligned}
= & a^{3}[(b+c)+(b-c)](b-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,813 |
Example 6 Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, and let $a=\min \left\{x_{1}, x_{2}, \cdots\right.$, $\left.x_{n}\right\}$. Prove that: $\sum_{j=1}^{n} \frac{1+x_{j}}{1+x_{j+1}} \leqslant n+\frac{1}{(1+a)^{2}} \sum_{j=1}^{n}\left(x_{j}-a\right)^{2}$ (let $x_{n+1}=$ $x_{1}$ ), and the equality ... | Prove that when $n=1$, $a=x_{1}$, the inequality is written as
$$\frac{1+x_{1}}{1+x_{1}} \leqslant 1+\frac{1}{\left(1+x_{1}\right)^{2}} \cdot\left(x_{1}-x_{1}\right)^{2}$$
It obviously holds, and is an equality.
Assume the proposition holds for $n-1$, consider the case for $n$. Since the inequality is cyclic symmetric... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,814 |
$\square$ Example 7 Given $a>2, \left\{a_{n}\right\}$ is defined as follows: $a_{0}=1, a_{1}=a, a_{n+1}=$ $\left(\frac{a_{n}^{2}}{a_{n-1}^{2}}-2\right) a_{n}$. Prove that for any $n \in \mathbf{N}^{\cdot}$, $\frac{1}{a_{0}}+\frac{1}{a_{1}}+\cdots+\frac{1}{a_{n}}<\frac{1}{2}(2+a$ $-\sqrt{a^{2}-4}$ ). (37th IMO Prelimina... | Proof: Let $f(x)=x^{2}-2$, then
$$\frac{a_{n+1}}{a_{n}}=f^{(1)}\left(\frac{a_{n}}{a_{n-1}}\right)=f^{(2)^{\circ}}\left(\frac{a_{n-1}}{a_{n-2}}\right)=\cdots=f^{(n)}\left(\frac{a_{1}}{a_{0}}\right)=f^{(n)}(a) .$$
Thus, $a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdot \cdots \cdot \frac{a_{1}}{a_{0}} \cd... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,815 |
Example 8 If the sequence $a_{0}, a_{1}, a_{2}, \cdots$ satisfies the conditions: $a_{1}=\frac{1}{2}, a_{k+1}=a_{k}+\frac{1}{n} a_{k}^{2}$ $(k=0,1,2, \cdots)$, where $n$ is some fixed positive integer. Prove: $1-\frac{1}{n}<a_{n}<$ 1. (1980 IMO Problem) | $$
\begin{aligned}
& \text{Since } a_{1}=a_{0}+\frac{1}{n} a_{0}^{2}=\frac{1}{2}+\frac{1}{4 n}=\frac{2 n+1}{4 n}, \text{ it is easy to prove that } \frac{n+1}{2 n+1} \\
& \frac{n+1}{2 n-k+2}+\frac{(n+1)^{2}}{n(2 n-k+2)^{2}} \\
& =\frac{n+1}{2 n-k+1}-\frac{n+1}{(2 n-k+1)(2 n-k+2)}+\frac{(n+1)^{2}}{n(2 n-k+2)^{2}} \\
& =... | 1-\frac{1}{n}<a_{n}<1 | Algebra | proof | Yes | Yes | inequalities | false | 733,816 |
$\square$ Example 9 Prove that for any $m, n \in \mathbf{N}^{*}$ and $x_{1}, x_{2}, \cdots, x_{n} ; y_{1}, y_{2}, \cdots, y_{n} \in(0,1]$ satisfying $x_{i}+y_{i}=1, i=1,2, \cdots, n$, we have $\left(1-x_{1} x_{2} \cdots x_{n}\right)^{m}+(1-$ $\left.y_{1}^{\prime \prime}\right)\left(1-y_{2}^{m}\right) \cdots\left(1-y_{n... | Prove for $n \in \mathbf{N}^{*}$ using induction.
First, because $\left(1-x_{1}\right)^{m}+\left(1-y_{1}^{m}\right)=y_{1}^{m}+\left(1-y_{1}^{m}\right)=1$, the conclusion is correct when $n=1$. Assume the conclusion holds for $n-1$, then
$$\begin{aligned}
& \left(1-x_{1} x_{2} \cdots x_{n}\right)^{m}+\left(1-y_{1}^{m}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,817 |
$\square$ Example 11 Given the sequence $\left\{r_{n}\right\}$ satisfies $r_{1}=2, r_{n}=r_{1} r_{2} \cdots r_{n-1}+1, n=2,3$, $\cdots$. If positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<1$, prove that: $\frac{1}{a_{1}}+$ $\frac{1}{a_{2}}+\cdots+\frac{1}{... | Proof First, it is easy to prove by induction that the sequence $\left\{r_{n}\right\}$ has the following property:
$$\frac{1}{r_{1}}+\frac{1}{r_{2}}+\cdots+\frac{1}{r_{n}}=1-\frac{1}{r_{1} r_{2} \cdots r_{n}}$$
We now prove the inequality by induction.
When $n=1$, the inequality obviously holds.
Now assume that the in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,819 |
$\square$
Example 1 Let $a, b, c$ be the lengths of the three sides of a triangle. If $a+b+c=1$, prove: $a^{2}+b^{2}+c^{2}+4 a b c \leqslant \frac{1}{2}$. (1990 Italian Mathematical Olympiad Problem) | Prove that the original inequality is equivalent to
$$\begin{aligned}
& (a+b+c)^{2}-2(a b+b c+c a)+4 a b c \leqslant \frac{1}{2} \\
\Leftrightarrow & 1-2(a b+b c+c a)+4 a b c \leqslant \frac{1}{2} \\
\Leftrightarrow & a b+b c+c a-2 a b c \geqslant \frac{1}{4}
\end{aligned}$$
Since $a, b, c$ are the lengths of the thre... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,821 |
$\square$ Example 5 Given the side lengths $a, b, c$ and the area $S$ of a triangle, prove the inequality:
$$\begin{array}{l}
\quad\left(a^{2}+b^{2}+c^{2}-4 \sqrt{3} S\right)\left(a^{2}+b^{2}+c^{2}\right) \geqslant 2\left[a^{2}(b-c)^{2}+b^{2}(c-a)^{2}\right. \\
\left.+c^{2}(a-b)^{2}\right] . \text { (31st IMO Shortlist... | Let $L=\left(a^{2}+b^{2}+c^{2}\right)^{2}-4 \sqrt{3} S\left(a^{2}+b^{2}+c^{2}\right)-4\left(a^{2} b^{2}+\right.$ $\left.b^{2} c^{2}+c^{2} a^{2}\right)+8 a b c p$, where $p=\frac{a+b+c}{2}$. By Heron's formula, we have
$$\begin{aligned}
16 S^{2} & =(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \\
& =\left[(b+c)^{2}-a^{2}\right]\left[a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,826 |
Example 6 Let real numbers $a, b, c$ be such that the sum of any two is greater than the third, then $\frac{2}{3}(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \geqslant a^{3}+b^{3}+c^{3}+3 a b c$, with equality if and only if $a=b=c$. (A generalization of a problem from the 13th Putnam Mathematical Competition) | It is not hard to verify that the above inequality is equivalent to
$$(-a+b+c)(a-b+c)(a+b-c) \geqslant(3 a-b-c)(3 b-c-a)(3 c-a-b).$$
Moreover, at least two of $3 a-b-c, 3 b-c-a, 3 c-a-b$ are positive (otherwise, suppose $3 a-b-c \leqslant 0, 3 b-c-a \leqslant 0$, then $b+c \leqslant a$, which contradicts the given cond... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,827 |
Example 7 In $\triangle A B C$, prove: $\frac{\sqrt{\sin A \sin B}}{\sin \frac{C}{2}}+\frac{\sqrt{\sin B \sin C}}{\sin \frac{A}{2}}+$ $\frac{\sqrt{\sin C \sin A}}{\sin \frac{B}{2}} \geqslant 3 \sqrt{3}$. (2005 Jiangsu Province Mathematical Winter Camp) | Prove that from the Law of Sines and the Law of Cosines,
$$\begin{aligned}
\frac{\sqrt{\sin A \sin B}}{\sin \frac{C}{2}} & =\frac{\sqrt{\sin A \sin B}}{\sin C} \cdot 2 \cos \frac{C}{2}=\frac{\sqrt{a b}}{c} \cdot \sqrt{2(1+\cos C)} \\
& =\frac{\sqrt{2 a b+2 a b \cos C}}{c}=\frac{\sqrt{(a+b)^{2}-c^{2}}}{c} \\
& =\frac{\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,828 |
Example 8 Prove that in an acute triangle $ABC$, $\frac{abc}{\sqrt{2\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}}$ $\geqslant \frac{r}{2R}$, where $r, R$ are the radii of the incircle and circumcircle of $\triangle ABC$, respectively. (2005 Jiangsu Province Mathematical Winter Camp) | To prove in $\triangle ABC$, $\frac{r}{R}=4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$, we need to prove
$$\frac{a b c}{\sqrt{2\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}} \geqslant \frac{r}{2 R}$$
it suffices to prove
$$\frac{a b c}{\sqrt{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,829 |
$$\square$$ Example 11 In a non-obtuse triangle $ABC$, prove the inequality:
$$\begin{array}{c}
\frac{(1-\cos 2A)(1-\cos 2B)}{1-\cos 2C}+\frac{(1-\cos 2C)(1-\cos 2A)}{1-\cos 2B}+ \\
\frac{(1-\cos 2B)(1-\cos 2C)}{1-\cos 2A} \geqslant \frac{9}{2} .(2006 \text{ China National Training Team Problem })
\end{array}$$ | Prove that if $x=\cot A, y=\cot B, z=\cot C$, then
$$x y+y z+z x=1$$
and $x, y, z \geqslant 0, 1+x^{2}=(x+y)(x+z), 1+y^{2}=(x+y)(y+z)$, $1+z^{2}=(x+z)(y+z)$, and
$$1-\cos 2 A=2 \sin ^{2} A=\frac{2}{1+x^{2}}=\frac{2}{(x+y)(x+z)}$$
Similarly, $1-\cos 2 B=\frac{2}{(x+y)(y+z)}, 1-\cos 2 C=\frac{2}{(x+z)(y+z)}$. Substitut... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,832 |
$\square$ Example 1 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy the following two conditions:
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.
Find the maximum value of $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$. (1997 C... | Solve for any set of $x_{1}, x_{2}, \cdots, x_{1997}$ that satisfies the given conditions. If there exist such $x_{i}$ and $x_{j}$ such that $\sqrt{3}>x_{i} \geqslant x_{j}>-\frac{1}{\sqrt{3}}$, then let $m=\frac{1}{2}\left(x_{i}+x_{j}\right), h_{0}=$ $\frac{1}{2}\left(x_{i}-x_{j}\right)=x_{i}-m=m-x_{j}$
We observe th... | 189548 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,833 |
■ Example 2 For non-negative real numbers $x_{1}$, $x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$. (40th IMO China National Team Selection Exam Question) | Solve using the adjustment method to explore the maximum value of the sum in the problem.
(1) First, for $x, y>0$, we compare $(x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}$ with $x^{4}-x^{5}+y^{4}-y^{5}$:
$$\begin{aligned}
& (x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}-\left(x^{4}-x^{5}+y^{4}-y^{5}\right) \\
= & x y\left(4 x^{2}+6 x y+4 y^{2}\r... | \frac{1}{12} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,834 |
$\square$ Example 3 Let $n$ be a fixed integer, $n \geqslant 2$.
(1) Determine the smallest constant $c$ such that the inequality $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j}^{2}\right) \leqslant$ $c\left(\sum_{1 \leqslant i \leqslant n} x_{i}\right)^{4}$ holds for all non-negative numbers;
(2) ... | Since the inequality is homogeneous and symmetric, we can assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$ $\geqslant 0$, and $\sum_{i=1}^{n} x_{i}=1$. At this point, we only need to discuss $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=$ $\sum_{1 \leqslant i < j \leqslant n} x_{i} x_{j}\left(x_{i}^{2}+x_{j... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,836 |
$\qquad$
Example 4 Given that $a, b, c$ are non-negative real numbers, and $a+b+c=1$, prove: (1$\left.a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2} \geqslant 2$ . (2000 Poland-Austria Mathematical Competition Question) | Proof: Suppose $a, b, c$ satisfy the given conditions, then $0, a+b, c$ are also three numbers that satisfy the conditions. Let the left side of the original inequality be $f(a, b, c)$, then
$$\begin{aligned}
& f(a, b, c)-f(0, a+b, c) \\
= & \left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2}-\le... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 733,837 |
$\square$ Example 2 Proof: Given two positive numbers $p, q$ and $p \leqslant q$, then for any $\alpha, \beta, \gamma$, $\delta, \varepsilon \in [p, q]$, we have
$$(\alpha+\beta+\gamma+\delta+\varepsilon)\left(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta}+\frac{1}{\varepsilon}\right) \leqslant 25+6... | Prove that for given positive numbers $u, v$, considering the function $f(x)=(u+x)\left(v+\frac{1}{x}\right), 0 \leqslant p \leqslant x \leqslant q$, it can be shown that for any $x \in [p, q]$, we have $f(x) \leqslant \max \{f(p), f(q)\}$.
In fact, without loss of generality, assume $p < q$, and let $\lambda = \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,839 |
$\square \square$ Example 3 (Kantorovich Inequality) Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, $\lambda_{1}, \lambda_{2}, \cdots, \lambda_{n}$ are real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1, 0<\lambda_{1} \leqslant \lambda_{2} \leqslant \cdots \leqslant \lambda_{n}$. Prove: $\sum_{i=1}^{n} \frac... | To prove that the conclusion of the problem has a discriminant structure, we construct the corresponding quadratic function, let
$$f(x)=\left(\sum_{i=1}^{n} \frac{a_{i}}{\lambda_{i}}\right) x^{2}-\left(\frac{\lambda_{1}+\lambda_{2}}{\sqrt{\lambda_{1} \lambda_{2}}}\right) x+\left(\sum_{i=1}^{n} a_{i} \lambda_{i}\right) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,840 |
Example 4 Find all real numbers $k$ such that $a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant k(a+b+c+d)$ holds for all $a, b, c, d \in[-1,+\infty$ ).
(2004 China Western Mathematical Olympiad) | When $a=b=c=d=-1$, we have $-3 \geqslant k(-4)$, so $k \geqslant \frac{3}{4}$. When $a=b=c=d=\frac{1}{2}$, we have $4 \times \frac{1}{8}+1 \geqslant k\left(4 \times \frac{1}{2}\right)$, so $k \leqslant \frac{3}{4}$. Therefore, $k=\frac{3}{4}$. Now we prove that
$$a^{3}+b^{3}+c^{3}+d^{3}+1 \geqslant \frac{3}{4}(a+b+c+d)... | \frac{3}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,841 |
$\square$ Example 5 Given that $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}$ are positive real numbers, prove: $\left(a_{1} b_{2}+a_{2} b_{1} + a_{2} b_{3}+a_{3} b_{2}+a_{3} b_{1}+a_{1} b_{3}\right)^{2} \geqslant 4\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right)\left(b_{1} b_{2}+b_{2} b_{3}+ b_{3} b_{1}\right)$. And prove... | Prove that
$$
f(x)=\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{1}\right) x^{2}-\left(a_{1} b_{2}+a_{2} b_{1}+a_{2} b_{3}\right.
$$
$$\begin{aligned}
& \left.+a_{3} b_{2}+a_{3} b_{1}+a_{1} b_{3}\right) x+\left(b_{1} b_{2}+b_{2} b_{3}+b_{3} b_{1}\right) \\
= & \left(a_{1} x-b_{1}\right)\left(a_{2} x-b_{2}\right)+\left(a_{2} x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,842 |
$\square$ Example 6 Let $a_{1}, a_{2}, \cdots, a_{n}$ and $b_{1}, b_{2}, \cdots, b_{n}$ be real numbers, satisfying $\left(a_{1}^{2}+a_{2}^{2}+\right.$ $\left.\cdots+a_{n}^{2}-1\right)\left(b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}-1\right)>\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}-1\right)^{2}$. Prove that $a_{1}^{2... | Proof by contradiction. Assume $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}<1$ and $b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}<1$, construct the quadratic function
$$\begin{aligned}
f(x) & =(x-1)^{2}-\sum_{k=1}^{n}\left(a_{k} x-b_{k}\right)^{2} \\
& =\left(1-\sum_{k=1}^{n} a_{k}^{2}\right) x^{2}-2\left(1-\sum_{k=1}^{n} a_{k} b_{k}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,843 |
$\square$ Example 9 Let $x, y, z$ be positive real numbers, and $x+y+z=1$. Find the minimum value of the function $f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}$, and provide a proof. (2003 Hunan Province Mathematics Competition Question) | Consider the function $g(t)=\frac{t}{1+t^{2}}$, we know that $g(t)$ is an odd function. Since when $t>0$, $\frac{1}{t}+t$ is decreasing in $(0,1)$, it is easy to see that $g(t)=\frac{1}{t+\frac{1}{t}}$ is increasing in $(0,1)$. For $t_{1}, t_{2} \in(0,1)$ and $t_{1} \leqslant t_{2}$, we have
$$\left(t_{1}-t_{2}\right)\... | 0 | Inequalities | proof | Yes | Yes | inequalities | false | 733,847 |
$\square$ Example 10 Let $a, b, c, d$ be non-negative real numbers satisfying $a b+b c+c d+d a=1$. Prove that: $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{c+d+a}+\frac{c^{3}}{d+a+b}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}$. (31st IMO Shortlist Problem) | Let $S=a+b+c+d$, then $S>0$. Construct the function $f(x)=\frac{x^{2}}{S-x}$. Since this function is increasing on $[0, S)$, for any $x \in[0, S)$, we have
$$\left(x-\frac{S}{4}\right)\left[f(x)-f\left(\frac{S}{4}\right)\right] \geqslant 0$$
Thus, $\frac{x^{3}}{S-x}-\frac{S x^{2}}{4(S-x)}-\frac{S x}{12}+\frac{S^{2}}{4... | \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 733,848 |
Example 12 Let $x_{1}, x_{2}, \cdots, x_{n} \geqslant 0$ satisfy $\sum_{i=1}^{n} \frac{1}{1+x_{i}}=1$, prove that: $\sum_{i=1}^{n} \frac{x_{i}}{n-1+x_{i}^{2}} \leqslant 1$. (2004 Chinese National Training Team Problem) | Prove that for $y_{i}=\frac{1}{1+x_{i}}(i=1,2, \cdots, n)$, thus $x_{i}=\frac{1}{y_{i}}-1$, and $\sum_{i=1}^{n} y_{i}=1$, we have
$$\begin{aligned}
& \sum_{i=1}^{n} \frac{x_{i}}{n-1+x_{i}^{2}}=\sum_{i=1}^{n} \frac{\frac{1}{y_{i}}-1}{n-1+\left(\frac{1}{y_{i}}-1\right)^{2}} \\
= & \sum_{i=1}^{n} \frac{y_{i}-y_{i}^{2}}{n ... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 733,850 |
$\square$ Example 14 Let $x, y, z$ be positive real numbers, and $x y z=1$, prove:
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geqslant \frac{3}{4} .$$
(IMO Shortlist) | Prove that the original inequality is equivalent to
$$x^{3}+x^{4}+y^{3}+y^{4}+z^{3}+z^{4} \geqslant \frac{3}{4}(1+x)(1+y)(1+z)$$
Since for any positive numbers $u, v, w$, we have $u^{3}+v^{3}+w^{3} \geqslant 3 u v w$, we will prove the stronger inequality
$$x^{3}+x^{4}+y^{3}+y^{4}+z^{3}+z^{4} \geqslant \frac{1}{4}\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,852 |
Example 15 Let $a, b, c$ be positive real numbers, and satisfy $a b c=1$, try to prove: $\frac{1}{a^{3}(b+c)}+$ $\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. (Problem 36th IMO) | Let \( f(a, b, c) = \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \). Clearly, when \( a = b = c = 1 \), \( f(a, b, c) = \frac{3}{2} \). Since \( a, b, c \) are positive real numbers and satisfy \( abc = 1 \), one of \( a, b, c \) must be no greater than 1. Without loss of generality, assume \( 0 < a \le... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 733,853 |
$\square$ Example 16 Given that $\alpha, \beta$ are the two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, and the function $f(x)=\frac{2 x-t}{x^{2}+1}$ has a domain of $[\alpha, \beta]$.
(1) Find $g(t)=\max f(x)-\min f(x)$;
(2) Prove that for $u_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2,3)... | (1) $f^{\prime}(x)=\frac{2\left(x^{2}+1\right)-2 x(2 x-t)}{\left(x^{2}+1\right)^{2}}=\frac{2 x^{2}+2 t x+2}{\left(x^{2}+1\right)^{2}}$.
Given that $\alpha, \beta$ are the two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, we know $4 x^{2}-4 t x-1=4(x-\alpha)(x-\beta)$. Since $x \in[\alpha, \... | \frac{9}{7} \sqrt{2}<\frac{3}{4} \sqrt{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,854 |
Example 17 Let $0<\alpha, \beta, \gamma<\frac{\pi}{2}$, and $\sin ^{3} \alpha+\sin ^{3} \beta+\sin ^{3} \gamma=1$, prove that: $\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma \geqslant \frac{3}{\sqrt[3]{9}-1}$. (2005 China Southeast Mathematical Olympiad problem strengthened) | To prove that for $x=\sin ^{3} \alpha, y=\sin ^{3} \beta, z=\sin ^{3} \gamma$, the inequality is equivalent to proving under the conditions $x, y, z>0$, and $x+y+z=1$:
$$\frac{\sqrt[3]{x^{2}}}{1-\sqrt[3]{x^{2}}}+\frac{\sqrt[3]{y^{2}}}{1-\sqrt[3]{y^{2}}}+\frac{\sqrt[3]{z^{2}}}{1-\sqrt[3]{z^{2}}} \geqslant \frac{3}{\sqrt... | \frac{3}{\sqrt[3]{9}-1} | Inequalities | proof | Yes | Yes | inequalities | false | 733,855 |
Example 18 For each positive integer $n$, let $p_{n}=\left(1+\frac{1}{n}\right)^{n}, P_{n}=\left(1+\frac{1}{n}\right)^{n+1}, h_{n}=\frac{2 p_{n} P_{n}}{p_{n}+P_{n}}$, prove: $h_{1}<h_{2}<\cdots<h_{n}<h_{n+1}$. (6th Annual Putnam Mathematical Competition) | It is easy to derive that $h_{n}=\frac{2(n+1)^{n+1}}{n^{n}(2 n+1)}$. Consider the function $g(x)$ defined by:
$$g(x)=\ln 2+(x+1) \ln (x+1)-x \ln x-\ln (2 x+1)$$
Then, for $0<x<+\infty$, we have
$$\begin{array}{c}
g^{\prime}(x)=\ln (x+1)-\ln x-\frac{2}{2 x+1} \\
g^{\prime \prime}(x)=\frac{1}{x+1}-\frac{1}{x}+\frac{4}{(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,856 |
Example 19 Let $n \in \mathbf{N}^{*}, x_{0}=0, x_{i}>0, i=1,2,3, \cdots, n$, and $\sum_{i=1}^{n} x_{i}$ $=1$, prove that:
$$1 \leqslant \sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{0}+x_{1}+x_{2}+\cdots+x_{i-1}} \cdot \sqrt{x_{i}+\cdots+x_{n}}}<\frac{\pi}{2} .$$
(1996 China Mathematical Olympiad Problem) | Prove: First, state a fact: Let $f(x)$ be a non-negative function defined on $[a, b]$ and monotonic increasing on $[a, b]$. Let $x_{1}, x_{2}, \cdots, x_{n}$ be the lengths of $n$ subintervals, which form a partition of $[a, b]$. Corresponding to each subinterval is a rectangle, and the sum of the areas of these rectan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,858 |
$\square$ Example 1 Let $n(n \geqslant 3)$ be an integer, and $t_{1}, t_{2}, \cdots, t_{n}$ be positive real numbers, satisfying $n^{2}+$ $1>\left(t_{1}+t_{2}+\cdots+t_{n}\right)\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}+\cdots+\frac{1}{t_{n}}\right)$. Prove: For all integers $i, j, k$ satisfying $1 \leqslant i<j<k$ $\leqsl... | Assume that among $t_{1}, t_{2}, \cdots, t_{n}$, there are three lengths that cannot form a triangle, let's say $t_{1}, t_{2}, t_{3}$, and $t_{1}+t_{2} \leqslant t_{3}$. Then,
$$\begin{aligned}
& \left(t_{1}+t_{2}+\cdots+t_{n}\right)\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}+\cdots+\frac{1}{t_{n}}\right) \\
= & \sum_{1<i<j<... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,859 |
Example 2 For all positive real numbers $a, b, c$, prove: $\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+$ $\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1$
(42nd IMO problem) | Let \( x = \frac{a}{\sqrt{a^{2} + 8 b c}}, y = \frac{b}{\sqrt{b^{2} + 8 c a}}, z = \frac{c}{\sqrt{c^{2} + 8 a b}} \), then \( x, y, z \) are positive numbers, because
\[ x^{2} = \frac{a^{2}}{a^{2} + 8 b c}, y^{2} = \frac{b^{2}}{b^{2} + 8 c a}, z^{2} = \frac{c^{2}}{c^{2} + 8 a b} \]
Thus, \(\frac{1}{x^{2}} - 1 = \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,860 |
$\square$ Example 4 Let $m$ and $n$ be positive integers, and $a_{1}, a_{2}, \cdots, a_{m}$ be distinct elements of $\{1,2, \cdots, n\}$. When $a_{i}+a_{j} \leqslant n, 1 \leqslant i \leqslant j \leqslant m$, there always exists $k(1 \leqslant k \leqslant m)$ such that $a_{i}+a_{j}=a_{k}$. Prove that $\frac{a_{1}+a_{2}... | To prove: For any $i$, when $1 \leqslant i \leqslant m$, we have
$$a_{i}+a_{m+1-i} \geqslant n+1$$
We use proof by contradiction. Suppose there exists some $i, 1 \leqslant i \leqslant m$ such that
$$a_{i}+a_{m+1-i} \leqslant n .$$
By $a_{1}>a_{2}>\cdots>a_{m}$, we get
$$a_{i}<a_{i}+a_{m}<a_{i}+a_{m-1}<\cdots<a_{i}+a_... | \frac{a_{1}+a_{2}+\cdots+a_{m}}{m} \geqslant \frac{n+1}{2} | Combinatorics | proof | Yes | Yes | inequalities | false | 733,862 |
$\square$ Example 5 Let $a, b, c \in \mathbf{R}^{+}$, when $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $a+b+c \leqslant 3$. (2003 Iran Mathematical Olympiad Problem) | Proof: We use proof by contradiction to prove: If $a+b+c>3$, then $a^{2}+b^{2}+c^{2}+$ $a b c>4$. $\square$
By Schur's inequality, we have
$$\begin{aligned}
& 2(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)+9 a b c-(a+b+c)^{3} \\
= & a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b) \geqslant 0
\end{aligned}$$
Thus, $2(a+b+c)\left(a^{2}+b^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,863 |
$\square \square$ Example 2 Let $0<a_{i} \leqslant a, i=1,2, \cdots, n$, prove:
(1) When $n=4$, we have
$$\frac{\sum_{i=1}^{4} a_{i}}{a}-\frac{a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+a_{4} a_{1}}{a^{2}} \leqslant 2 ;$$
(2) When $n=6$, we have
$$\frac{\sum_{i=1}^{6} a_{i}}{a}-\frac{a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{6} a_{1}... | Prove (1) Rewrite the inequality as
$$a_{1}\left(a-a_{2}\right)+a_{2}\left(a-a_{3}\right)+a_{3}\left(a-a_{4}\right)+a_{4}\left(a-a_{1}\right) \leqslant 2 a^{2} .$$
Each term on the left side represents the area of one of the rectangles in the figure. These 4 rectangles can cover the square with side $a$ at most twice,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,865 |
Example 3 Let $0 \leqslant a, b, c \leqslant 1$, prove: $\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+$ $(1-a)(1-b)(1-c) \leqslant 1$. (9th USA Mathematical Olympiad Problem) | To prove that if we directly find a common denominator, the problem will become very complicated. Without loss of generality, we can assume \(0 \leqslant a \leqslant b \leqslant c \leqslant 1\), 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 | 733,868 |
For three distinct real numbers $a_{1}, a_{2}, a_{3}$, define three real numbers $b_{1}, b_{2}, b_{3}$ as follows:
$b_{j}=\left(1+\frac{a_{j} a_{i}}{a_{j}-a_{i}}\right)\left(1+\frac{a_{j} a_{k}}{a_{j}-a_{k}}\right)$, where $\{i, j, k\}=\{1,2,3\}$.
Prove: $1+\left|a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}\right| \leqslant\lef... | Let \( A = \frac{a_{1} a_{2}}{a_{1} - a_{2}}, B = \frac{a_{1} a_{3}}{a_{1} - a_{3}}, C = \frac{a_{2} a_{3}}{a_{2} - a_{3}} \), then
\[
\begin{aligned}
& a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} \\
= & a_{1}(1 + A)(1 + B) + a_{2}(1 - A)(1 + C) + a_{3}(1 - B)(1 - C) \\
= & a_{1} + a_{2} + a_{3} + (a_{1} - a_{2}) A + (a_{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,870 |
■ Example 2 Let $P(x)$ be a polynomial with real coefficients $P(x)=a x^{3}+b x^{2}+c x+d$. Prove: If for any $|x|<1$, we have $|P(x)| \leqslant 1$, then $|a|+|b|+|c|+|d|$ $\leqslant 7$. (37th IMO Preliminary Problem) | Prove that if $P(x)$ is a continuous function and $|x|<1$, then $|P(x)| \leqslant 1$, hence $|x| \leqslant 1$ implies $|P(x)| \leqslant 1$. By setting $x=\lambda$ and $\frac{\lambda}{2}$ (where $\lambda= \pm 1$), we get $\mid \lambda a+b+\lambda c+d|\leqslant 1,| \frac{\lambda}{8} a+\frac{1}{4} b+\frac{\lambda}{2} c+d|... | 7 | Algebra | proof | Yes | Yes | inequalities | false | 733,872 |
- Example 4 Given $x_{i} \in \mathbf{R}(i=1,2, \cdots, n)$, satisfying $\sum_{i=1}^{n}\left|x_{i}\right|=1, \sum_{i=1}^{n} x_{i}$ $=0$, prove: $\left|\sum_{i=1}^{n} \frac{x_{i}}{i}\right| \leqslant \frac{1}{2}-\frac{1}{2 n}$. (1989 National High School Mathematics League Question) | Proof: Let $x_{1}, x_{2}, \cdots, x_{n}$ be real numbers where the positive ones are $x_{k_{1}}, x_{k_{2}}, \cdots, x_{k_{l}}$, and the non-positive ones are $x_{k_{l+1}}, x_{k_{l+2}}, \cdots, x_{k_{n}}$. From the given conditions, we have
$$\sum_{i=1}^{l} x_{k_{i}}=\frac{1}{2}, \sum_{i=l+1}^{n} x_{k_{i}}=-\frac{1}{2}$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,874 |
Example 5 Given the functions $F(x)=a x^{2}+b x+c$ and $G(x)=c x^{2}+b x+a$, where $|F(0)| \leqslant 1,|F(1)| \leqslant 1,|F(-1)| \leqslant 1$, prove that for $|x| \leqslant 1$, we have $|F(x)| \leqslant \frac{5}{4},|G(x)| \leqslant 2$. (26th IMO Shortlist) | Prove (1) $F(0)=c, F(-1)=a-b+c, F(1)=a-b+c$, so
$$F(x)=\frac{x(x-1)}{2} F(-1)-\left(x^{2}-1\right) F(0)+\frac{x(x+1)}{2} F(1),$$
Hence $2|F(x)|=|x(x-1)| \cdot|F(-1)|+2\left|x^{2}-1\right| \cdot|F(0)|$
$$\begin{aligned}
& +|x(x+1)| \cdot|F(1)| \\
\leqslant & |x(x-1)|+2\left|x^{2}-1\right|+|x(x+1)|
\end{aligned}$$
For ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,875 |
$\square$ Example 6 Given $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 3)$ are all greater than 1, and $\left|a_{k+1}-a_{k}\right|<1, k=1,2, \cdots, n-1$, prove: $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{a_{1}}<2 n-1$. 21st All-Russian Mathematical Olympiad Problem | Prove that it is easy to get $a_{n}-a_{1}=\sum_{j=1}^{n-1}\left(a_{j+1}-a_{j}\right)$, the original inequality is equivalent to
$$\sum_{j=1}^{n-1} \frac{a_{j}-a_{j+1}}{a_{j+1}}+\frac{a_{n}-a_{1}}{a_{1}}<n-1 \text { or } \sum_{j=1}^{n-1}\left(a_{j}-a_{j+1}\right)\left(\frac{1}{a_{j+1}}-\frac{1}{a_{1}}\right)<n-1 .$$
Ho... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,876 |
$\square$ Example 8 Let $a_{1}, a_{2}, \cdots, a_{n}$ be real numbers, $s$ be a non-negative number, satisfying $a_{1} \leqslant a_{2} \leqslant \cdots$ $\leqslant a_{n}, a_{1}+a_{2}+\cdots+a_{n}=0,|a_{1}|+|a_{2}|+\cdots+|a_{n}|=s$, prove that: $a_{n}-$ $a_{1} \geqslant \frac{2 s}{n}$. (1996 Australian Mathematical Oly... | Proof: When $s=0$, the inequality obviously holds. Therefore, we may assume $s>0$. From $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right|=s$, we know that at least one of $a_{1}, a_{2}, \cdots, a_{n}$ is not zero. Since $a_{1}+a_{2}+\cdots+a_{n}=0$, there are at least two non-zero terms among $a_{1}, a_{... | \delta \geqslant \frac{2 s}{n} | Inequalities | proof | Yes | Yes | inequalities | false | 733,878 |
Example 1
(1) Let $x, y, z$ be positive numbers, then $\frac{y^{2}-x^{2}}{z+x}+\frac{z^{2}-y^{2}}{x+y}+\frac{x^{2}-z^{2}}{y+z} \geqslant 0$. (W. Janous Conjecture)
(2) Let $x, y, z$ be positive numbers, then $\frac{y^{2}-z x}{z+x}+\frac{z^{2}-x y}{x+y}+\frac{x^{2}-y z}{y+z} \geqslant 0$. (Problem from the Chinese journ... | Proof (1) Let $z+x=a, x+y=b, y+z=c$, then
$$\begin{array}{c}
x+y+z=\frac{1}{2}(a+b+c) \\
x=\frac{1}{2}(a+b-c), y=\frac{1}{2}(b+c-a), z=\frac{1}{2}(c+a-b)
\end{array}$$
Thus, the original inequality can be transformed into
$$\frac{b c}{a}+\frac{c a}{b}+\frac{a b}{c} \geqslant a+b+c$$
By the AM-GM inequality, we have
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,879 |
$\square$ Example 2 For real numbers $x_{1}, y_{1}$, $z_{1}, x_{2}, y_{2}, z_{2}$ satisfying $x_{i}>0, y_{i}>0, x_{i} y_{i}-z_{i}^{2}>0(i=1,2)$, the following inequality holds:
$$\frac{8}{\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2}} \leqslant \frac{1}{x_{1} y_{1}-z_{1}^{2}}+\frac{1}{x_... | Proof: Let $a=x_{1} y_{1}-z_{1}^{2}, b=x_{2} y_{2}-z_{2}^{2}$, then
$$\begin{aligned}
& \left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)-\left(z_{1}+z_{2}\right)^{2} \\
= & a+b+x_{1} y_{2}+x_{2} y_{1}-2 z_{1} z_{2} \\
= & a+b+2 \sqrt{a b}+\left(\frac{x_{1}}{x_{2}} b+\frac{x_{2}}{x_{1}} a-2 \sqrt{a b}\right)+\frac{x_{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,881 |
Example 3 If $x_{i}>0(i=1,2, \cdots, n), n \geqslant 2$, and $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+$ $\frac{1}{1+x_{n}}=1$, prove that: $x_{1} x_{2} \cdots x_{n} \geqslant(n-1)^{n}$. | Proof: Let $\frac{1}{1+x_{1}}=\frac{a_{1}}{a_{1}+a_{2}+\cdots+a_{n}}, \frac{1}{1+x_{2}}=\frac{a_{2}}{a_{1}+a_{2}+\cdots+a_{n}}$, $\cdots, \frac{1}{1+x_{n}}=\frac{a_{n}}{a_{1}+a_{2}+\cdots+a_{n}}$, where $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers, then
$$\begin{array}{c}
x_{1}=\frac{a_{2}+a_{3}+\cdots+a_{n}}{a_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,882 |
$\square$ Example 4 Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$, prove: $0 \leqslant y z+$ $z x+x y-2 x y z \leqslant \frac{7}{27}$. (25th IMO Problem) | Assume without loss of generality that $x \geqslant y \geqslant z \geqslant 0$, from $x+y+z=1$ we know $z \leqslant \frac{1}{3}, x+y$ $\geqslant \frac{2}{3}$, thus $2 x y z \leqslant \frac{2}{3} x y \leqslant x y$, so $y z+z x+x y-2 x y z \geqslant 0$. To prove the right side of the inequality, let $x+y=\frac{2}{3}+\de... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,883 |
D Example 5 Let $0 \leqslant \alpha \leqslant 1,0 \leqslant x \leqslant \pi$, prove: $(2 \alpha-1) \sin x+(1-$ $\alpha) \sin (1-\alpha) x \geqslant 0$. (1983 Swiss Mathematical Olympiad Problem) | Prove that when $\alpha=0,1$ and $x=0$, the inequality obviously holds.
When $0<\alpha<1$ and $x>0$, then $\frac{\sin x}{x} \leqslant$ $\frac{\sin (1-\alpha) x}{(1-\alpha) x}$; i.e., $\square$
$$(1-\alpha) \sin x \leqslant \sin (1-\alpha) x$$
Thus,
$$(1-\alpha) \sin (1-\alpha) x \geqslant(1-\alpha)^{2} \sin x=\left(1-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,884 |
Example 6 Given that $a, b, c$ are all positive numbers, and satisfy $\frac{a^{2}}{1+a^{2}}+\frac{b^{2}}{1+b^{2}}+\frac{c^{2}}{1+c^{2}}=$ 1, prove: $a b c \leqslant \frac{\sqrt{2}}{4}$. (31st IMO Chinese National Training Team Problem) | Let $a=\tan \alpha, b=\tan \beta, c=\tan \gamma, \alpha, \beta, \gamma \in\left(0, \frac{\pi}{2}\right)$, then the given condition can be transformed into $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1$, so
$$\cos ^{2} \alpha=\sin ^{2} \beta+\sin ^{2} \gamma \geqslant 2 \sin \beta \sin \gamma$$
$$\begin{array}{l}... | a b c \leqslant \frac{\sqrt{2}}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 733,885 |
$\square$
Example 7 Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$, prove: for every $n \in$ $\mathbf{N}^{*},(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1}$. (1988 National High School Mathematics League Exam Question) | Prove that if $a=\sec ^{2} \theta, b=\csc ^{2} \theta, 0<\theta<\frac{\pi}{2}$, then $\frac{1}{a}+\frac{1}{b}=1$, the original inequality is equivalent to
i.e., $\square$
$$\begin{array}{c}
\left(a^{n}-1\right)\left(b^{n}-1\right) \geqslant\left(2^{n}-1\right)^{2} \\
\left(\sec ^{2 n} \theta-1\right)\left(\csc ^{2 \pi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,886 |
$\square$ Example 8 Given real numbers $a, b, c, d, e$ satisfy the equations $a+b+c+d+e=8$, $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=16$, prove: $0 \leqslant e \leqslant \frac{16}{5}$.
(7th USAMO Problem) | Prove that considering $e$ as a constant, let $a=\frac{8-e}{4}+t_{1}, b=\frac{8-e}{4}+t_{2}, c=\frac{8-e}{4}$ $+t_{3}, d=\frac{8-e}{4}+t_{4}$, where $t_{1}+t_{2}+t_{3}+t_{4}=0$.
From $16-e^{2}=a^{2}+b^{2}+c^{2}+d^{2}=\frac{(8-e)^{2}}{4}+t_{1}^{2}+t_{2}^{2}+t_{3}^{2}+t_{4}^{2} \geqslant$ $\frac{(8-e)^{2}}{4}$, i.e., $1... | 0 \leqslant e \leqslant \frac{16}{5} | Algebra | proof | Yes | Yes | inequalities | false | 733,887 |
$\square$ Example 9 Let $x, y, z$ be non-negative real numbers, and $x+y+z=1$, prove: $0 \leqslant y z+$ $z x+x y-2 x y z \leqslant \frac{7}{27}$. (25th IMO Problem) | Proof: By symmetry, we may assume $x \geqslant y \geqslant z$, thus $1=x+y+z \geqslant 3 z$, which means $z \leqslant \frac{1}{3}$. Therefore, $2 x y z \leqslant \frac{2}{3} x y \leqslant x y$, hence
$$y z+z x+x y-2 x y z \geqslant 0$$
For the right inequality, we can consider using the mean substitution: Let $x+y=\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,888 |
$\square$ Example 11 Prove: For any positive real numbers $a, b, c$, we have $\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+\right.$ $2) \geqslant 9(a b+b c+c a) .(2004$ Asia Pacific Mathematical Olympiad Problem) | Let $a=\sqrt{2 \tan A}, b=\sqrt{2 \tan B}, c=\sqrt{2 \tan C}$, where $A, B, C \in \left(0, \frac{\pi}{2}\right)$. Since $1+\tan ^{2} \theta=\sec ^{2} \theta$, the original inequality is equivalent to $\cos A \cos B \cos C(\cos A \sin B \sin C+\sin A \cos B \sin C+\sin A \sin B \cos C) \leqslant \frac{4}{9}$.
Because $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,891 |
■ Example 13 Let $x, y, z$ be positive real numbers. Prove: $(x y+y z+z x)$ $\left[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right] \geqslant \frac{9}{4}$. (1996 Iran Mathematical Olympiad Problem) | Let $p=x+y+z, q=xy+yz+zx, r=xyz$, then by Schur's inequality we have
$$x(x-y)(x-z)+y(y-x)(y-z)+z(z-y)(z-x) \geqslant 0$$
which is equivalent to $(x y+z)^{v}-4(x+y+z)(yz+zx+xy)+9xyz \geqslant 0$, or
$$p^{3}-4pq+9r \geq 0$$
Again by Schur's inequality we have
$$x^{2}(x-y)(x-z)+y^{2}(y-x)(y-z)+z^{2}(z-y)(z-x) \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,893 |
Example 2 Let $a, b, c$ be positive numbers, prove that: $\frac{a b}{(a+c)(b+c)}+\frac{b c}{(b+c)(c+a)}$ $+\frac{c a}{(c+b)(a+b)} \geqslant \frac{3}{4}$. (30th IMO Shortlist Problem) | $$\begin{array}{l}
4[a b(a+b)+b c(b+c)+c a(c+a)] \geqslant 3(a+b)(b+c)(c+a) \\
\Leftrightarrow 4\left[a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right)\right] \\
\geqslant 3\left[a\left(b^{2}+c^{2}\right)+b\left(c^{2}+a^{2}\right)+c\left(a^{2}+b^{2}\right)+2 a b c\right] \\
\Leftrightarrow ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,895 |
- Example 3 Positive numbers $a, b, c$ satisfy $a+b+c=1$, prove: $\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}$ $\leqslant 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)$ (2004 Japan Mathematical Olympiad Problem) | Prove that the original inequality is equivalent to
$$\begin{aligned}
& \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \geqslant \frac{3}{2}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \\
\Leftrightarrow & \frac{b}{a}-\frac{b}{c+a}+\frac{c}{b}-\frac{c}{a+b}+\frac{a}{c}-\frac{a}{b+c} \geqslant \frac{3}{2} \\
\Leftrightarrow & \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,896 |
Example 4 Let $x, y$ be two distinct real numbers, $R=\sqrt{\frac{x^{2}+y^{2}}{2}}, A=\frac{x+y}{2}, G$ $=\sqrt{x y}, H=\frac{2 x y}{x+y}$, determine which of $R-A, A-G, G-H$ is the largest and which is the smallest. (30th IMO Canadian Training Problem) | Solve $A-G$ is maximum, $G-H$ is minimum. Because
$$\begin{array}{l}
\frac{x+y}{2}-\sqrt{x y} \geqslant \sqrt{\frac{x^{2}+y^{2}}{2}}-\frac{x+y}{2} \\
\Leftrightarrow x+y \geqslant \sqrt{\frac{x^{2}+y^{2}}{2}}+\sqrt{x y} \\
\Leftrightarrow(x+y)^{2} \geqslant \frac{x^{2}+y^{2}}{2}+x y+\sqrt{2 x y\left(x^{2}+y^{2}\right)... | A-G \text{ is maximum, } G-H \text{ is minimum} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,897 |
$\square$ Example 5 Prove or disprove: If $x, y$ are real numbers, $y \geqslant 0, y(y+1) \leqslant(x+$ $1)^{2}$, then $y(y-1) \leqslant x^{2}$. (30th IMO Canadian Training Problem) | We prove that when $y \geqslant 0, y(y+1) \leqslant(x+1)^{2}$, we have
$$y(y-1) \leqslant x^{2}$$
If $y \leqslant 1$, then $y(y-1) \leqslant 0$ and (1) is obviously true.
If $y>1$, from $y \geqslant 0$ and $y(y+1) \leqslant(x+1)^{2}$, we get
$$y \leqslant \sqrt{\frac{1}{4}+(x+1)^{2}}-\frac{1}{2}$$
And (1) $\Leftright... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,898 |
Example 6 Prove that for all positive numbers $a, b, c$, we have $\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}$ $+\frac{1}{c^{3}+a^{3}+a b c} \leqslant \frac{1}{a b c}$. (26th USAMO Problem) | Prove that by eliminating the denominator and simplifying, the original inequality is equivalent to
$$a^{6}\left(b^{3}+c^{3}\right)+b^{6}\left(c^{3}+a^{3}\right)+c^{6}\left(a^{3}+b^{3}\right) \geqslant 2 a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right)$$
Since \(2 a^{2} b^{2} c^{2}\left(a^{3}+b^{3}+c^{3}\right) \leqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,899 |
Example 7 Let $a, b, c, d$ be positive real numbers, and satisfy $a b c d=1$, prove that: $\frac{1}{(1+a)^{2}}$ $+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geqslant 1$. (2005 China National Training Team for IMO) | Proof: First, we prove a lemma using the analytical method: Let $a, b$ be positive numbers, then
$$\frac{1}{(1+a)^{2}} + \frac{1}{(1+b)^{2}} \geqslant \frac{1}{1+a b}$$
In fact, $\frac{1}{(1+a)^{2}} + \frac{1}{(1+b)^{2}} \geqslant \frac{1}{1+a b}$
$$\begin{array}{l}
\Leftrightarrow (1+a b)\left[(1+a)^{2}+(1+b)^{2}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,900 |
Example 1
(1) Let $x, y, z$ be three positive real numbers, and $x y z=1$. Prove that: $x^{2}+y^{2}+z^{2}+x y+y z+z x \geqslant 2(\sqrt{x}+\sqrt{y}+\sqrt{z})$.
(2) Let $a, b, c$ be three positive real numbers, and $a b c=1$. Prove that: $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1} \leqslant 1$. | Prove (1) By the binary mean inequality, we have
$$\begin{array}{l}
x^{2}+y z \geqslant 2 \sqrt{x^{2} y z}=2 \sqrt{x \cdot x y z}=2 \sqrt{x}, \\
y^{2}+z x \geqslant 2 \sqrt{y^{2} z x}=2 \sqrt{y \cdot x y z}=2 \sqrt{y}, \\
z^{2}+x y \geqslant 2 \sqrt{z^{2} x y}=2 \sqrt{z \cdot x y z}=2 \sqrt{z},
\end{array}$$
Adding th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,901 |
$\square$ Example 8 Given that $a, b, c, d$ are positive real numbers, prove: $\sqrt{(a+c)^{2}+(b+d)^{2}} \leqslant$ $\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \leqslant \sqrt{(a+c)^{2}+(b+d)^{2}}+\frac{2|a d-b c|}{\sqrt{(a+c)^{2}+(b+d)^{2}}}$. | Proof: Let $\boldsymbol{u}=(a, b), \boldsymbol{v}=(c, d)$, then the inequality
$$\sqrt{(a+c)^{2}+(b+d)^{2}} \leqslant \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}$$
is a triangle inequality constructed by vectors $\boldsymbol{u}$ and $\boldsymbol{v}$, which is obviously true. Therefore, to prove the original inequality, it s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,902 |
$\square$ Example 9 Let $a, b, c, d$ be positive real numbers, and satisfy $a+b+c+d=1$. Prove that: $(1-\sqrt{a})(1-\sqrt{b})(1-\sqrt{c})(1-\sqrt{d}) \geqslant \sqrt{a b c d}$. (2005 Jiangsu Province Mathematical Winter Camp Lecture Question) | To prove the following inequality:
$$(1-\sqrt{a})(1-\sqrt{b}) \geqslant \sqrt{c d}$$
To prove (D), it suffices to prove
$$(1-\sqrt{a})(1-\sqrt{b}) \geqslant \frac{c+d}{2} .$$
And (2) $\Leftrightarrow 2+2 \sqrt{a b}-2 \sqrt{a}-2 \sqrt{b} \geqslant 1-a-b$.
$$\begin{array}{l}
\Leftrightarrow 1+a+b-2 \sqrt{a}-2 \sqrt{b}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,903 |
Example 12 Let positive real numbers $a, b, c$ satisfy $a b c \geqslant 2^{9}$, prove: $\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}$ $+\frac{1}{\sqrt{1+c}} \geqslant \frac{3}{\sqrt{1+\sqrt[3]{a b c}}} \cdot(2004$ China Taiwan Mathematical Olympiad Problem) | We prove its equivalent proposition: Let positive real numbers $a, b, c$ satisfy $abc = k^3$, and $k \geq 8$, then
$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \geqslant \frac{3}{\sqrt{1+k}}$$
From the given, we have
$$\begin{array}{c}
a+b+c \geqslant 3 \cdot \sqrt[3]{abc}=3k \\
ab+bc+ca \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,906 |
$\square$
Example 13 If $a, b, c \in \mathbf{R}$, prove: $\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+\right.$ $\left.c a+a^{2}\right) \geqslant(a b+b c+c a)^{3}$. And determine when equality holds. (31st IMO Preliminary Problem) | To prove that since
$$\begin{array}{l}
a^{2}+a b+b^{2} \geqslant \frac{3}{4}(a+b)^{2}, \\
b^{2}+b c+c^{2} \geqslant \frac{3}{4}(b+c)^{2}, \\
c^{2}+c a+a^{2} \geqslant \frac{3}{4}(c+a)^{2},
\end{array}$$
we only need to prove the inequality
$$27(a+b)^{2}(b+c)^{2}(c+a)^{2} \geqslant 64(a b+b c+c a)^{3}$$
which is equiv... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,907 |
$$\square$$ Example 15 Given that $a, b, c, d, k$ are all positive real numbers, and $a, b, c, d \leqslant k$, prove the inequality:
$$\begin{aligned}
& \frac{a^{4}+b^{4}+c^{4}+d^{4}}{(2 k-a)^{4}+(2 k-b)^{4}+(2 k-c)^{4}+(2 k-d)^{4}} \\
\geqslant & \frac{a b c d}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)} .(2002 \text { China Taiwan... | Prove that the original inequality is equivalent to
$$\frac{a^{4}+b^{4}+c^{4}+d^{4}}{a b c d} \geqslant \frac{(2 k-a)^{4}+(2 k-b)^{4}+(2 k-c)^{4}+(2 k-d)^{4}}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)}$$
is equivalent to $\frac{\left(a^{2}-b^{2}\right)^{2}+\left(c^{2}-d^{2}\right)^{2}+2\left(a^{2} b^{2}+c^{2} d^{2}\right)}{a b c d... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,909 |
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