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Example 2 Let real numbers $a, b$ satisfy $a b>0$, prove that $\sqrt[3]{\frac{a^{2} b^{2}(a+b)^{2}}{4}} \leqslant$ $\frac{a^{2}+10 a b+b^{2}}{12}$, and specify the condition for equality.
In general, prove that for any real numbers $a, b$, $\sqrt[3]{\frac{a^{2} b^{2}(a+b)^{2}}{4}} \leqslant$ $\frac{a^{2}+a b+b^{2}}{3}... | Prove that when $ab > 0$, we have
$$\begin{aligned}
\frac{a^{2}+10ab+b^{2}}{12} & =\frac{1}{3}\left[\frac{(a+b)^{2}}{4}+2ab\right] \\
& =\frac{1}{3}\left[\frac{(a+b)^{2}}{4}+ab+ab\right] \geqslant \sqrt[3]{\frac{a^{2}b^{2}(a+b)^{2}}{4}}.
\end{aligned}$$
Equality holds if and only if $\frac{(a+b)^{2}}{4}=ab$, i.e., $a=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,911 |
\square \square Example 17 Let $a, b, c$ be positive real numbers. Prove that:
$\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(a+2 b+c)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(a+b+2 c)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8$. (2003 USA Mathematical Olympiad Problem) | Prove that for a function $f$ of $n$ variables, the symmetric sum is defined as $\sum_{\text {sym }} f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{\sigma} f\left(x_{\sigma(1)}, x_{\sigma(2)}, \cdots, x_{\sigma(n)}\right)$. Here, $\sigma$ is a permutation of $1,2, \cdots, n$, and sym denotes the symmetric sum. For exa... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,912 |
$\square$ Example 1 Let real numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{100} \geqslant 0, a_{1}+$ $a_{2} \leqslant 100, a_{3}+a_{4}+\cdots+a_{100} \leqslant 100$, determine the maximum value of $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}$, and find the sequence $a_{... | $$\begin{aligned}
& \text{Solution: Since } a_{1}+a_{2}+a_{3}+a_{4}+\cdots+a_{100} \leqslant 200, \text{ then} \\
& a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2} \leqslant\left(100-a_{2}\right)^{2}+a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2} \\
= & 100^{2}-200 a_{2}+2 a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2} \\
\leqslant & 100^{2}-\... | 10000 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,913 |
$\square$ Example 2 Real numbers $a, b, c$ and a positive number $\lambda$ make $f(x)=x^{3}+a x^{2}+b x+c$ have three real numbers $x_{1}, x_{2}, x_{3}$, and satisfy:
(1) $x_{2}-x_{1}=\lambda$;
(2) $x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)$.
Find the maximum value of $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$. (2002 Na... | Let $x_{1}=m-\frac{\lambda}{2}, x_{2}=m+\frac{\lambda}{2}, x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)=m$, and let $x_{3}=m+t(\lambda>0, t>0)$.
By Vieta's formulas, we have
$$\left\{\begin{array}{l}
a=-\left(x_{1}+x_{2}+x_{3}\right)=-(3 m+t) \\
b=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=3 m^{2}+2 m t-\frac{\lambda^{2}}{4} \\... | \frac{3 \sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,914 |
$\square$ Example 5 For a fixed $\theta \in\left(0, \frac{\pi}{2}\right)$, find the smallest positive number $a$ that satisfies the following two conditions:
(1) $\frac{\sqrt{a}}{\cos \theta}+\frac{\sqrt{a}}{\sin \theta}>1$;
(2) There exists $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right... | From (1) we get
$$\sqrt{a}>\frac{\sin \theta \cos \theta}{\sin \theta+\cos \theta}$$
Without loss of generality, assume $\frac{a}{\sin ^{2} \theta}+\frac{a}{\cos ^{2} \theta} \leqslant 1$. Condition (2) is equivalent to the existence of $x \in\left[1-\frac{\sqrt{a}}{\sin \theta}, \frac{\sqrt{a}}{\cos \theta}\right]$, ... | \frac{3 \sin ^{2} \theta \cos ^{2} \theta}{1+\sqrt{3} \sin \theta \cos \theta} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,917 |
Example 6 Real numbers $x_{1}, x_{2}, \cdots, x_{2001}$ satisfy $\sum_{k=1}^{2000}\left|x_{k}-x_{k+1}\right|=2001$, let $y_{k}$ $=\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right), k=1,2, \cdots, 2001$, find the maximum possible value of $\sum_{k=1}^{2000} \mid y_{k}-y_{k+1}$. (2001 Shanghai High School Mathematics Comp... | For $k=1,2, \cdots, 2000$, we have
$$\begin{aligned}
& \quad\left|y_{k}-y_{k+1}\right| \\
& =\left|\frac{1}{k}\left(x_{1}+x_{2}+\cdots+x_{k}\right)-\frac{1}{k+1}\left(x_{1}+x_{2}+\cdots+x_{k+1}\right)\right| \\
& \quad=\left|\frac{x_{1}+x_{2}+\cdots+x_{k}-k x_{k+1}}{k(k+1)}\right| \\
& =\frac{\left|\left(x_{1}-x_{2}\ri... | 2000 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,918 |
$\square$ Example 7 Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x^{5}}{y^{2}+z^{2}-y z}+$ $\frac{y^{5}}{z^{2}+x^{2}-z x}+\frac{z^{5}}{x^{2}+y^{2}-x y}$. (2006 Turkish Mathematical Olympiad Problem) | By Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\quad\left(\frac{x^{6}}{x y^{2}+x z^{2}-x y z}+\frac{y^{6}}{z^{2} y+x^{2} y-x y z}+\frac{z^{6}}{x^{2} z+y^{2} z-x y z}\right)\left(x y^{2}\right. \\
\left.+x z^{2}+z^{2} y+x^{2} y+x^{2} z+y^{2} z-3 x y z\right) \geqslant\left(x^{3}+y^{3}+z^{3}\right)^{2},
\end{ar... | \frac{\sqrt{3}}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,919 |
$\square$ Example 8 Given an integer $n$ greater than 3, let real numbers $x_{1}, x_{2}, \cdots, x_{n}, x_{n+1}, x_{n+2}$ satisfy $0<x_{1}<x_{2}<\cdots<x_{n}<x_{n+1}<x_{n+2}$. Try to find the minimum value of
$$\frac{\left(\sum_{i=1}^{n} \frac{x_{i+1}}{x_{i}}\right)\left(\sum_{j=1}^{n} \frac{x_{j+2}}{x_{j+1}}\right)}{\... | Let $t_{i}=\frac{x_{i+1}}{x_{i}}(>1), 1 \leqslant i \leqslant n$, the expression in the problem can be written as
$$\frac{\left(\sum_{i=1}^{n} t_{i}\right)\left(\sum_{i=1}^{n} t_{i+1}\right)}{\left(\sum_{i=1}^{n} \frac{t_{i} t_{i+1}}{t_{i}+t_{i+1}}\right)\left[\sum_{i=1}^{n}\left(t_{i}+t_{i+1}\right)\right]}$$
By the ... | 1 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,920 |
(1) Example 3 $x, y, z$ are positive numbers, prove that: $\frac{x y z\left(x+y+z+\sqrt{x^{2}+y^{2}+z^{2}}\right)}{\left(x^{2}+y^{2}+z^{2}\right)(y z+z x+x y)} \leqslant$ $\frac{3+\sqrt{3}}{9}$. (1997 Hong Kong Mathematical Olympiad Training Team Question) | Prove that first transform the left side of the inequality into an average form.
$$\begin{aligned}
\text { Left }= & \frac{x+y+z+\sqrt{x^{2}+y^{2}+z^{2}}}{\left(x^{2}+y^{2}+z^{2}\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)} \\
= & \frac{x+y+z}{\sqrt{x^{2}+y^{2}+z^{2}}} \cdot \frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}... | \frac{3+\sqrt{3}}{9} | Inequalities | proof | Yes | Yes | inequalities | false | 733,922 |
$\square$ Example 10 Determine the smallest real number $M$ such that the inequality $\mid a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}\right.$ $\left.-c^{2}\right)+c a\left(c^{2}-a^{2}\right) \mid \leqslant M\left(a^{2}+b^{2}+c^{2}\right)^{2}$ holds for all real numbers $a, b, c$. (47th IMO Problem) | Solving because
$$\begin{aligned}
& a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right) \\
= & a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left[\left(c^{2}-b^{2}\right)+\left(b^{2}-a^{2}\right)\right] \\
= & \left(a^{2}-b^{2}\right)(a b-c a)+\left(b^{2}-c^{2}\right)(b c-c... | \frac{9 \sqrt{2}}{32} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,923 |
Example 1 Let the sequence of positive terms $\left\{a_{n}\right\}$ satisfy for any $n \in \mathbf{N}^{\cdot}$, $\sum_{i=1}^{n} a_{i} \geqslant \sqrt{n}$. Prove that for any $n \in \mathbf{N}^{*}$, $\sum_{i=1}^{n} a_{i}^{2} \geqslant \frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)$. (1994 USA Mathe... | Prove that if $b_{i}=a_{i}-(\sqrt{i}-\sqrt{i-1}), i=1,2, \cdots$, then for any $n \in \mathbf{N}^{*}, \sum_{i=1}^{n} a_{i} \geqslant \sqrt{n}$ if and only if for any $n \in \mathbf{N}^{*}, \sum_{i=1}^{n} b_{i}=0$. Clearly,
$$\begin{aligned}
\sum_{i=1}^{n} a_{i}^{2} & =\sum_{i=1}^{n}\left[(\sqrt{i}-\sqrt{i-1})+b_{i}\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,924 |
Example 2: Prove that for any positive integer $n, \frac{2 n}{3} \sqrt{n}<1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n}<\frac{4 n+3}{6} \sqrt{n}$. (13th Putnam Mathematical Competition Question) | We prove the strengthened inequality:
$$\frac{2 n+1}{3} \sqrt{n} - \left(\frac{1}{\sqrt{1}+\sqrt{1}}+\frac{2}{\sqrt{2}+\sqrt{2}}+\cdots+\frac{n-1}{\sqrt{n-1}+\sqrt{n-1}}\right) + n \sqrt{n} \\
= \left(-\frac{1}{2}\right)(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n-1}) + n \sqrt{n} \\
= \left(-\frac{1}{2}\right)(1+\sqrt{2}+\sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,925 |
$\square$ Example 3 Let $a_{1}, a_{2}, \cdots$ be a sequence of positive real numbers, and for all $i, j=1,2, \cdots$, satisfy $a_{i+j} \leqslant a_{i}+a_{j}$. Prove that for any positive integer $n$, we have $a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdots+\frac{a_{n}}{n} \geqslant a_{n}$. (1999 Asia Pacific Mathematical... | Let $S_{i}=a_{1}+a_{2}+\cdots+a_{i}, i=1,2, \cdots, n$, and define $S_{0}=0$, then
$$2 S_{i}=\left(a_{1}+a_{i}\right)+\left(a_{2}+a_{i-1}\right)+\cdots+\left(a_{i}+a_{1}\right) \geqslant i a_{i+1},$$
i.e., $S_{i} \geqslant \frac{i}{2} a_{i+1}$. Therefore,
$$\begin{aligned}
& a_{1}+\frac{a_{2}}{2}+\frac{a_{3}}{3}+\cdot... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,926 |
$\square$ Example 4 For $1 \leqslant i \leqslant n$, let $a_{i}$ and $b_{i}$ be real numbers, satisfying $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant$ $0, b_{1} \geqslant a_{1}, b_{1} b_{2} \geqslant a_{1} a_{2}, b_{1} b_{2} b_{3} \geqslant a_{1} a_{2} a_{3}, \cdots, b_{1} b_{2} \cdots b_{n} \geqsl... | Prove that for the definition $c_{i}=\frac{b_{i}}{a_{i}}, 1 \leqslant i \leqslant n$. Given $c_{1} \geqslant 1, c_{1} c_{2} \geqslant 1, \cdots, c_{1} c_{2} \cdots c_{n} \geqslant 1$, we need to prove $\left(c_{1}-1\right) a_{1}+\left(c_{2}-1\right) a_{2}+\cdots+\left(c_{n}-1\right) a_{n} \geqslant 0$, which is
$$d_{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,927 |
$\square$ Example 5 Given two positive integers $n \geqslant 2$ and $T \geqslant 2$, find all positive integers $a$, such that for any positive numbers $a_{1}, a_{2}, \cdots, a_{n}$, we have $\sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}<T^{2} \sum_{k=1}^{n} \frac{1}{a_{k}}$, where $S_{k}$ $=a_{1}+a_{2}+\cdots+a_{k}... | $$\text { Solve } \begin{aligned}
& \sum_{k=1}^{n} \frac{a k+\frac{a^{2}}{4}}{S_{k}}=\sum_{k=1}^{n} \frac{1}{S_{k}}\left[\left(k+\frac{a}{2}\right)^{2}-k^{2}\right] \\
= & \frac{1}{S_{1}}\left[\left(1+\frac{a}{2}\right)^{2}-1^{2}\right]+\frac{1}{S_{2}}\left[\left(2+\frac{a}{2}\right)^{2}-2^{2}\right] \\
& +\frac{1}{S_{... | 1,2,3, \cdots, 2(T-1) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,928 |
Example 2 The sequence of real numbers $\left\{a_{n}\right\}$ satisfies $a_{1}=\frac{1}{2}, a_{k+1}=-a_{k}+\frac{1}{2-a_{k}}, k=1,2$, $\cdots$, prove the inequality:
$$\left[\frac{n}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}-1\right]^{n} \leqslant\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right)^{n}\left(\frac{1}{a_{1}}-1\... | Prove first by mathematical induction that $0 < a_n \leq \frac{1}{2}$ for $n > 0$, i.e., the proposition holds for $n+1$.
By mathematical induction, $0 < a_n \leq \frac{1}{2}, n=1,2, \cdots$. The original proposition is equivalent to
$$\begin{aligned}
\left(\frac{n}{a_{1}+a_{2}+\cdots+a_{n}}\right)^{n} & \leqslant\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,930 |
$\square$
Example 3 Let $n(n \geqslant 2)$ be an integer, prove: $\sum_{i=1}^{n-1} \frac{n}{n-k} \cdot \frac{1}{2^{k-1}}<4$. (1992 Japan Mathematical Olympiad) | Proof. Let $a_{n}=\sum_{i=1}^{n-1} \frac{n}{n-k} \cdot \frac{1}{2^{k-1}}$, then
$$\begin{aligned}
a_{n+1} & =\sum_{i=1}^{n} \frac{n+1}{n+1-k} \cdot \frac{1}{2^{k-1}}=\frac{n+1}{n}+\frac{n+1}{n-1} \cdot \frac{1}{2}+\cdots+\frac{n+1}{2^{n}} \\
& =\frac{n+1}{n}+\frac{1}{2}\left(\frac{n}{n-1}+\cdots+\frac{n}{2^{n-2}}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,931 |
$\square$ Example 4 Given that $a, b$ are positive numbers, $n \in \mathbf{N}^{*}$ and $n \geqslant 2$, prove: $\frac{a^{n}+a^{n-1} b+a^{n-2} b^{2}+\cdots+a b^{n-1}+b^{n}}{n+1} \geqslant\left(\frac{a+b}{2}\right)^{n}$. (1988 Hunan Province Middle School Mathematics Summer Camp Mathematics Competition Question) | Let $P_{n}=a^{n}+a^{n-1} b+a^{n-2} b^{2}+\cdots+a b^{n-1}+b^{*}, Q_{n}=$ $\left(\frac{a+b}{2}\right)^{n}$. Then prove the original inequality is equivalent to proving $P_{n} \geqslant(n+1) Q_{n}$.
Since $(a+b) P_{n-1}+a^{n}+b^{n}=2 P_{n}$, i.e., $P_{n}=\frac{a+b}{2} P_{n-1}+\frac{a^{n}+b^{n}}{2}$. By the mean value in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,932 |
D Example 4 Given that $a, b, c$ are positive numbers, prove: $\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2} \geqslant(a+b+c) \left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right) \cdot(2005$ UK Mathematical Olympiad problem) | Prove that from $x^{2}+1 \geqslant 2 x$ we get $x^{2} \geqslant 2 x-1$, thus
$$\left(\frac{a}{b}\right)^{2} \geqslant 2 \cdot \frac{a}{b}-1,\left(\frac{b}{c}\right)^{2} \geqslant 2 \cdot \frac{b}{c}-1,\left(\frac{c}{a}\right)^{2} \geqslant 2 \cdot \frac{c}{a}-1$$
Adding the three inequalities, we get
$$\left(\frac{a}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,933 |
■ Example 6 Let $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ be a sequence of positive real numbers satisfying $a_{i-1} a_{i+1} \leqslant a_{i}^{2}(i=1,2, \cdots)$. Prove that for all $n>1$, $\frac{a_{0}+a_{1}+a_{2}+\cdots+a_{n}}{n+1} \cdot \frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1} \geqslant \frac{a_{0}+a_{1}+a_{2}+\cdots... | Given the conditions, we have $\frac{a_{0}}{a_{1}} \leqslant \frac{a_{1}}{a_{2}} \leqslant \cdots \leqslant \frac{a_{n-2}}{a_{n-1}} \leqslant \frac{a_{n-1}}{a_{n}}$; thus,
$$a_{0} a_{n} \leqslant a_{1} a_{n-1} \leqslant a_{2} a_{n-2} \leqslant \cdots$$
Let $S=a_{1}+a_{2}+\cdots+a_{n-1}$, the inequality to be proven ca... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,935 |
$\square$ Example 8 Given the sequence $\left\{a_{n}\right\}$ satisfies ${ }^{\circ} a_{1}=\frac{21}{16}, 2 a_{n}-3 a_{n-1}=\frac{3}{2^{n+1}}, n \geqslant 2$, let $m$ be a positive integer, $m \geqslant 2$. Prove: when $n \leqslant m$, we have $\left(a_{n}+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}$. $\left[m-\left(\frac{2... | From the given conditions, we have \(2^n a_n = 3 \cdot 2^{n-1} a_{n-1} + \frac{3}{4}\). Let \(b_n = 2^n a_n\), for \(n = 1, 2, 3, \cdots\), then \(b_n = 3 b_{n-1} + \frac{3}{4}\), which means
\[ b_n + \frac{3}{8} = 3 \left( b_{n-1} + \frac{3}{8} \right) \]
Since \(b_1 = 2 a_1 = \frac{21}{8}\), we have \(b_n + \frac{3}... | proof | Algebra | proof | Yes | Yes | inequalities | false | 733,937 |
$\square$ Example 1 Given that $D$ is any point on side $AB$ of $\triangle ABC$, and $E$ is any point on side $AC$, connect $DE$, and $F$ is any point on $DE$. Let $\frac{AD}{AB}=x, \frac{AE}{AC}=y, \frac{DF}{DE}=z$, prove:
(1) $S_{\triangle BDF}=(1-x) y z S_{\triangle ABC}, S_{\triangle CEF}=x(1-y)(1-z) S_{\triangle A... | Prove (1) As shown in the figure, $S_{\triangle B D F}=z S_{\triangle B O E}=$
$$\begin{array}{l}
z(1-x) S_{\triangle A B E}=z(1-x) y S_{\triangle A B C} \\
S_{\triangle C E F}=(1-z) S_{\triangle C D E}=(1-z)(1-y) S_{\triangle A C D} \\
=(1-z)(1-y) x S_{\triangle A B C}
\end{array}$$ | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,938 |
(2) $\sqrt[3]{S_{\triangle B D F}}+\sqrt[3]{S_{\triangle C E F}} \leqslant \sqrt[3]{S_{\triangle A B C}}$. (2003 Girls' Mathematical Olympiad Problem) | $\begin{array}{l}\text { (2) } \sqrt[3]{S_{\triangle B D F}}+\sqrt[3]{S_{\triangle C E F}} \\ =[\sqrt[3]{(1-x) y z}+\sqrt[3]{x(1-y)(1-z)}] \sqrt[3]{S_{\triangle A B C}} \\ \leqslant\left[\frac{(1-x)+y+z}{3}+\frac{x+(1-y)+(1-z)}{3}\right] \sqrt[3]{S_{\triangle A B C}}=\sqrt[3]{S_{\triangle A B C}} .\end{array}$ | \sqrt[3]{S_{\triangle A B C}} | Inequalities | proof | Yes | Yes | inequalities | false | 733,939 |
■ Example 3 Prove: The perimeter of any convex quadrilateral with an area of 1, plus the sum of its two diagonals, is not less than $4+2 \sqrt{2}$. (1985 Austrian and Polish Joint Mathematical Competition Question) | Proof: Let quadrilateral $ABCD$ be any convex quadrilateral with an area of 1 (as shown in Figure (1)). Then we have
$$\begin{aligned}
1 & =\frac{1}{2}(eg+gf+fh+he) \sin a \leqslant \frac{1}{2}(eg+gf+fh+he) \\
& =\frac{1}{2}(e+f)(g+h) \leqslant \frac{1}{2}\left(\frac{e+f+g+h}{2}\right)^{2}
\end{aligned}$$
Thus, the su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,941 |
Example 4 (Erdös-Mordell Inequality) Let $P$ be any point inside $\triangle ABC$, and let the distances from $P$ to the sides $BC$, $CA$, and $AB$ be $PD=p$, $PE=q$, and $PF=r$, respectively. Denote $PA=x$, $PB=y$, and $PC=z$. Then $x+y+z \geqslant 2(p+q+r)$. Equality holds if and only if $\triangle ABC$ is an equilate... | Prove that, as shown in the figure, taking the angle bisector of $\angle B$ as the axis of symmetry, the symmetric points of $A$ and $C$ are $A'$ and $C'$, respectively. Connect $A'C'$, $PA'$, and $PC'$. In $\triangle BA'C'$, it is easy to obtain
$$S_{\triangle BA'P} + S_{\triangle BC'P} \leqslant \frac{1}{2} BP \cdot ... | x + y + z \geqslant 2(p + q + r) | Inequalities | proof | Yes | Yes | inequalities | false | 733,942 |
$\square$
Example 5 Let $a, b, c$ be positive numbers, and $a+b+c=1$, prove: $a^{2}+b^{2}+c^{2}+2 \sqrt{3 a b c} \leqslant 1$. (2000 Polish Mathematical Olympiad problem) | Notice that $a^{2} b^{2}+b^{2} c^{2} \geqslant 2 a b^{2} c, b^{2} c^{2}+b^{2} a^{2} \geqslant 2 a c b^{2}, b^{2} a^{2}+$ $a^{2} c^{2} \geqslant 2 a^{2} b c$, so
$$a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geqslant a^{2} b c+a b^{2} c+a b c^{2}$$
Therefore, $(a b+b c+c a)^{2} \geqslant 3\left(a^{2} b c+a b^{2} c+a b c^{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,944 |
$\square$ Example 6 Let quadrilateral $ABCD$ be a convex quadrilateral with an inscribed circle, and each of its interior and exterior angles is no less than $60^{\circ}$. Prove: $\frac{1}{3}\left|AB^{3}-AD^{3}\right| \leqslant\left|BC^{3}-CD^{3}\right| \leqslant 3\left|AB^{3}-AD^{3}\right|$. And determine when equalit... | Proof using the cosine theorem, we know
$$\begin{aligned}
B D^{2} & =A D^{2}+A B^{2}-2 A D \cdot A B \cos \angle D A B \\
& =C D^{2}+B C^{2}-2 C D \cdot B C \cos \angle D C B .
\end{aligned}$$
From the given conditions, we know \(60^{\circ} \leqslant \angle D A B, \angle D C B \leqslant 120^{\circ}\), hence \(-\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,945 |
$\square$ Example 8 Let circles $K$ and $K_{1}$ be concentric, with radii $R, R_{1}$ respectively, and $R_{1}>R$. Quadrilateral $ABCD$ is inscribed in circle $K$, and quadrilateral $A_{1}B_{1}C_{1}D_{1}$ is inscribed in circle $K_{1}$. Points $A_{1}, B_{1}, C_{1}, D_{1}$ lie on the rays $CD, DA, AB$, and $BC$ respectiv... | To make writing convenient, let $AB = a, BC = b, CD = c, DA = d, AB_1 = e, BC_1 = f, CD_1 = g, DA_1 = h$, then
$$S_{\triangle AB_1C_1} = \frac{1}{2}(a+f) e \sin \angle B_1AC_1$$
$\angle DCB = \angle B_1AC_1$
$$\begin{aligned}
S_{\text{quadrilateral } ABCD} & = S_{\triangle ABD} + S_{\triangle BDC} \\
& = \frac{1}{2} a ... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,947 |
$\square$ Example 9 Given a quadrilateral $A_{1} A_{2} A_{3} A_{4}$ that has both an inscribed circle and a circumscribed circle, the inscribed circle touches the sides $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{1}$ at points $B_{1}, B_{2}, B_{3}, B_{4}$ respectively. Prove: $\left(\frac{A_{1} A_{2}}{B_{1} B_{2}}... | Proof As shown in the figure, let the radius of the incircle of quadrilateral $A_{1} A_{2} A_{3} A_{4}$ be $r$, and let
$$\begin{array}{l}
\angle A_{1} A_{2} A_{3}=2 \alpha, \angle A_{2} A_{3} A_{4}=2 \beta \\
\angle A_{3} A_{4} A_{1}=2 \gamma, \angle A_{4} A_{1} A_{2}=2 \theta
\end{array}$$
Then, by the fact that poi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,948 |
$\square$ Example 11 In $\triangle ABC$, the circumcircle $K$ has a radius $R$, and the internal angle bisectors intersect the circle $K$ at $A_{1}, B_{1}, C_{1}$. Prove: $16 Q^{3} \geqslant 27 R^{4} P$. Here, $Q$ and $P$ are the areas of $\triangle A_{1} B_{1} C_{1}$ and $\triangle ABC$, respectively. (30th IMO Shortl... | Proof: Let the internal angles of $\triangle ABC$ be $\alpha, \beta, \gamma$, then
\[ P = \frac{1}{2} R^{2} (\sin 2 \alpha + \sin 2 \beta + \sin 2 \gamma). \]
Since the internal angles of $\triangle A_{1} B_{1} C_{1}$ are $\frac{\beta+\gamma}{2}, \frac{\gamma+\alpha}{2}, \frac{\alpha+\beta}{2}$, we have
\[ Q = \frac{1... | proof | Geometry | proof | Yes | Yes | inequalities | false | 733,950 |
$\square$ Example 13 Draw a line through the point $P(a, b)(a>0, b>0)$ that intersects the $x$-axis and $y$-axis at points $M, N$ respectively. Try to find the range of values for $O M+O N-M N$. | Let $\angle P M O=\theta\left(0-1\right)$ be monotonically decreasing on $\left(-1, \sqrt{\frac{2 a}{b}}-1\right]$ and monotonically increasing on $\left[\sqrt{\frac{2 a}{b}}-1,+\infty\right)$. Considering that $0<t<1$, we discuss in three cases:
(1) When $1<\frac{2 a}{b}<4$, we have $\frac{a}{2}<b<2 a$, thus $0<\sqrt{... | (b, a) \text{ or } (a, b) \text{ or } (\min (a, b), 2(a+b)-2 \sqrt{2 a b}] | Geometry | math-word-problem | Yes | Yes | inequalities | false | 733,952 |
$\square$ Example 6 Let $0<x, y, z<1$, prove: $\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \geqslant$ $\frac{3 \sqrt[3]{x y z}}{1-\sqrt[3]{x y z}}$. 2002 Irish Mathematical Olympiad Problem) | Prove that by the AM-GM inequality,
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \geqslant 3 \cdot \sqrt[3]{\frac{x y z}{(1-x)(1-y)(1-z)}}$$
Below, we prove that $\sqrt[3]{(1-x)(1-y)(1-z)} \leqslant 1-\sqrt[3]{x y z}$.
In fact, by the AM-GM inequality,
$$\begin{array}{c}
\sqrt[3]{x y z} \leqslant \frac{x+y+z}{3} \\
\sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,953 |
Example 4 Given that $a, b, c$ are positive numbers, prove:
(1) $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}$. (1963 Moscow Mathematical Olympiad Problem)
(2) $\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{a+b+c}{2}$.
(2nd World Friendly Standard Mathematics Competition Probl... | Proof 1 (1) Since
$$\begin{aligned}
& \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2} \\
= & \left(\frac{a}{b+c}-\frac{1}{2}\right)+\left(\frac{b}{c+a}-\frac{1}{2}\right)+\left(\frac{c}{a+b}-\frac{1}{2}\right) \\
= & \frac{1}{2}\left(\frac{a-b}{b+c}+\frac{a-c}{b+c}\right)+\frac{1}{2}\left(\frac{b-c}{c+a}+\frac{b-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,955 |
$\qquad$
Example 8 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 \( x = \frac{1}{a^2 + 2}, y = \frac{1}{b^2 + 2}, z = \frac{1}{c^2 + 2} \). By the AM-GM inequality, we have
\[
\begin{aligned}
2 \cdot \frac{1}{c^2 + 2} \left( \frac{a}{a^2 + 2} \cdot \frac{b}{b^2 + 2} \right) & \leq \frac{1}{c^2 + 2} \left[ \frac{a^2}{(a^2 + 2)^2} + \frac{b^2}{(b^2 + 2)^2} \right] \\
& = z \left( ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,956 |
Example 9 Let $a, b, c$ be positive real numbers. Prove that: $\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(a+2 b+c)^{2}}{2 b^{2}+(c+a)^{2}}+$ $\frac{(a+b+2 c)^{2}}{2 c^{2}+(a+b)^{2}} \leqslant 8$. (2003 USA Mathematical Olympiad Problem) | Prove that if $x=a+b, y=b+c, z=c+a$, then
$$\begin{array}{c}
2 a+b+c=z+x, a+2 b+c=x+y, a+b+2 c=y+z \\
2 a=z+x-y, 2 b=x+y-z, 2 c=y+z-x .
\end{array}$$
Thus, to prove the original inequality, we need to prove
$$\frac{2(z+x)^{2}}{2 y^{2}+(z+x-y)^{2}}+\frac{2(y+z)^{2}}{2 x^{2}+(y+z-x)^{2}}+\frac{2(x+y)^{2}}{2 z^{2}+(x+y-z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,957 |
Example 10 Let $a, b, c$ be positive real numbers, find the minimum value of $\frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c}$. (3rd China Girls Mathematical Olympiad problem) | Let
$$\left\{\begin{array}{l}x=a+2 b+c, \\ y=a+b+2 c \\ z=a+b+3 c .\end{array}\right.$$
Then we have $x-y=b-c, z-y=c$, from which we can derive
$$\left\{\begin{array}{l}
a+3 c=2 y-x \\
b=z+x-2 y \\
c=z-y
\end{array}\right.$$
Thus,
$$\begin{aligned}
& \frac{a+3 c}{a+2 b+c}+\frac{4 b}{a+b+2 c}-\frac{8 c}{a+b+3 c} \\
= &... | -17+12 \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,958 |
$\square$ Example 11 An infinite sequence of positive numbers $\left\{x_{n}\right\}$ has the property: $x_{0}=1, x_{i+1} \leqslant x_{i}(i \geqslant 0)$.
(1) Prove that for any sequence with the above property, there exists an $n \geqslant 1$ such that the following inequality holds:
$$\frac{x_{0}^{2}}{x_{2}}+\frac{x_{... | Prove (1) by repeatedly applying the binary mean inequality:
$$\begin{array}{l}
\frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}} \geqslant \frac{x_{0}^{2}}{x_{1}}+\frac{x_{1}^{2}}{x_{2}}+\cdots+\frac{x_{n-2}^{2}}{x_{n-1}}+\frac{x_{n-1}^{2}}{1} \\
\geqslant \frac{x_{0}^{2}}{x_{1}}+\frac{x... | 3.999 | Inequalities | proof | Yes | Yes | inequalities | false | 733,959 |
Example 12 Given that $a, b, c$ are non-negative real numbers, prove: $\left(\frac{a+2 b}{a+2 c}\right)^{3}+\left(\frac{b+2 c}{b+2 a}\right)^{3}+\left(\frac{c+2 a}{c+2 b}\right)^{3} \geqslant 3$. (2004 MOP Problem) | Prove: If $x, y, z$ are non-negative real numbers, then
$$x^{3}+y^{3}+z^{3} \geqslant 3\left(\frac{x+y+z}{3}\right)^{3}$$
That is, $\square$
In fact, let $t=\frac{x+y+z}{3}$, then
$$\begin{array}{c}
x^{3}+t^{3}+t^{3} \geqslant 3 x t t=3 x t^{2}, \\
x^{3}+2 t^{3} \geqslant 3 x t^{2} .
\end{array}$$
Similarly,
$$y^{3}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,960 |
$\square$
Example 1 Let the length, width, and height of a rectangular prism be $a, b, c$, and its diagonal length be $l$. Prove: $\left(l^{4}-a^{4}\right)\left(l^{4}-b^{4}\right)\left(l^{4}-c^{4}\right) \geqslant 512 a^{4} b^{4} c^{4}$. (2002 Hunan Mathematical Olympiad) | Prove that the original inequality is equivalent to
$$\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$$
Let $x=\frac{a^{2}}{l^{2}}, y=\frac{b^{2}}{l^{2}}, z=\frac{c^{2}}{l^{2}}$, then $x+y+z=1$, and the original inequality can be written as
$$\begin{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,964 |
[1 Example 2 Given that $a, b, c$ are positive numbers, and $a b c \leqslant 1$, prove: $\frac{a+b}{c}+\frac{b+c}{a}+$ $\frac{c+a}{b} \geqslant 2(a+b+c) .($ Mathematical Bulletin Problem 1171) | Prove by changing the structure and applying the AM-GM inequality:
$$\begin{aligned}
& \frac{a+b}{c(a+b+c)}+\frac{b+c}{a(a+b+c)}+\frac{c+a}{b(a+b+c)} \\
= & \frac{(a+b+c)-c}{c(a+b+c)}+\frac{(a+b+c)-a}{a(a+b+c)}+\frac{(a+b+c)-b}{b(a+b+c)} \\
= & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{3}{a+b+c} \geqslant 3 \cdot \sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,965 |
Example 3 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$, prove: $\prod_{k=1}^{n}\left(1+\frac{1}{a_{k}}\right) \geqslant\left(1+\frac{1}{n}\right)^{n}$. | Prove that by the AM-GM inequality,
$$\frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \geqslant \sqrt[n]{a_{1} a_{2} \cdots a_{n}}$$
Therefore,
$$\prod_{k=1}^{n} \frac{1}{a_{k}} \geqslant \frac{1}{n^{n}}$$
Considering the equality holds if and only if \(a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}\), we can decompose \(\frac{1}{a_{k}}\) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,967 |
Let $x, y, z$ be 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
$$\begin{aligned}
& \frac{x^{4}}{y(1-y)(1+y)}+\frac{y}{8}+\frac{1-y}{16}+\frac{1+y}{32} \\
\geqslant & 4 \cdot \sqrt[4]{\frac{x^{4}}{y(1-y)(1+y)} \cdot \frac{y}{8} \cdot \frac{1-y}{16} \cdot \frac{1+y}{32... | \frac{1}{8} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,969 |
Example 7 Proof: For all positive real numbers $a, b, c$, we have $\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+$ $\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1$. (42nd IMO Problem) | $$\begin{array}{l}
\left(a^{\frac{4}{3}}+b^{\frac{1}{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{1}{3}}+a^{\frac{4}{3}}+b^{\frac{4}{3}}+c^{\frac{4}{3}}\right) \\
\geqslant 2 b^{\frac{2}{3}} c^{\frac{2}{3}} \cdot 4 a^{\frac{2}{3}} b^{\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,971 |
Example 8 Let $x, y, z, w$ be real numbers, not all zero, find $\frac{x y+2 y z+z w}{x^{2}+y^{2}+z^{2}+w^{2}}$ | Solution: It is clear that we only need to consider the case where $x \geqslant 0, y \geqslant 0, z \geqslant 0, w \geqslant 0$. Introducing undetermined positive constants $\alpha, \beta, \gamma$, we have
$$a^{2} x^{2}+y^{2} \geqslant 2 \alpha x y, \beta^{2} y^{2}+z^{2} \geqslant 2 \beta y z, \gamma^{2} z^{2}+w^{2} \g... | \frac{\sqrt{2}+1}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,972 |
■ Example 9 Let $a, b, c$ be positive real numbers, and satisfy $abc=1$. Prove: $\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1$. (41st IMO Problem) | Prove that for $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, where $x, y, z$ are positive real numbers, the original inequality is equivalent to
$$(x+y-z)(y+z-x)(x+z-y) \leqslant x y z .$$
Let $u=x+z-y, v=x+y-z, w=y+z-x$. Since the sum of any two of these numbers is positive, at most one of them can be non-positive. ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,973 |
$\square$ Example 10 Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, and satisfy $a_{1}+a_{2}+\cdots+a_{n} < 1$, prove: $\frac{a_{1} a_{2} \cdots a_{n}\left[1-\left(a_{1}+a_{2}+\cdots+a_{n}\right)\right]}{\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(1-a_{1}\right)\left(1-a_{2}\right) \cdots\left(1-a_{... | Proof: Let $a_{n+1}=1-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$, obviously $a_{n+1}>0$, thus we obtain $n+1$ positive numbers $a_{1}, a_{2}, \cdots, a_{n}, a_{n+1}$ whose sum is 1, and the inequality becomes
$$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 | 733,974 |
$\square$ Example 11 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 Question) | Prove that since $a, b \in \mathbf{R}^{+}$, and $\frac{1}{a}+\frac{1}{b}=1 \geqslant 2 \sqrt{\frac{1}{a} \cdot \frac{1}{b}}$, it follows that $a b \geqslant 4$. By the binomial theorem, we have $(a+b)^{n}=\sum_{k=0}^{n} \mathrm{C}_{n}^{k} a^{n-k} b^{k}$, so
$$(a+b)^{n}-a^{n}-b^{n}=\sum_{k=1}^{n-1} \mathrm{C}_{n}^{k} a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,975 |
Example 12 Let $a, b, c$ be positive real numbers, and $a+b+c=3$, prove: $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}$ $\geqslant a^{2}+b^{2}+c^{2}$. (2006 Romanian National Training Team Problem) | To prove that $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \geqslant a^{2}+b^{2}+c^{2}$ is equivalent to proving
$$a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \geqslant a^{4} b^{2} c^{2}+b^{4} c^{2} a^{2}+c^{4} a^{2} b^{2}$$
Now we prove
$$(a+b+c)^{4}\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \geqslant 81\left(a^{4}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,976 |
Example 1 Let $x_{1}, x_{2}, \cdots, x_{n}$ be any real numbers, prove: $\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}$ $+\cdots+\frac{x_{n}}{1+x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}<\sqrt{n}$. (42nd IMO Shortlist Problem) | Prove that by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(\frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}\right)^{2} \\
\leqslant & {\left[\left(\frac{x_{1}}{1+x_{1}^{2}}\right)^{2}+\left(\frac{x_{2}}{1+x_{1}^{2}+x_{2}^{2}}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,980 |
$\square$ Example 2 Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, let $S=a_{1}+a_{2}+\cdots+a_{n}$,
then $\sum_{k=1}^{n} \frac{a_{k}}{S-a_{k}} \geqslant \frac{n}{n-1}$. (1976 British Mathematical Olympiad) | Proof Consider $(n-1) S=n S-S=n S-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$ $=\left(S-a_{1}\right)+\left(S-a_{2}\right)+\cdots+\left(S-a_{n}\right)$ and $n^{2}=(1+1+\cdots+1)^{2}$. By Corollary 1, we have
i.e. $\square$
$$\sum_{k=1}^{n}\left(S-a_{k}\right) \sum_{k=1}^{n} \frac{1}{S-a_{k}} \geqslant n^{2}$$
Rearranging g... | \sum_{k=1}^{n} \frac{a_{k}}{S-a_{k}} \geqslant \frac{n}{n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 733,981 |
$\square$ Example 3 Let $a_{1}>a_{2}>\cdots>a_{n}>a_{n+1}$, prove that: $\frac{1}{a_{1}-a_{2}}+\frac{1}{a_{2}-a_{3}}+\cdots$ $+\frac{1}{a_{n}-a_{n+1}}+\frac{1}{a_{n+1}-a_{1}}>0$ | To prove that changing the structure of the original inequality transforms the proof of the original inequality into proving
$$\left(a_{1}-a_{n+1}\right)\left(\frac{1}{a_{1}-a_{2}}+\frac{1}{a_{2}-a_{3}}+\cdots+\frac{1}{a_{n}-a_{n+1}}\right)>1$$
To apply the Cauchy-Schwarz inequality, write \(a_{1}-a_{n+1}\) in the fol... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,982 |
$$\square$$ Example 4 Let $a, b, c, d$ be positive real numbers satisfying $a b + c d = 1$. Points $P_{i}(x_{i}, y_{i})$ $(i=1,2,3,4)$ are four points on the unit circle centered at the origin. Prove:
$$\begin{array}{l}
\quad\left(a y_{1} + b y_{2} + c y_{3} + d y_{4}\right)^{2} + \left(a x_{4} + b x_{3} + c x_{2} + d ... | Let $\alpha=a y_{1}+b y_{2}+c y_{3}+d y_{4}, \beta=a x_{4}+b x_{3}+c x_{2}+d x_{1}$. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\quad\left[\left(\sqrt{a d} y_{1}\right)^{2}+\left(\sqrt{b c} y_{2}\right)^{2}+\left(\sqrt{b c} y_{3}\right)^{2}+\left(\sqrt{a d} y_{4}\right)^{2}\right] \cdot \\
{\left[\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,983 |
$\qquad$
Example 5 Let $a, b, x, y, k$ be positive numbers, and $k<2, a^{2}+b^{2}-k a b=x^{2}+y^{2}$ $-k x y=1$, prove:
$$|a x-b y| \leqslant \frac{2}{\sqrt{4-k^{2}}},|a y+b x-k b y| \leqslant \frac{2}{\sqrt{4-k^{2}}}$$ | Prove that because $a^{2}+b^{2}-k a b=1$, so
$$\left(a-\frac{k b}{2}\right)^{2}+\left(\frac{\sqrt{4-k^{2}}}{2} b\right)^{2}=1$$
Similarly, $\left(\frac{\sqrt{4-k^{2}}}{2} x\right)^{2}+\left(\frac{k x}{2}-y\right)^{2}=1$.
Applying the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& {\left[\left(a-\frac{k b}{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,984 |
$\square$
Example 6 Prove: For any positive real numbers $a, b, c$, we have $\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+\right.$ 2) $\geqslant 9(a b+b c+c a) .(2004$ Asia Pacific Mathematical Olympiad Problem) | Prove that by the Cauchy-Schwarz inequality,
$$\left(a^{2}+2\right)\left(b^{2}+2\right)=\left(a^{2}+1+1\right)\left(1+b^{2}+1\right) \geqslant(a+b+1)^{2},$$
Similarly,
$$\begin{array}{l}
\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant(b+c+1)^{2}, \\
\left(c^{2}+2\right)\left(a^{2}+2\right) \geqslant(c+a+1)^{2}
\end... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,985 |
$\square$ Example 8 Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, prove that $\frac{a_{1}}{1+a_{2}+a_{3}+\cdots+a_{n}}+\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+\cdots+\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a_{n-1}}$ has a minimum value, and calculate it. (1982 Federal German M... | Solve
$$\begin{array}{l}
\frac{a_{1}}{1+a_{2}+a_{3}+\cdots+a_{n}}+1 \\
\quad=\frac{1+\left(a_{1}+a_{2}+a_{3} \cdots+a_{n}\right)}{2-a_{1}} \\
\quad=\frac{2}{2-a_{1}}
\end{array}$$
Similarly,
$$\begin{array}{c}
\frac{a_{2}}{1+a_{1}+a_{3}+\cdots+a_{n}}+1=\frac{2}{2-a_{2}} \\
\cdots \\
\frac{a_{n}}{1+a_{1}+a_{2}+\cdots+a... | \frac{n}{2 n-1} | Inequalities | proof | Yes | Yes | inequalities | false | 733,987 |
Example 8 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$, prove: $\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+$ $\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$. (46th IMO problem) | Prove that because $x y z \geqslant 1$, we have
$$\begin{aligned}
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} & \geqslant \frac{x^{5}-x^{2} \cdot x y z}{x^{5}+\left(y^{2}+z^{2}\right) \cdot x y z} \\
& =\frac{x^{4}-x^{2} y z}{x^{4}+y z\left(y^{2}+z^{2}\right)} \\
& \geqslant \frac{2 x^{4}-x^{2}\left(y^{2}+z^{2}\right)}{2 x^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,988 |
■Example 9 Given that $a, b$ are positive real numbers, and $\frac{1}{a}+\frac{1}{b}=1$, prove: for every $n \in \mathbf{N}^{\cdot}$, we have $(a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1}$. (1988 National High School Mathematics League Question) | Prove that because $\frac{1}{a}+\frac{1}{b}=1$, so
$$a+b=a b, \quad (a-1)(b-1)=1.$$
Since $\frac{1}{a}+\frac{1}{b}=1 \geqslant 2 \sqrt{\frac{1}{a b}}$, it follows that $a b \geqslant 4$, thus
$$\begin{aligned}
& (a+b)^{n}-a^{n}-b^{n}+1 \\
= & (a b)^{n}-a^{n}-b^{n}+1=\left(a^{n}-1\right)\left(b^{n}-1\right) \\
= & (a-1... | (a+b)^{n}-a^{n}-b^{n} \geqslant 2^{2 n}-2^{n+1} | Inequalities | proof | Yes | Yes | inequalities | false | 733,989 |
$\square$ Example 10 Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}^{+}$, prove that: $\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n-1}^{2}}{x_{n}}+\frac{x_{n}^{2}}{x_{1}}$ $\geqslant x_{1}+x_{2}+\cdots+x_{n-1}+x_{n}$. (1984 National High School Mathematics League Question) | Prove that multiplying the left side of the inequality by the factor $x_{2}+x_{3}+\cdots+x_{n}+x_{1}$, it is equivalent to multiplying by the factor $x_{1}+x_{2}+\cdots+x_{n}$. Applying the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n-1}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,990 |
$\square$ Example 11 Positive integers $n \geqslant 3, x_{1}, x_{2}, \cdots, x_{n}$ are positive real numbers, $x_{n+j}=x_{j}(1 \leqslant$ $j \leqslant n-1)$, find the minimum value of $\sum_{i=1}^{n} \frac{x_{j}}{x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}}$. (1995 Chinese National Mathematical Olympiad Training Team Que... | Let $a_{j}^{2}=\frac{x_{j}}{x_{i+1}+2 x_{i+2}+\cdots+(n-1) x_{i+1}}$,
$$b_{j}^{2}=x_{j}\left[x_{j+1}+2 x_{j+2}+\cdots+(n-1) x_{j+n-1}\right]$$
Then $a_{j}, b_{j}$ are positive numbers, and $a_{j} b_{j}=x_{j}$. The problem is to find the minimum value of $S=\sum_{j=1}^{n} a_{j}^{2}$. By the Cauchy-Schwarz inequality, $... | \frac{2}{n-1} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 733,991 |
$\square$ Example 12 Let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$, prove that: $\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)\left(\frac{a_{1}}{a_{2}^{2}+a_{2}}+\frac{a_{2}}{a_{3}^{2}+a_{3}}+\cdots+\frac{a_{n}}{a_{1}^{2}+a_{1}}\right) \geqslant$ $\frac{n}{n+1}$.... | Prove that by the Cauchy-Schwarz inequality,
$$\begin{aligned}
& \left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)\left(\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n}}{a_{1}}\right) \\
\geqslant & \left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}=1
\end{aligned}$$
Thus, $\frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,992 |
Example 13 (1) Let three positive real numbers $a, b, c$ satisfy $\left(a^{2}+b^{2}+c^{2}\right)^{2}>2\left(a^{4}+\right.$ $\left.b^{4}+c^{4}\right)$, prove that $a, b, c$ must be the lengths of the sides of some triangle.
(2) Let $n$ positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\left(a_{1}^{2}+a_{2}^{2... | Prove (2) by Cauchy-Schwarz inequality:
$$\begin{aligned}
& (n-1)\left(a_{1}^{4}+a_{2}^{4}+\cdots+a_{n}^{4}\right)<\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)^{2} \\
= & \left(\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{2}+a_{4}^{2}+\cdots+a_{n}^{2}\right)^{2} \\
\leqslant & (n-1)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,993 |
Example 14 Let $a_{1}, a_{2}, \cdots, a_{n}(n>1)$ be real numbers, and $A+\sum_{i=1}^{n} a_{i}^{2}<$ $\frac{1}{n-1}\left(\sum_{i=1}^{n} a_{i}\right)^{2}$, prove that for $1 \leqslant i<j \leqslant n$, we have $A<2 a_{i} a_{j}$. (38th USA Putnam Mathematical Competition Question) | Prove that by Cauchy-Schwarz inequality,
$$\begin{aligned}
& {\left[\left(a_{1}+a_{2}\right)+a_{3}+\cdots+a_{n}\right]^{2} } \\
\leqslant & \left(1^{2}+1^{2}+\cdots+1^{2}\right) \cdot\left[\left(a_{1}+a_{2}\right)^{2}+a_{3}^{2}+\cdots+a_{n}^{2}\right]
\end{aligned}$$
Therefore,
$$\frac{1}{n-1}\left(\sum_{i=1}^{n} a_{i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,994 |
$\square$ Example 15 Let $\frac{3}{2} \leqslant x \leqslant 5$, prove the inequality: $2 \sqrt{x+1}+\sqrt{2 x-3}+$ $\sqrt{15-3 x}<2 \sqrt{19}$. (2003 National High School Mathematics Competition Problem) | Prove that for $\lambda, \mu \in \mathbf{R}^{+}$, and satisfying $\lambda+2 \mu=3$, by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& (2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x})^{2} \\
= & {\left[2 \frac{1}{\sqrt{\lambda}} \sqrt{\lambda(x+1)}+\frac{1}{\sqrt{\mu}} \sqrt{\mu(2 x-3)}+\sqrt{15-3 x}\right]^{2} } \... | 2 \sqrt{x+1}+\sqrt{2 x-3}+\sqrt{15-3 x}<2 \sqrt{19} | Inequalities | proof | Yes | Yes | inequalities | false | 733,995 |
$\square$ Example 17 Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$ and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} \sqrt{\frac{i}{j}} x_{i} x_{j}$ $=1$, find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$. (2001 National High School Mathematics League Additional Question) | Solve for the minimum value first. Since
$$\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \geqslant \sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant n} \sqrt{\frac{i}{j}} x_{i} x_{j}=1$$
Therefore, $\sum_{i=1}^{n} x_{i} \geqslant 1$, equality holds if and only if there exists an $i... | \left[\sum_{k=1}^{n}(\sqrt{k}-\sqrt{k-1})^{2}\right]^{\frac{1}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,997 |
$\square$ Example 18 Let $2 n$ real numbers $a_{1}, a_{2}, \cdots, a_{2 n}$ satisfy the condition $\sum_{i=1}^{2 n-1}\left(a_{i+1}-a_{i}\right)^{2}=$ 1, find the maximum value of $\left(a_{n+1}+a_{n+2}+\cdots+a_{2 n}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)$. (2003 China Western Mathematical Olympiad Problem) | First, when $n=1$, $\left(a_{2}-a_{1}\right)^{2}=1$, so $a_{2}-a_{1}= \pm 1$. It is easy to see that the maximum value required at this time is 1.
When $n \geqslant 2$, let $x_{1}=a_{1}, x_{i+1}=a_{i+1}-a_{i}, i=1,2, \cdots, 2 n-1$, then $\sum_{i=2}^{2 \pi} x_{i}^{2}=1$, and $a_{k}=x_{1}+x_{2}+\cdots+x_{k}, k=1,2, \cd... | \sqrt{\frac{n\left(2 n^{2}+1\right)}{3}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 733,998 |
$\square$ Example 9 Let $x, y, z$ be positive real numbers, prove that: $(x y+y z+z x)\left[\frac{1}{(x+y)^{2}}+\right.$ $\left.\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}\right] \geqslant \frac{9}{4} .$(1996 Iranian Mathematical Olympiad) | Given $x \geqslant y \geqslant z > 0$, then
$$\begin{aligned}
& (x y + y z + z x)\left[\frac{1}{(x+y)^{2}} + \frac{1}{(y+z)^{2}} + \frac{1}{(z+x)^{2}}\right] - \frac{9}{4} \\
= & \frac{x y + z(x+y)}{(x+y)^{2}} + \frac{y z + x(y+z)}{(y+z)^{2}} + \frac{z x + y(z+x)}{(z+x)^{2}} - \frac{9}{4} \\
= & \frac{x}{y+z} + \frac{y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 733,999 |
$\square$ Example 19 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$, prove: $\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+$ $\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geqslant 0$. (46th IMO Shortlist Problem) | Prove that the original inequality can be transformed into
$$\frac{x^{2}+y^{2}+z^{2}}{x^{5}+y^{2}+z^{2}}+\frac{x^{2}+y^{2}+z^{2}}{y^{5}+z^{2}+x^{2}}+\frac{x^{2}+y^{2}+z^{2}}{z^{5}+x^{2}+y^{2}} \leqslant 3 .$$
By the Cauchy-Schwarz inequality and the condition \(xyz \geqslant 1\), we get
$$\begin{aligned}
& \left(x^{5}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,000 |
$\square$ Example 20 Let $x_{1}, x_{2}, \cdots, x_{n}$ all be positive numbers $(n \geqslant 2)$, and $\sum_{i=1}^{n} x_{i}=1$, prove that: $\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \frac{\sum_{i=1}^{n} \sqrt{x_{i}}}{\sqrt{n-1}}$. (4th CMO Problem) | Prove that let $y_{i}=1-x_{i}(i=1,2, \cdots, n)$, by Cauchy-Schwarz inequality we have
i.e.,
$$\begin{array}{c}
\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2} \leqslant n \sum_{i=1}^{n} x_{i}=n \\
\sum_{i=1}^{n} \sqrt{x_{i}} \leqslant \sqrt{n}
\end{array}$$
Similarly, $\left(\sum_{i=1}^{n} \sqrt{y_{i}}\right)^{2} \leqs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,001 |
Example 3 Let $a, b, c, d$ be positive real numbers, and satisfy $a+b+c+d=1$. Prove that: $6\left(a^{3}+b^{3}+c^{3}+d^{3}\right) \geqslant\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+\frac{1}{8}$. (8th China Hong Kong Mathematical Olympiad Problem) | Prove that by the AM-GM inequality,
$$\begin{array}{l}
a^{3}+\left(\frac{a+b+c+d}{4}\right)^{3}+\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant 3 a\left(\frac{a+b+c+d}{4}\right)^{2} \\
b^{3}+\left(\frac{a+b+c+d}{4}\right)^{3}+\left(\frac{a+b+c+d}{4}\right)^{3} \geqslant 3 b\left(\frac{a+b+c+d}{4}\right)^{2} \\
c^{3}+\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,004 |
$\square$
Example 4 Let $\left\{a_{1}, a_{2}, a_{3}, \cdots\right\}$ be an infinite sequence of positive numbers. Prove the inequality $\sum_{n=1}^{N} a_{n}^{2} \leqslant 4 \sum_{n=1}^{N} \alpha_{n}^{2}$ for any positive integer $N$. Here $\alpha_{n}$ is the average of $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$, i.e., $\alph... | Prove that $\alpha_{n}=\frac{a_{1}+a_{2}+a_{3}+\cdots+a_{n}}{n}$ satisfies
$$\begin{aligned}
\alpha_{n}^{2}-2 \alpha_{n} a_{n} & =a_{n}^{2}-2 \alpha_{n}\left[n \alpha_{n}-(n-1) \alpha_{n-1}\right] \\
& =(1-2 n) \alpha_{n}^{2}+2(n-1) \alpha_{n} \alpha_{n-1} \\
& \leqslant(1-2 n) \alpha_{n}^{2}+(n-1)\left(\alpha_{n}^{2}+... | \sum_{n=1}^{N} \alpha_{n}^{2} \leqslant 4 \sum_{n=1}^{N} a_{n}^{2} | Inequalities | proof | Yes | Yes | inequalities | false | 734,005 |
Example 5 Given $a, b, c>0$, prove: $\sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \geqslant$ $a b c+\sqrt[3]{\left(a^{3}+a b c\right)\left(b^{3}+a b c\right)\left(c^{3}+a b c\right)}$. (2001 China Mathematical Olympiad Problem) | $$\begin{aligned}
& \sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \\
= & \frac{1}{2} \sqrt{\left[b\left(a^{2}+b c\right)+c\left(b^{2}+c a\right)+a\left(c^{2}+a b\right)\right] \cdot\left[c\left(a^{2}+b c\right)+a\left(b^{2}+c a\right)+b\left(c^{2}+a b\right)\right]} \\
\geqslant & \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,006 |
$\square$ Example 6 Let $a, b, c, \lambda>0, a^{n-1}+b^{n-1}+c^{n-1}=1(n \geqslant 2)$, prove: $\frac{a^{n}}{b+\lambda c}+\frac{b^{n}}{c+\lambda a}+\frac{c^{n}}{a+\lambda b} \geqslant \frac{1}{1+\lambda}$. (2006 China National Training Team Test Question) | Prove that by Cauchy-Schwarz inequality,
$$\begin{array}{l}
\quad\left(\frac{a^{n}}{b+\lambda c}+\frac{b^{n}}{c+\lambda a}+\frac{c^{n}}{a+\lambda b}\right)\left[a^{n-2}(b+\lambda c)+b^{n-2}(c+\lambda a)+c^{n-2}(a\right. \\
+\lambda b)] \geqslant\left(a^{n-1}+b^{n-1}+c^{n-1}\right)^{2}=1 .
\end{array}$$
By the AM-GM in... | \frac{a^{n}}{b+\lambda c}+\frac{b^{n}}{c+\lambda a}+\frac{c^{n}}{a+\lambda b} \geqslant \frac{1}{1+\lambda} | Inequalities | proof | Yes | Yes | inequalities | false | 734,007 |
Example 7 Let $x, y, z$ be positive real numbers, and satisfy $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$. Prove the inequality: $\frac{x^{2}+y z}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}+z x}{\sqrt{2 y^{2}(z+x)}}+\frac{x^{2}+x y}{\sqrt{2 z^{2}(x+y)}} \geqslant 1$. (2007 Asia Pacific Mathematical Olympiad Problem) | Prove that by Cauchy-Schwarz inequality,
$$\begin{array}{l}
{\left[\frac{x^{2}}{\sqrt{2 x^{2}(y+z)}}+\frac{y^{2}}{\sqrt{2 y^{2}(z+x)}}+\frac{z^{2}}{\sqrt{2 z^{2}(x+y)}}\right][\sqrt{2(y+z)}+} \\
\sqrt{2(z+x)}+\sqrt{2(x+y)}] \geqslant(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=1 \\
\quad\left[\frac{y z}{\sqrt{2 x^{2}(y+z)}}+\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,008 |
Theorem 1 Let $a_{i j} \in \mathbf{R}^{+} \quad(i=1,2, \cdots, n ; j=1,2, \cdots, m)$, then
$$\begin{array}{l}
\left(a_{11}^{m}+a_{21}^{m}+\cdots+a_{n 1}^{m}\right)\left(a_{12}^{m}+a_{22}^{m}+\cdots+a_{n 2}^{m}\right) \cdots\left(a_{1 m}^{m}+a_{2 m}^{m}+\cdots+a_{m n}^{m}\right) \\
\geqslant\left(a_{11} a_{12} \cdots a... | Proof: Let $\left.A_{j}=\sqrt[m]{a_{1 j}^{m}+a_{2 j}^{m}+\cdots+a_{m j}^{m}}, j=1,2, \cdots, m\right)$, (1) is equivalent to
$$A_{1} A_{2} \cdots A_{m} \geqslant a_{11} a_{12} \cdots a_{1 m}+a_{21} a_{22} \cdots a_{2 m}+\cdots+a_{n 1} a_{n 2} \cdots a_{m m}$$
By the AM-GM inequality, we have
$$\frac{a_{i 1} a_{i 2} \c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,009 |
Example 1 Let $a_{i}>0(i=1,2, \cdots, n), A_{n}=\frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)$, $G_{n}=\sqrt[n]{a_{1} a_{2} \cdots a_{n}}$, prove the Popovic inequality: $\left(\frac{G_{n+1}}{A_{n+1}}\right)^{n+1} \geqslant\left(\frac{G_{n}}{A_{n}}\right)^{n}$. | Prove that by the generalization of the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
& \left(n A_{n}+a_{n+1}\right)^{n+1} \\
= & \left(A_{n}+A_{n}+\cdots+A_{n}+a_{n+1}\right)\left(A_{n}+A_{n}+\cdots+a_{n+1}+A_{n}\right) \cdots \\
& \left(a_{n+1}+A_{n}+\cdots+A_{n}\right) \\
\geqslant & \left(\sqrt[n+1]{A_{n}^{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,012 |
Example 2 Let $x_{i}>0, x_{i} y_{i}-z_{i}^{2}>0(i=1,2, \cdots, n)$, then
$$\frac{n^{3}}{\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} y_{i}-\left(\sum_{i=1}^{n} z_{i}\right)^{2}} \leqslant \sum_{i=1}^{n} \frac{1}{x_{i} y_{i}-z_{i}^{2}}$$
holds. Equality occurs if and only if $x_{1}=x_{2}=\cdots=x_{n} ; y_{1}=y_{2}=\cdots=y_{n} ... | Let \( A_{i}=\sqrt{x_{i} y_{i}}+z_{i}, B_{i}=\sqrt{x_{i} y_{i}}-z_{i} (i=1,2, \cdots, n) \). Using the Cauchy-Schwarz inequality and its generalization, we have:
\[
\begin{aligned}
& \left[\sum_{i=1}^{n} x_{i} \cdot \sum_{i=1}^{n} y_{i}-\left(\sum_{i=1}^{n} z_{i}\right)^{2}\right] \cdot \sum_{i=1}^{n} \frac{1}{x_{i} y_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,013 |
$\square$ Example 3 If $x, y, z$ are all positive real numbers, find the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)}$. (2003 Singapore Mathematical Olympiad problem) | Solve: By the generalization of Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
(1+5 x)(4 x+3 y)(5 y+6 z)(z+18) \geqslant(\sqrt[4]{1 \cdot 4 x \cdot 5 y \cdot z}+ \\
\sqrt[4]{5 x \cdot 3 y \cdot 6 z \cdot 18})^{4}=5120 x y z
\end{array}$$
Therefore, the maximum value of $\frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+... | \frac{1}{5120} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,014 |
$\square$ Example 4 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
$$\begin{array}{l}
R=\left(\frac{1}{n} \sum_{i=1}^{n} r_{i}\right), S=\left(\frac{1}{n} \sum_{i=1}^{n} s_{i}\right), T=\left(\frac{1}{n} \sum_{i=1}^{n} t_{i}\right), U=\left(\frac{1}{n} ... | Proof: Let $x_{1}, x_{2}, \cdots, x_{n} \in(1,+\infty)$, then by the generalized Cauchy inequality we have
$$\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\right) \geqslant\left(1+\sqrt[n]{x_{1} x_{2} \cdots x_{n}}\right)^{n}$$
i.e., $\sqrt[n]{\left(1+x_{1}\right)\left(1+x_{2}\right) \cdots\left(1+x_{n}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,015 |
$\square$ Example 5 Let $a_{1}, a_{2}, \cdots, a_{n}$ be given non-zero real numbers. If the inequality
$$r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+r_{n}\left(x_{n}-a_{n}\right) \leqslant \sqrt[m]{x_{1}^{m}+x_{2}^{m}+\cdots+x_{n}^{m}}$$
$-\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$ (where $... | Substitute $x_{i}=0(i=1,2, \cdots, n)$ into the original inequality, we get
$$-\left(r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}\right) \leqslant-\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$
That is,
$$r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \geqslant \sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$
Substitute ... | r_{i}=\left(\frac{a_{i}}{\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}}\right)^{m-1}(i=1,2, \cdots, n) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 734,016 |
$\square$ Example 7 Given that $a, b$ are positive constants, and $x$ is an acute angle, find the minimum value of the function $y=\frac{a}{\sin ^{n} x}+$ $\frac{b}{\cos ^{n} x}$. | By Hölder's inequality, we have
$$\begin{aligned}
& \left(\frac{a}{\sin ^{n} x}+\frac{b}{\cos ^{n} x}\right)^{\frac{2}{n+2}}\left(\sin ^{2} x+\cos ^{2} x\right)^{\frac{n}{n+2}} \\
\geqslant & \left(\frac{a}{\sin ^{n} x}\right)^{\frac{2}{n+2}}\left(\sin ^{2} x\right)^{\frac{n}{n+2}}+\left(\frac{b}{\cos ^{n} x}\right)^{\... | y_{\min }=\left(a^{\frac{2}{n+2}}+b^{\frac{2}{n+2}}\right)^{\frac{n+2}{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 734,018 |
$\square$ Example 8 Given $a_{1}, a_{2}, \cdots, a_{m} \in \mathbf{R}^{+}$, and $a_{1}+a_{2}+\cdots+a_{m}=1, m$, $n \in \mathbf{N}^{\cdot}$, prove: $\left(\frac{1}{a_{1}^{n}}-1\right)\left(\frac{1}{a_{2}^{n}}-1\right) \cdots\left(\frac{1}{a_{m}^{n}}-1\right) \geqslant\left(m^{n}-1\right)^{m}$. | Prove that because $a_{1}+a_{2}+\cdots+a_{m} \geqslant m \sqrt[m]{a_{1} a_{2} \cdots a_{m}}$, so
$$\frac{1}{a_{1} a_{2} \cdots a_{m}} \geqslant m^{m}$$
Also, because
$$\begin{array}{c}
\frac{1}{a_{1}}-1=\frac{a_{2}+a_{3}+\cdots+a_{m}}{a_{1}} \geqslant(m-1) \frac{\sqrt[m-1]{a_{2} a_{3} \cdots a_{m}}}{a_{1}} \\
\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,019 |
Example 9 Let $a_{1}, a_{2}, \cdots, a_{m}$ be positive real numbers, and $\sum_{i=1}^{m} \frac{1}{a_{i}}=1$. Then for every $n \in \mathbf{N}^{\cdot}$, we have $\left(\sum_{i=1}^{m} a_{i}\right)^{n}-\sum_{i=1}^{m} a_{i}^{n} \geqslant m^{2 n}-m^{\kappa+1}$. | Prove that because $\sum_{i=1}^{m} \frac{1}{a_{i}}=1 \geqslant m \sqrt[m]{\frac{1}{a_{1} a_{2} \cdots a_{m}}}$, so
$$a_{1} a_{2} \cdots a_{m} \geqslant m^{m}$$
According to the polynomial expansion theorem, we have
$$\left(\sum_{i=1}^{m} a_{i}\right)^{n}=\sum \frac{n!}{n_{1}!n_{2}!\cdots n_{m}!} a_{1}^{n_{1}} a_{2}^{n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,020 |
Example 11 Given $n$ real numbers $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n-1} \leqslant a_{n}$, let $x=\frac{1}{n}\left(a_{1}+\right.$ $\left.a_{2}+\cdots+a_{n}\right), y=\frac{1}{n}\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)$, prove that: $2 \sqrt{y-x^{2}} \leqslant a_{n}-a_{1}$ $\leqslant \sqrt{2 ... | Prove that given $n x=a_{1}+a_{2}+\cdots+a_{n}, n y=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}$, then
$$\begin{aligned}
n^{2}\left(y-x^{2}\right)= & n\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)-\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \\
= & (n-1) \sum_{i=1}^{n} a_{i}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} a_{i} a_{... | 2 \sqrt{y-x^{2}} \leqslant a_{n}-a_{1} \leqslant \sqrt{2 n\left(y-x^{2}\right)} | Inequalities | proof | Yes | Yes | inequalities | false | 734,021 |
$\square$ Example 10 Given that $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers, $k \geqslant 0, n$ is a positive integer, then
$$\prod_{i=1}^{n}\left(a_{i}^{n+k}-a_{i}^{k}+n\right) \geqslant\left(\sum_{i=1}^{n} a_{i}\right)^{n} .$$ | Prove that since $a_{i}$ are positive real numbers $(i=1,2, \cdots, n)$, we have
that is
$$\begin{array}{c}
\left(a_{i}^{k}-1\right)\left(a_{i}^{n}-1\right) \geqslant 0, \\
a_{i}^{n+k}-a_{i}^{k} \geqslant a_{i}^{n}-1, \\
a_{i}^{n+k}-a_{i}^{k}+n \geqslant a_{i}^{n}+n-1,
\end{array}$$
thus
$$\begin{aligned}
& \prod_{i=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,022 |
$\square$ Example 11 Let positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $a_{1}+a_{2}+\cdots+a_{n}=1$. Prove that: $\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)\left(\frac{a_{1}}{a_{2}^{2}+a_{2}}+\frac{a_{2}}{a_{3}^{2}+a_{3}}+\cdots+\frac{a_{n}}{a_{1}^{2}+a_{1}}\right) \geqslant$ $\frac{n}{n+1}$.... | Prove that from the generalization of the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\quad\left(a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{1}\right)\left(\frac{a_{1}}{a_{2}^{2}+a_{2}}+\frac{a_{2}}{a_{3}^{2}+a_{3}}+\cdots+\frac{a_{n}}{a_{1}^{2}+a_{1}}\right)\left[a _ { 1 } \left(a_{2}+\right.\right. \\
\left.1)+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,023 |
Example 1 Prove that for all positive numbers $a, b, c$, we have $\frac{1}{a^{3}+b^{3}+a b c}+\frac{1}{b^{3}+c^{3}+a b c}$ $+\frac{1}{c^{3}+a^{3}+a b c} \leqslant \frac{1}{a b c}$. (26th United States of America Mathematical Olympiad problem) | $$\begin{array}{l}
\text { Prove that since } a, b, c \text { are positive real numbers, } \\
a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, c^{3}+a^{3} \geqslant c^{2} a+c a^{2}, \\
\text { hence } \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} \\
\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,024 |
$\square$ Example $2 a, b, c$ are positive real numbers, and $a b c=1$, prove:
$\frac{a b}{a^{5}+b^{5}+a b}+\frac{b c}{b^{5}+c^{5}+b c}+\frac{c a}{c^{5}+a^{5}+c a} \leqslant 1$. (37th IMO Shortlist) | Prove that since $a, b, c$ are positive real numbers, we have
$$\begin{aligned}
a^{5}+b^{5} & \geqslant a^{3} b^{2}+a^{2} b^{3}, \\
b^{5}+c^{5} & \geqslant b^{3} c^{2}+b^{2} c^{3}, \\
c^{5}+a^{5} & \geqslant c^{3} a^{2}+c^{2} a^{3} .
\end{aligned}$$
Also, since $a b c=1$, we have
$$\begin{array}{c}
a^{5}+b^{5}+a b=a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,025 |
$\square$ Example 3 Given that $a, b$ are positive numbers, and $n$ is a positive integer, prove: $\frac{a^{n}+b^{n}}{2} \geqslant\left(\frac{a+b}{2}\right)^{n}$. (1975 Soviet University Mathematics Competition) | Prove that by the binomial theorem,
$$(a+b)^{n}=\sum_{k=0}^{n} a^{k} b^{n-k}=\sum_{k=0}^{n} \mathrm{C}_{n}^{k} a^{n k} b^{k},$$
Therefore,
$$\begin{aligned}
2(a+b)^{n} & =\sum_{k=0}^{n} \mathrm{C}_{n}^{k}\left(a^{k} b^{n-k}+a^{n-k} b^{k}\right) \leqslant \sum_{k=0}^{n} \mathrm{C}_{n}^{k}\left(a^{n}+b^{n}\right) \\
& =... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,026 |
$\square$ Example 4 Given that $a, b$ are positive numbers, $n \in \mathbf{N}^{*}$ and $n \geqslant 2$, prove:
$\frac{a^{n}+a^{n-1} b+a^{n-2} b^{2}+\cdots+a b^{n-1}+b^{n}}{n+1} \geqslant\left(\frac{a+b}{2}\right)^{n}$. (1988 Hunan Province Middle School Mathematics Summer Camp Mathematics Competition Question) | Prove by mathematical induction:
(1) When $n=2$, since $a^{2}+b^{2} \geqslant 2 a b$, adding $3\left(a^{2}+b^{2}\right)+$ $4 a b$ to both sides gives $4\left(a^{2}+b^{2}+a b\right) \geqslant 3(a+b)^{2}$, i.e., $\frac{a^{2}+b^{2}+a b}{3} \geqslant\left(\frac{a+b}{2}\right)^{2}$, the inequality holds, with equality if an... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,027 |
$\square$ Example 5 A geometric sequence with the first term and common ratio both being positive numbers has the same first and last terms as an arithmetic sequence. Then the sum of this geometric sequence is not greater than the sum of this arithmetic sequence. (1979 Shandong Province Mathematics Competition Question... | Proof: Let the first term of a geometric sequence be $a$, the common ratio be $q$, and the number of terms be $n$, then the last term and the sum of this sequence are
$$a_{n}=a q^{n-1}, S=a\left(1+q+q^{2}+\cdots+q^{n-1}\right) .$$
For an arithmetic sequence with the first term $a$ and the last term $a q^{n-1}$, its su... | proof | Algebra | proof | Yes | Yes | inequalities | false | 734,028 |
Example 6 Prove that for any real numbers $a, b$ we have $\left(\frac{a+b}{2}\right)\left(\frac{a^{2}+b^{2}}{2}\right)\left(\frac{a^{3}+b^{3}}{2}\right) \leqslant$ $\frac{a^{6}+b^{6}}{2}$. (1963 Polish Mathematical Olympiad Problem) | Prove that since $a^{6}+b^{6} \geqslant a^{4} b^{2}+a^{2} b^{4}$, we have
$$2\left(a^{6}+b^{6}\right) \geqslant\left(a^{4}+b^{4}\right)\left(a^{2}+b^{2}\right)$$
Thus,
$$\left(\frac{a^{2}+b^{2}}{2}\right)\left(\frac{a^{4}+b^{4}}{2}\right) \leqslant \frac{a^{6}+b^{6}}{2}$$
For any $a, b$,
So,
$$a^{4}+b^{4} \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,029 |
$\square$ Example 7 Given positive real numbers $a, b, c, d$, prove: $\frac{a^{3}+b^{3}+c^{3}}{a+b+c}+\frac{b^{3}+c^{3}+d^{3}}{b+c+d}+$ $\frac{c^{3}+d^{3}+a^{3}}{c+d+a}+\frac{d^{3}+a^{3}+b^{3}}{d+a+b} \geqslant a^{2}+b^{2}+c^{2}+d^{2}$. (US College Mathematics Competition Problem) | Prove that since \(a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, c^{3}+a^{3} \geqslant c^{2} a + c a^{2}\), adding these three inequalities and adding \(a^{3}+b^{3}+c^{3}\) gives
\[
\left(a^{3}+b^{3}+c^{3}\right)(1+1+1) \geqslant\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)
\]
Thus,
\[
\frac{a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,030 |
Example 8 Given that $a, b, c$ are positive numbers, prove: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant \frac{a^{8}+b^{8}+c^{8}}{a^{3} b^{3} c^{3}}$. (1967 IMO Shortlist) | Prove that because $a^{8}+b^{8} \geqslant a^{6} b^{2}+a^{2} b^{6}, b^{8}+c^{8} \geqslant b^{6} c^{2}+b^{2} c^{6}, c^{8}+a^{8}$ $\geqslant c^{6} a^{2}+a^{2} c^{6}$, so
$$2\left(a^{8}+b^{8}+c^{8}\right) \geqslant a^{2}\left(b^{6}+c^{6}\right)+b^{2}\left(c^{6}+a^{6}\right)+c^{2}\left(a^{6}+b^{6}\right),$$
Adding $a^{8}+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,031 |
Example 1 Given $a^{2}+b^{2}+c^{2}+d^{2}=1$. Prove:
$$\begin{aligned}
& (a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4} \\
+ & (b+d)^{4}+(c+d)^{4} \leqslant 6
\end{aligned}$$ | Prove that
$$\begin{aligned}
& (a+b)^{4}+(a+c)^{4}+(a+d)^{4}+(b+c)^{4}+(b+d)^{4} \\
+ & (c+d)^{4}+(a-b)^{4}+(a-c)^{4}+(a-d)^{4}+(b-c)^{4} \\
+ & (b-d)^{4}+(c-d)^{4} \\
= & 6\left(a^{2}+b^{2}+c^{2}+d^{2}\right)^{2}
\end{aligned}$$
From equation (3), equation (2) immediately follows. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,032 |
Example 2 Let $a, b, c, d>0$, and $d=\max \{a, b, c, d\}$, prove: $a(d-b)+b(d-c)+c(d-a)<d^{2}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Prove that the right side of equation (4) minus the left side, and rearranging with respect to $d$, yields
$$d^{2}-d(a+b+c)+(a b+b c+c a)$$
Recalling the identity
$$\begin{aligned}
& (d-a)(d-b)(d-c) \\
= & d^{3}-d^{2}(a+b+c)+d(a b+b c+c a)-a b c
\end{aligned}$$
By comparing (5) and (6), we can see that (4) holds. | null | Inequalities | proof | Yes | Yes | inequalities | false | 734,033 |
Example 11 (An Zhenping Inequality)
Let the side lengths, semi-perimeters, and areas of $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ be $a_{1}, a_{2}, a_{3}, p_{1}, S_{1}$ and $b_{1}, b_{2}, b_{3}, p_{2}, S_{2}$, respectively. Prove:
$$\begin{aligned}
& b_{1}\left(p_{2}-b_{1}\right)\left(p_{1}-a_{2}\... | Prove that by introducing 6 positive numbers in the following way:
$$\left\{\begin{array} { l }
{ x = a _ { 2 } + a _ { 3 } - a _ { 1 } , } \\
{ y = a _ { 3 } + a _ { 1 } - a _ { 2 } , } \\
{ z = a _ { 1 } + a _ { 2 } - a _ { 3 } , } \\
{ p = b _ { 2 } + b _ { 3 } - b _ { 1 } , } \\
{ q = b _ { 3 } + b _ { 1 } - b _ {... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,034 |
Example 7 Proof: For positive real numbers $a, b, c$, we have
$$\begin{aligned}
& \sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \\
\geqslant & a b c+\sqrt[3]{\left(a^{3}+a b c\right)\left(b^{3}+a b c\right)\left(c^{3}+a b c\right)}
\end{aligned}$$ | $$\begin{aligned}
& \sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \\
= & \frac{1}{2} \sqrt{\left[b\left(a^{2}+b c\right)+c\left(b^{2}+r a\right)+a\left(c^{2}+a b\right)\right]\left[c\left(a^{2}+b c\right)+a\left(b^{2}+c a\right)+b\left(c^{2}+a b\right)\right]} \\
\geqslant & \frac{1}{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,035 |
Example 9 Let $a_{ij} (i, j=1,2, \cdots, n)$ be positive real numbers. Prove:
$$\begin{aligned}
& \left(a_{11}^{n}+a_{12}^{n}+\cdots+a_{1 n}^{n}\right) \cdot\left(a_{21}^{n}+a_{22}^{n}+\cdots+a_{2 n}^{n}\right) \cdot \cdots \cdot\left(a_{n 1}^{n}+a_{n 2}^{n}\right. \\
& \left.+\cdots+a_{n n}^{n}\right) \\
\geqslant & \... | Prove that according to the homogeneity of equation (11), we can set
$\left(a_{i 1}^{n}+a_{i 2}^{n}+\cdots+a_{i n}^{n}\right)^{\frac{1}{n}}=1$ or $a_{i 1}^{n}+a_{i 2}^{n}+\cdots+a_{i n}^{n}=1(i=1,2, \cdots$,
n),
Thus, we only need to prove
$$a_{11} a_{21} \cdots a_{n 1}+a_{12} a_{22} \cdots a_{n 2}+\cdots+a_{1 n} a_{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,037 |
Example 10 Let non-negative real numbers $x, y$ satisfy $x^{2}+y^{3} \geqslant x^{3}+y^{4}$. Prove:
$$x^{3}+y^{3} \leqslant 2$$ | Prove that using the Cauchy-Schwarz inequality and the Arithmetic Mean-Geometric Mean inequality, we can obtain
$$\begin{aligned}
x^{3}+y^{3} & \leqslant \sqrt{\left(x^{3}+y^{4}\right)\left(x^{3}+y^{2}\right)} \leqslant \sqrt{\left(x^{2}+y^{3}\right)\left(x^{3}+y^{2}\right)} \\
& \leqslant \frac{x^{2}+y^{3}+x^{3}+y^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,038 |
Example 11 Let non-negative real numbers $x, y, z$ satisfy $xyz=1$. Prove:
$$\sqrt{4+9 x^{2}}+\sqrt{4+9 y^{2}}+\sqrt{4+9 z^{2}} \leqslant \sqrt{13}(x+y+z) .$$ | Prove using the Cauchy-Schwarz inequality:
$$\begin{aligned}
& \sum_{\mathrm{cyc}} \sqrt{4+9 x^{2}}=\sum_{\mathrm{cyc}} \sqrt{4 \sqrt[3]{x^{2} y^{2} z^{2}}+9 x^{2}} \equiv \sum_{\mathrm{cyc}} \sqrt{4 u^{2} v^{2} w^{2}+9 u^{6}} \\
= & \sum_{\mathrm{cyc}} u \sqrt{4 v^{2} w^{2}+9 u^{4}} \leqslant \sqrt{\sum_{\mathrm{cyc}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 734,039 |
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