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Example 2: Let the three sides of $\triangle A B C$ be $a, b, c$, and the area be $\Delta$, then
$$\min \left\{a^{4}+b^{4}, b^{4}+c^{4}, c^{4}+a^{4}\right\} \geq 8 \Delta^{2}$$ | Proof: Without loss of generality, let $a \geq b \geq c$. Then, by the two-variable mean inequality and the boundedness of the sine function, we get
$$\begin{aligned}
\min \left\{a^{4}+b^{4}, b^{4}+c^{4}, c^{4}\right. & \left.+a^{4}\right\} \geq b^{4}+c^{4} \\
& \geq 2 b^{2} c^{2} \geq 8\left(\frac{1}{2} b c \sin A\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,582 |
Given: $a, b \in R^{+}, a+b=1$, prove:
$$\frac{3}{2}<\frac{1}{a^{4}+1}+\frac{1}{b^{4}+1} \leq \frac{32}{17}$$ | Prove: From the given, we have $a^{4}\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}$
$\therefore$ It suffices to prove $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}>\frac{3}{2}$
$\Leftrightarrow 2\left(a^{2}+b^{2}\right)+4>3\left(a^{2}+1\right)\left(b^{2}+1\right)$
$\Leftrightarrow 1>a^{2}+b^{2}+3 a^{2} b^{2}$
$\Leftrightarrow 1>(a+b)^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,583 |
Example 20: (Problem 73 of Issue 3, 1985) If $x, y, z$ are positive numbers, prove:
$$x^{3}+y^{3}+z^{3} \geq 3 x y z+x(y-z)^{2}+y(z-x)^{2}+z(x-y)^{2} .$$ | This is clearly a strengthening of the 3-variable mean inequality.
The transformation of this inequality leads to the 1975 All-Soviet Union Mathematical Olympiad, Grade 10, Problem 2:
For positive numbers $x, y, z$, the following inequality holds:
$$x^{3}+y^{3}+z^{3}+3 x y z \geq x y(x+y)+y z(y+z)+z x(z+x)$$
A very si... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,585 |
Example 21: Let $\alpha, \beta, \gamma$ be acute angles, and $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1$, then
$$\frac{1}{\cos ^{2} \alpha+\cos ^{2} \beta}+\frac{1}{\cos ^{2} \beta+\cos ^{2} \gamma}+\frac{1}{\cos ^{2} \gamma+\cos ^{2} \alpha} \leq \frac{1}{4}\left(\frac{1}{\sin \alpha \sin \beta}+\frac{1}{\si... | Prove: By applying the binary mean inequality and the quaternary mean inequality, we get
$$\begin{array}{l}
\frac{1}{\cos ^{2} \alpha+\cos ^{2} \beta} \\
\leq \frac{1}{2-\left(\sin ^{2} \alpha+\sin ^{2} \beta\right)} \\
=\frac{1}{\sin ^{2} \alpha+\sin ^{2} \beta+2 \sin ^{2} \gamma} \\
\leq \frac{1}{4 \sqrt[4]{\sin ^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,586 |
Example 22: Given $a, b \in R, \left|a^{2}-b^{2}\right|+2|a b|=1$, prove: $\frac{\sqrt{2}}{2} \leq a^{2}+b^{2} \leq 1$. | Proof: Let $a^{2}+b^{2}=r^{2}(r \geq 0)$, and let $a=r \cos \theta, b=r \sin \theta$, then we have
$$\begin{aligned}
1 & =\left|a^{2}-b^{2}\right|+2|a b| \\
& =r^{2}|\cos 2 \theta|+|\sin 2 \theta| \\
& =r^{2} \sqrt{(|\cos 2 \theta|+|\sin 2 \theta|)^{2}} \\
& =r^{2} \sqrt{1+|\sin 4 \theta|}
\end{aligned}$$
On one hand,... | \frac{\sqrt{2}}{2} \leq a^{2}+b^{2} \leq 1 | Inequalities | proof | Yes | Yes | inequalities | false | 736,587 |
Example 23: (2005 National Women's Mathematical Olympiad Problem) Let positive real numbers $x, y$ satisfy $x^{3}+y^{3}=x-y$. Prove:
$$x^{2}+4 y^{2}<1$$ | Proof: By the mean inequality
$$5 y^{3}+x^{2} y \geq 2 \sqrt{5 x^{2} y^{4}}>4 x y^{2}$$
Therefore, $\left(x^{2}+4 y^{2}\right)(x-y)0)$, isn't this the equation of an ellipse? Then naturally, we can think of its transformed form, which is the parametric equation:
$$\left\{\begin{array}{l}
x=\rho \cos \theta \\
y=\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,588 |
Example 24: Given $x, y>0, n \in N^{*}$, prove: $\frac{x^{n}}{1+x^{2}}+\frac{y^{n}}{1+y^{2}} \leq \frac{x^{n}+y^{n}}{1+x y}$. | Prove: From the 2-variable mean inequality and the $n$-variable mean inequality, we get
that is,
$$\left(1+x^{2}\right)\left(1+y^{2}\right)=1+x^{2}+y^{2}+x^{2} y^{2} \geq 1+2 x y+x^{2} y^{2}=(1+x y)^{2}$$
that is,
$$\begin{array}{l}
\left(1+x^{2}\right)\left(1+y^{2}\right) \geq(1+x y)^{2} \text {. } \\
x^{n-1} y+x y^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,589 |
Example 25: In an acute triangle $\triangle A B C$, prove that
$$\frac{3}{2} \leq \frac{\cos A}{\cos (B-C)}+\frac{\cos B}{\cos (C-A)}+\frac{\cos C}{\cos (A-B)}<2 \text {. }$$ | $$\begin{array}{l}
\frac{\cos A}{\cos (B-C)}=\frac{-\cos (B+C)}{\cos (B-C)} \\
=\frac{\sin B \sin C-\cos B \cos C}{\sin B \sin C+\cos B \cos C} \\
=\frac{1-\cot B \cot C}{1+\cot B \cot C} \\
=-1+\frac{2}{1+\cot B \cot C}
\end{array}$$
Similarly,
$$\begin{array}{l}
\frac{\cos B}{\cos (C-A)}=-1+\frac{2}{1+\cot C \cot A}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,590 |
Class Problem: (Mathematical Problems in "Mathematics Bulletin" $1144,1998,7$) In an acute triangle $\triangle A B C$, we have $\frac{\cos (\mathrm{B}-\mathrm{C}) \cos (\mathrm{C}-\mathrm{A}) \cos (\mathrm{A}-\mathrm{B})}{\cos A \cos \mathrm{B} \cos C} \geq 8$. | Prove: Since
$$\frac{\cos (B-C)}{\cos A}=\frac{2 \sin (B+C) \cos (B-C)}{2 \sin A \cos A}=\frac{\sin 2 B+\sin 2 C}{\sin 2 A}$$
Thus, by the AM-GM inequality, we get
$$\begin{array}{l}
\frac{\cos (B-C)}{\cos A}=\frac{\sin 2 B+\sin 2 C}{\sin 2 A} \geq \frac{2 \sqrt{\sin 2 B \sin 2 C}}{\sin 2 A} \\
\frac{\cos (B-C)}{\cos ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,591 |
Example 26: Given $x, y, z \in R^{+}, x+y+z=1$, then
$$\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right) \geq\left(\frac{8}{3}\right)^{3}$$ | $$\begin{array}{l}
\because 1=x+y+z \geq 3 \cdot \sqrt[3]{x y z}, \\
\therefore x y z \leq \frac{1}{27}, \frac{1}{x y z} \geq 27 . \text { and note that } a^{2}+b^{2}+c^{2} \geq a b+b c+c a, \text { we get } \\
\therefore\left(\frac{1}{x}-x\right)\left(\frac{1}{y}-y\right)\left(\frac{1}{z}-z\right) \\
=\frac{\left(1-x^... | \left(\frac{8}{3}\right)^{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,592 |
Example 28: Given $x, y, z \in R, A, B, C$ are the three interior angles of a triangle, prove:
$$x^{2}+y^{2}+z^{2} \geq 2 y z \cos A+2 z x \cos B+2 x y \cos C \text {. }$$
Equality holds if and only if $\frac{x}{\sin A}=\frac{y}{\sin B}=\frac{z}{\sin C}$.
Focusing on the condition for equality in the inequality:
$$\l... | Proof: Using the triangle angle sum theorem and applying $2 p q \leq p^{2}+q^{2}$ twice, we get
$$\begin{aligned}
& 2 y z \cos A+2 z x \cos B+2 x y \cos C \\
= & 2 y z \cos [\pi-(B+C)]+2 x(z \cos B+2 y \cos C) \\
\leq & -2 y z \cos (B+C)+x^{2}+(z \cos B+2 y \cos C)^{2} \\
= & x^{2}+2 y z \sin B \sin C+z^{2} \cos ^{2} B... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,595 |
Example 29: Given that $a, b, c$ are the lengths of the sides of an acute triangle, prove:
$$b c \sqrt{b^{2}+c^{2}-a^{2}}+c a \sqrt{c^{2}+a^{2}-b^{2}}+a b \sqrt{a^{2}+b^{2}-c^{2}}>2 \sqrt{2} a b c$$ | Prove: By transforming the inequality, we can obtain
$$\sqrt{\frac{b^{2}+c^{2}-a^{2}}{2 a^{2}}}+\sqrt{\frac{c^{2}+a^{2}-b^{2}}{2 b^{2}}}+\sqrt{\frac{a^{2}+b^{2}-c^{2}}{2 c^{2}}}>2$$
If we let $x=b^{2}+c^{2}-a^{2}, y=c^{2}+a^{2}-b^{2}, z=a^{2}+b^{2}-c^{2}$, then the above inequality is equivalent to proving that given ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,597 |
Example 32: Let $a>0, b>0$, prove: $\frac{a^{2}+b^{2}}{2 a b}+\frac{2 \sqrt{a b}}{a+b} \geq 2$. | Prove: By the 2-variable mean inequality, we get
$$\frac{a^{2}+b^{2}}{2 a b}+\frac{2 \sqrt{a b}}{a+b} \geq 2 \sqrt{\frac{a^{2}+b^{2}}{2 a b} \cdot \frac{2 \sqrt{a b}}{a+b}}=2 \sqrt{\frac{a^{2}+b^{2}}{\sqrt{a^{3} b}+\sqrt{a b^{3}}}}$$
Therefore, it suffices to prove $\frac{a^{2}+b^{2}}{\sqrt{a^{3} b}+\sqrt{a b^{3}}} \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,599 |
Example 33: Given $a, b, c, d \in R^{+}, a b c d=1$, prove:
$$a^{3}+b^{3}+c^{3}+d^{3} \geq a+b+c+d .$$ | Proof: By the 6-term mean inequality, we get
$$\begin{array}{l}
\frac{a^{3}+a^{3}+a^{3}+b^{3}+c^{3}+d^{3}}{6} \geq \sqrt[6]{a^{9} b^{3} c^{3} d^{3}}=a \sqrt{a b c d}=a \\
\text { i.e., } \frac{3 a^{3}+b^{3}+c^{3}+d^{3}}{6} \geq a \\
\text { Similarly, } \quad \frac{a^{3}+3 b^{3}+c^{3}+d^{3}}{6} \geq b, \frac{a^{3}+b^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,600 |
Example 35: (2007 China Girls' Mathematical Olympiad) Given $a, b, c \geq 0$, and $a+b+c=1$, prove:
$$\sqrt{a+\frac{1}{4}(b-c)^{2}}+\sqrt{b}+\sqrt{c} \leq \sqrt{3} .$$ | $$\begin{array}{l}
\left(\sqrt{a+\frac{1}{4}(b-c)^{2}}+\sqrt{b}+\sqrt{c}\right)^{2} \\
=\left(\sqrt{a+\frac{1}{4}(b-c)^{2}}+\frac{\sqrt{b}+\sqrt{c}}{2}+\frac{\sqrt{b}+\sqrt{c}}{2}\right)^{2} \\
\leq 3\left[a+\frac{1}{4}(b-c)^{2}+\left(\frac{\sqrt{b}+\sqrt{c}}{2}\right)^{2}+\left(\frac{\sqrt{b}+\sqrt{c}}{2}\right)^{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,602 |
Example 36: Given $x, y, z>1$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$, prove:
$$\sqrt{x+y+z} \geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$ | Prove: The conditional equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ can be equivalently transformed into $\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1$. Therefore, by the Cauchy-Schwarz inequality, we have
$$\begin{aligned}
\sqrt{x+y+z} & =\sqrt{(x+y+z)\left(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}\right)} \\
& \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,603 |
Example 4: Let $0 \leq a, b, c \leq \lambda(\lambda>0)$, prove:
$$\frac{a}{\lambda^{2}+b c}+\frac{b}{\lambda^{2}+c a}+\frac{c}{\lambda^{2}+a b} \leq \frac{2}{\lambda} .$$ | Proof: We first prove
$$\frac{a}{\lambda^{2}+b c} \leq \frac{2}{\lambda} \cdot \frac{a}{a+b+c}$$
This inequality is equivalent to
$$\lambda(a+b+c) \leq 2 \lambda^{2}+2 b c,$$
which is
$$(b-\lambda)(c-\lambda)+b c+\lambda^{2} \geq \lambda a$$
Noting the condition $0 \leq a, b, c \leq \lambda(\lambda>0)$, it is clear ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,604 |
Promotion: Given $x_{1}, x_{2}, \cdots, x_{n} \in R^{+}$, prove:
$$\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n}^{2}}{x_{1}} \geq x_{1}+x_{2}+\cdots+x_{n}+\frac{4\left(x_{1}-x_{2}\right)^{2}}{x_{1}+x_{2}+\cdots+x_{n}}$$ | Prove: Since $x^{2}+y^{2}-2xy=(x-y)^{2}$, we have
$$\frac{x^{2}}{y}=2x-y+\frac{(x-y)^{2}}{y}.$$
Therefore, we have the relations
$$\begin{aligned}
\frac{x_{1}^{2}}{x_{2}} & =2x_{1}-x_{2}+\frac{\left(x_{1}-x_{2}\right)^{2}}{x_{2}} \\
\frac{x_{2}^{2}}{x_{3}} & =2x_{2}-x_{3}+\frac{\left(x_{2}-x_{3}\right)^{2}}{x_{3}} \\
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,605 |
Class problem: (Math problem $946,1995,4$ Su Xiaoyang) Let $a, b, c \in R^{+}$, prove:
$$\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a} \geq \frac{a+b+c}{2}$$ | Prove: Let $a=b+\alpha, b=c+\beta, c=a+\gamma$, then $\alpha+\beta+\gamma=0$, thus
$$\begin{aligned}
& 4\left(\frac{a^{2}}{a+b}+\frac{b^{2}}{b+c}+\frac{c^{2}}{c+a}\right) \\
= & \frac{(a+b+\alpha)^{2}}{a+b}+\frac{(b+c+\beta)^{2}}{b+c}+\frac{(c+a+\gamma)^{2}}{c+a} \\
= & 2(a+b+c)+2(\alpha+\beta+\gamma)+\frac{\alpha^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,606 |
Example 38: Let $a, b, c>0, a+b+c=\lambda$, then
$$\frac{a b^{2}+\lambda^{2} b}{c+a}+\frac{b c^{2}+\lambda^{2} c}{a+b}+\frac{c a^{2}+\lambda^{2} a}{b+c} \geq \frac{5 \lambda^{2}}{3}$$ | Prove: Using the three-variable mean inequality, we get
$$\begin{array}{l}
\frac{a b+\lambda b}{c+a}+\frac{b c+\lambda c}{a+b}+\frac{c a+\lambda a}{b+c} \\
= \frac{a(\lambda-c-a)+\lambda b}{c+a}+\frac{b(\lambda-a-b)+\lambda c}{a+b}+\frac{c(\lambda-b-c)+\lambda a}{b+c} \\
= \lambda\left(\frac{a+b}{c+a}+\frac{b+c}{a+b}+... | \frac{5 \lambda^{2}}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,607 |
Example 39: Let $a, b, c$ be positive numbers, and satisfy $a b c=1$. Prove:
$$\frac{a}{a^{2}+a b+b c}+\frac{b}{b^{2}+b c+c a}+\frac{c}{c^{2}+c a+a b} \leq 1$$ | Proof: First, we prove: $\frac{a}{a^{2}+a b+b c} \leq \frac{1}{b c+b+1}$, which is equivalent to $a(b c+b+1) \leq a^{2}+a b+b c$, or
$$1+a \leq a^{2}+\frac{1}{a}$$
By the two-variable and three-variable mean inequalities, we get
$$\begin{array}{l}
a^{2}+1 \geq 2 a \\
a^{2}+\frac{1}{a}+\frac{1}{a} \geq 3
\end{array}$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,608 |
Example 40: Given real numbers $a, b, c$ satisfy $a+b+c=1, a^{2}+b^{2}+c^{2}=1$, prove: $a^{5}+b^{5}+c^{5} \leq 1$. | Prove: For the equation $a+b+c=1$, squaring both sides, we get
$$\begin{array}{l}
1=(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \\
=1+2(a b+b c+c a) \\
a b+b c+c a=0
\end{array}$$
That is, $\square$
Squaring both sides of the above equation and transforming, we get
$$\begin{aligned}
a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,609 |
Example 41: Let $a, b, c$ be positive numbers, and satisfy $abc=1$. Prove:
$$\sqrt{\frac{a}{2a^2 + b^2 + 3}} + \sqrt{\frac{b}{2b^2 + c^2 + 3}} + \sqrt{\frac{c}{2c^2 + a^2 + 3}} \leq \frac{\sqrt{6}}{2}$$ | Proof: First, we prove that under the condition $abc=1$, the following identity holds:
$$\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1$$
In fact, from the condition $abc=1$, we know $1=abc$. Substituting this, we get
$$\begin{aligned}
& \frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1} \\
= & \frac{a}{ab+a+1}+... | \frac{\sqrt{6}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,610 |
Example 42: Let the three sides of $\triangle ABC$ be $\mathrm{a}, \mathrm{b}, \mathrm{c}$, and the semi-perimeter be $s$. Prove:
$$b \cos \frac{C}{2} + c \cos \frac{B}{2} \leq \sqrt{s(b+c)} .$$ | Proof: By Cauchy-Schwarz inequality and the projection theorem, we have
$$\begin{aligned}
b \cos \frac{C}{2}+c \cos \frac{B}{2} & =\sqrt{b} \cdot \sqrt{b} \cos \frac{C}{2}+\sqrt{c} \cdot \sqrt{c} \cos \frac{B}{2} \\
& \leq \sqrt{(b+c)\left[b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2}\right]} \\
& =\sqrt{(b+c)\left[b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,611 |
Example 43: Let $x, y, z \in(-1,1)$, prove:
$$\frac{1}{\sin [(1-x)(1-y)(1-z)]}+\sin \frac{1}{(1+x)(1+y)(1+z)} \geq 2$$ | Prove: Since $x, y, z \in (-1,1)$, we have $1 - x^2 \leq 1$, which implies $\frac{1}{1+x} \geq 1-x > 0$. Similarly, $\frac{1}{1+y} \geq 1-y > 0$, and $\frac{1}{1+z} \geq 1-z > 0$. Multiplying these three inequalities, we get
$$1 > \frac{1}{(1+x)(1+y)(1+z)} \geq (1-x)(1-y)(1-z) > 0,$$
Since the function $y = \sin \thet... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 736,612 |
Example 44: If $a, b, c$ are positive numbers, prove:
$$\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{c}^{2}}{\mathrm{~b}^{2}}+\frac{\mathrm{a}^{2}}{\mathrm{c}^{2}}+3 \geq 2\left(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}\right)$$ | Prove: Let $\frac{a}{b}=x, \frac{b}{c}=y, \frac{c}{a}=z$, then the original inequality is equivalent to:
If positive numbers $x, y, z$ satisfy $xyz=1$, then the inequality holds:
$$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+3 \geq 2(x+y+z)$$
That is,
$$x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+3 \geq 2(x+y+z)$$
Notin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,613 |
Example 45: Given that $n$ is a positive integer, and the positive number $x$ satisfies the relation $x^{n+1}-x-n=0$, prove:
$$\sqrt[n+2]{n+1}<x<\sqrt[n+1]{n+2}$$ | Proof: Obviously, for a positive number $x \neq 1$, applying the $(n+1)$-element mean inequality:
On one hand, we have
$$x^{n+1}=x+n=x+1+1+\cdots+1>(n+1) \sqrt[n+1]{x}$$
which implies $\square$
$$x^{(n+1)^{2}}>(n+1)^{n+1} x$$
which is equivalent to
$$x^{(n+1)^{2}-1}=x^{n(n+2)}>(n+1)^{n+1}>(n+1)^{n}$$
Thus, we have $... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,614 |
Example 46: Let $a, b, c \geq 0, a+b+c=1$, prove that:
$$\sqrt{9-8 a^{2}}+\sqrt{9-8 b^{2}}+\sqrt{9-8 c^{2}} \geq 7$$ | Proof: First, we prove that if $x \geq 0, y \geq 0, x+y \leq 9$, then the inequality holds:
$$\sqrt{9-x}+\sqrt{9-y} \geq 3+\sqrt{9-x-y}$$
Indeed,
$$\begin{aligned}
\sqrt{9-x}+\sqrt{9-y} & =\sqrt{(\sqrt{9-x}+\sqrt{9-y})^{2}} \\
& =\sqrt{18-x-y+2 \sqrt{9-x} \sqrt{9-y}} \\
& =\sqrt{18-x-y+2 \sqrt{9^{2}-9 x-9 y+x y}} \\
&... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,616 |
Example 47: Given that $a, b, \mathrm{c}$ are positive numbers satisfying $\mathrm{abc}=1$, prove:
$$\frac{a}{2 a^{2}+b^{2}+3}+\frac{b}{2 b^{2}+c^{2}+3}+\frac{c}{2 c^{2}+a^{2}+3} \leq \frac{1}{2} \text {. }$$ | Proof: Since $2a \leq 1 + a^2$, $2b \leq 1 + b^2$, $2c \leq 1 + c^2$, to prove inequality (2), it suffices to prove the following inequality:
$$\frac{1 + a^2}{2(1 + a^2) + (1 + b^2)} + \frac{1 + b^2}{2(1 + b^2) + (1 + c^2)} + \frac{1 + c^2}{2(1 + c^2) + (1 + a^2)} \leq 1$$
Let $x = \frac{1 + b^2}{1 + a^2}$, $y = \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,617 |
Example 48: (Mathematical Problems in "Mathematics Bulletin" 497, 1987, 10, Sun Weixin) Let x, y, z all be positive numbers, prove that $\left|\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-x}{z+x}\right|<1$. | Prove: Let \( M=\frac{x-y}{x+y}+\frac{y-z}{y+z}+\frac{z-x}{z+x} \), by replacing \( z-x \) with \( (z-y)+(y-x) \), we get
\[
\begin{aligned}
M & =(x-y)\left(\frac{1}{x+y}-\frac{1}{z+x}\right)+(y-z)\left(\frac{1}{y+z}-\frac{1}{z+x}\right) \\
& =\frac{(x-y)(y-z)}{z+x} \cdot\left(-\frac{1}{x+y}+\frac{1}{y+z}\right) \\
& =... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,618 |
Example 50: (2008, IMO-2) If $\mathrm{x}, \mathrm{y}$ and z are 3 real numbers not equal to 1, such that $x y z=1$, prove:
$$\frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geq 1$$ | Proof: Let $\frac{x}{x-1}=a, \frac{y}{y-1}=b, \frac{z}{z-1}=c$, then $x=\frac{a}{a-1}, y=\frac{b}{b-1}, z=\frac{c}{c-1}$, thus, we have $\frac{a b c}{(a-1)(b-1)(c-1)}=1$, simplifying, we get $a b+b c+c a=a+b+c-1$. (*) The inequality to be proven is equivalent to $a^{2}+b^{2}+c^{2} \geq 1$.
This inequality is equivalen... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,620 |
4. (IMO-46 Problem) Given $x, y, z \in \mathbb{R}^{+}, xyz \geq 1$, prove that
$$\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}-y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}-z^{2}}{z^{5}+x^{2}+y^{2}} \geq 0 \text {. }$$ | Prove that first, we need to show
\[
\frac{x^{5}-x^{2}}{x^{5}+y^{2}+z^{2}} \geq \frac{x^{2}-\frac{1}{x}}{x^{2}+y^{2}+z^{2}}
\]
This inequality is equivalent to
\[
\left(x^{5}-x^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) \geq \left(x^{2}-\frac{1}{x}\right)\left(x^{5}+y^{2}+z^{2}\right)
\]
Using the difference method, we ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,621 |
5. Given $x, y, z \in R^{+}, x^{2}+y^{2}+z^{2} \leq \sqrt{3} x y z$, prove: $x+y+z \leq x y z$. | We can easily prove that $a b+b c+c a \leq \frac{1}{3}(a+b+c)^{2}, a b+b c+c a \leq a^{2}+b^{2}+c^{2}$, thus we have
$$\begin{array}{l}
\frac{x+y+z}{x y z}=\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \leq \frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^{2} \\
=\frac{1}{3}\left(\frac{x y+y z+z x}{x y z}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,622 |
6. Given $x, y, z \in R^{+}, x+y+z \geq x y z$, prove that $x^{4}+y^{4}+z^{4} \geq x^{2} y^{2} z^{2}$. | Proof: From the inequality $a^{2}+b^{2}+c^{2} \geq a b+b c+c a$,
$$a^{2}+b^{2}+c^{2} \geq \frac{1}{3}(a+b+c)^{2} \geq a b+b c+c a$$
and the given conditions, we get
$$\begin{array}{c}
\frac{x^{4}+y^{4}+z^{4}}{x^{2} y^{2} z^{2}} \geq \frac{x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}{x^{2} y^{2} z^{2}} \\
=\frac{1}{x^{2}}+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,623 |
8. If $a, b, c \in R^{+}$, the cubic equation $x^{3}-a x^{2}+b x-c=0$ has 3 real roots, prove that
$$\frac{3}{a}-\frac{6}{b}+\frac{1}{c} \geq 0 .$$ | Proof: Let the 3 real roots of the known equation be $p, q, r$. By the relationship between roots and coefficients, we have
$$\left\{\begin{array}{c}
p+q+r=a \\
p q+q r+r p=b \\
p q r=c
\end{array}\right.$$
Thus, we have
$$\left\{\begin{array}{c}
p+q+r>0 \\
p q+q r+r p>0 \\
p q r>0
\end{array}\right.$$
From this, we ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,625 |
Example 6: If $a, b$ are positive numbers, and satisfy $a^{4}+b^{4}=1$, prove: $\sqrt[3]{a}+b \geq \sqrt{a^{2}}+b^{2}$. | To prove: The inequality to be proved is equivalent to
$$(a+b)^{2} \geq\left(a^{2}+b^{2}\right)^{3}$$
which is
$$(a+b)^{2} \geq\left(a^{2}+b^{2}\right)^{2}\left(a^{2}+b^{2}\right)$$
equivalent to
$$a^{2}+b^{2}+2 a b \geq\left(a^{4}+b^{4}+2 a^{2} b^{2}\right)\left(a^{2}+b^{2}\right)$$
equivalent to
$$2 a b \geq 2 a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,626 |
9. Let $a, b, c \in R^{+}$, prove:
$$\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{3}{1+a b c}$$ | Prove: The inequality to be proved is equivalent to
$$\frac{a+a b+1+a b c}{a(1+b)}+\frac{b+b c+1+a b c}{b(1+c)}+\frac{c+c a+1+a b c}{c(1+a)} \geq 6$$
is equivalent to $\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}+\frac{1+b}{b(1+c)}+\frac{c(1+a)}{1+c}+\frac{1+c}{c(1+a)}+\frac{a(1+b)}{1+a} \geq 6$,
which can be proven using th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,627 |
10. (Mathematics Teaching, Issue 6, 2001, by An Zhenping) In $\triangle ABC$, the sides opposite to angles A, B, and C are a, b, and c, respectively. If $A + C \leq 2B$, prove that $a^{4} + c^{4} \leq 2b^{4}$. | Prove $\because A+C=\pi-B \leq 2 B$,
$$\therefore B \geq \frac{\pi}{3}, \quad \cos B \leq \frac{1}{2} \text {. }$$
By the cosine rule, we get
$$b^{2}=a^{2}+c^{2}-2 a c \cos B \geq a^{2}+c^{2}-a c,$$
i.e., $a^{2}+c^{2} \leq b^{2}+a c$, thus
$$\begin{aligned}
a^{4}+c^{4} & =\left(a^{2}+c^{2}\right)^{2}-2 a^{2} c^{2} \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,628 |
11. (Mathematical Problems 999, 1996, 2) Let the side lengths of $\triangle ABC$ be $a, b, c$, and the radii of the circumcircle, incircle, and excircles be $R, r, r_{a}, r_{b}, r_{c}$, respectively. Prove that: $\frac{a^{2}}{r_{b} r_{c}}+\frac{b^{2}}{r_{c} r_{a}}+\frac{c^{2}}{r_{a} r_{b}} \geq \frac{2 R}{r}$. Equality... | Proof: From the trigonometric identities
$$R=\frac{a b c}{4 R}, r=\frac{\Delta}{s}, r_{a}=\frac{\Delta}{s-a}$$
where \( s, \triangle \) represent the semi-perimeter and area of \( \triangle ABC \), respectively. Thus, the inequality to be proven is equivalent to
$$a^{2}(s-b)(s-c)+b^{2}(s-c)(s-a)+c^{2}(s-a)(s-b) \geq \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,629 |
13. Given that $a, b, c$ are positive real numbers, $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $a+b+c \leq 3$. | Proof: Without loss of generality, let $(b-1)(c-1) \geq 0$, then $b c \geq b+c-1$.
$$4-a^{2}=b^{2}+c^{2}+a b c \geq 2 b c+a b c=b c(2+a),$$
i.e., $2-a \geq b c \geq b+c-1$, so $a+b+c \leq 3$. | a+b+c \leq 3 | Inequalities | proof | Yes | Yes | inequalities | false | 736,631 |
14. Real numbers $x, y, z$ satisfy $x y z=1$, prove that $x^{2}+y^{2}+z^{2}+3 \geq 2(x y+y z+z x)$. | Proof: Given that $x^{2} y^{2} z^{2}=1$, there must exist two of them that are both not greater than 1 or both not less than 1. Without loss of generality, assume
$$\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right) \geq 0$$
Then
$$\begin{array}{l}
x^{2}+y^{2}+z^{2}+3-2(x y+y z+z x) \\
=(x-y)^{2}+\left(\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,632 |
New problem Given $x, y \in R_{+}$, and $x \neq y$, prove:
$$\left|\frac{1}{1+x^{3}}-\frac{1}{1+y^{3}}\right|<|x-y| \text {. }$$ | Prove by imitating the proof above, first transform the left side.
$$\left|\frac{1}{1+x^{3}}-\frac{1}{1+y^{3}}\right|=|x-y| \cdot \frac{x^{2}+x y+y^{2}}{\left(1+x^{3}\right)\left(1+y^{3}\right)}<|x-y| \cdot \frac{x^{2}+x y+y^{2}}{1+x^{3}+y^{3}}$$
Thus, it suffices to prove that $\frac{x^{2}+x y+y^{2}}{1+x^{3}+y^{3}}<1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,635 |
Given: $a, b, c$ are the three sides of a triangle.
To prove: The equation $b^{2} x^{2}+\left(b^{2}+c^{2}-a^{2}\right) x+c^{2}=0$ has no real roots. | The discriminant of the equation $\Delta^{=}=\left(b^{2}+c^{2}-a^{2}\right)^{2}-4 b^{2} c^{2}$
$$=a^{4}+b^{4}+c^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2}$$
The problem is equivalent to:
Given: $a, b, c$ are the three sides of a triangle.
Prove: $a^{4}+b^{4}+c^{4}-2 a^{2} b^{2}-2 b^{2} c^{2}-2 c^{2} a^{2}<0$.
If yo... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,636 |
Example 1 (1996 Hong Kong International Mathematical Olympiad Problem) Given $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, and $\frac{\sin ^{4} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{4} \alpha}{\sin ^{2} \beta}=1$, prove: $\alpha$ $+\beta=\frac{\pi}{2}$. (Example 1 in [1]) | Note that $\sin ^{2} \alpha+\cos ^{2} \alpha=1, \sin ^{2} \beta+\cos ^{2} \beta=1$, then $1=\frac{\sin ^{4} \alpha}{\cos ^{2} \beta}+\frac{\cos ^{4} \alpha}{\sin ^{2} \beta} \geqslant \frac{\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}}{\cos ^{2} \beta+\sin ^{2} \beta}=1$, equality holds if and only if $\frac{\sin... | \alpha+\beta=\frac{\pi}{2} | Algebra | proof | Yes | Yes | inequalities | false | 736,637 |
Example 2 (1980 Leningrad Mathematical Olympiad Problem) Let $a$, $b$, $c$, $d \in \mathbf{R}^{+}$, and $a+b+c+d=1$, then $\sqrt{4 a+1}+$ $\sqrt{4 b+1}+\sqrt{4 c+1}+\sqrt{4 d+1}<6$. (Example 8 in Ref. [1], Example 3 in Ref. [3]) | $$\begin{array}{l}
\sqrt{4 a+1}+\sqrt{4 b+1}+\sqrt{4 c+1}+\sqrt{4 d+1} \\
=\frac{\sqrt{4 a+1}}{1^{-\frac{1}{2}}}+\frac{\sqrt{4 b+1}}{1^{-\frac{1}{2}}}+\frac{\sqrt{4 c+1}}{1^{-\frac{1}{2}}}+\frac{\sqrt{4 d+1}}{1^{-\frac{1}{2}}} \\
\leqslant \frac{\sqrt{(4 a+1)+(4 b+1)+(4 c+1)+(4 d+1)}}{(1+1+1+1)^{-\frac{1}{2}}} \\
=\fra... | 4 \sqrt{2} < 6 | Inequalities | proof | Yes | Yes | inequalities | false | 736,638 |
Example 11 Let real numbers $a, b$ satisfy $0<a<1,0<b<1$, and $ab=\frac{1}{36}$, find the minimum value of $u=\frac{1}{1-a}+\frac{1}{1-b}$. (Example 4 from [2]) | From the condition, we know that $a+b \geqslant 2 \sqrt{a b}=\frac{1}{3}, u=\frac{1}{1-a}+$ $\frac{1}{1-b}=\frac{1^{2}}{1-a}+\frac{1^{2}}{1-b} \geqslant \frac{(1+1)^{2}}{1-a+1-b}=\frac{4}{2-(a+b)} \geqslant$ $\frac{12}{5}$. The equality holds if and only if $\frac{1}{1-a}=\frac{1}{1-b}$ and $a+b=\frac{1}{3}$, i.e., $a=... | \frac{12}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,639 |
Example 12 Given $0<x<\frac{\pi}{2}, a, b$ are positive real numbers, $m$, $n \in \mathbf{N}_{+}$, find the minimum value of the function $y=\frac{a}{\sin ^{n} x}+\frac{b}{\cos n x}$. (Example 5 from [5]) | $$\begin{array}{l}
\text { Prove } y=\frac{a}{\sin ^{\frac{n}{} x}}+\frac{b}{\cos ^{n} x} \\
=\frac{\left(a^{\frac{2 m}{n+2 m}}\right)^{\frac{n+2 m}{2 m}}}{\left(\sin ^{2} x\right)^{\frac{n}{2 m}}}+\frac{\left(b^{\frac{2 m}{n+2 m}} \frac{n+2 m}{2 m}\right.}{\left(\cos ^{2} x\right)^{\frac{n}{2 m}}} \\
\geqslant \frac{\... | \left(a^{\frac{2 m}{2 m+n}}+b^{\frac{2 m}{2 m+n}}\right)^{\frac{2 m+n}{2 m}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,640 |
Example 3 (24th All-Soviet Union Mathematical Olympiad Problem) Given that $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$ are all positive numbers and their sum is 1, prove that: $\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2}$. (Example 9 in [1]) | Prove that from the problem, we have $\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+$ $\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}}{2\left(a_{1}+a_{2}+\cdots+a_{n}\right)}=\frac{1}{2}$. Equality holds if and only if $\frac{a_{1}}{a_{1}+a_{2}}=\frac{a_{2}}... | \frac{1}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,641 |
Example 4 Let positive numbers $a, b$ satisfy $a+b=1$, prove: $\frac{1}{a^{2}}+$ $\frac{1}{b^{2}} \geqslant 8$. (Example 1 from Text [2]) | Prove $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1^{3}}{a^{2}}+\frac{1^{3}}{b^{2}} \geqslant \frac{(1+1)^{3}}{(a+b)^{2}}=8$. Equality holds if and only if $\frac{1}{a}=\frac{1}{b}$, i.e., $a=b=\frac{1}{2}$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,642 |
Example 5 Let positive numbers $a, b$ satisfy $a+2b=1$, prove that: $\frac{1}{a}+$ $\frac{2}{b} \geqslant 9$. (Example 2 from [2]) | Prove that $\frac{1}{a}+\frac{2}{b}=\frac{1^{2}}{a}+\frac{2^{2}}{2 b} \geqslant \frac{(1+2)^{2}}{a+2 b}=9$. Equality holds if and only if $\frac{1}{a}=\frac{2}{2 b}$, i.e., $a=b=\frac{1}{3}$. | 9 | Inequalities | proof | Yes | Yes | inequalities | false | 736,643 |
Example 6 (22nd IMO problem) Let $P$ be any point inside triangle $ABC$, and the distances from $P$ to the three sides $BC$, $CA$, $AB$ are $d_{1}$, $d_{2}$, $d_{3}$ respectively. Denote $BC=a$, $CA=b$, $AB=c$. Find the minimum value of $u=\frac{a}{d_{1}} + \frac{b}{d_{2}} + \frac{c}{d_{3}}$. (Example 10 in [1]) | Let the area of $\triangle ABC$ be $S$, then $a d_{1}+b d_{2}+c d_{3} = 2 S$. Since $a, b, c, d_{1}, d_{2}, d_{3} \in \mathbf{R}^{+}$, then $u=\frac{a}{d_{1}}+\frac{b}{d_{2}} +\frac{c}{d_{3}}=\frac{a^{2}}{a d_{1}}+\frac{b^{2}}{b d_{2}}+\frac{c^{2}}{c d_{3}} \geqslant \frac{(a+b+c)^{2}}{a d_{1}+b d_{2}+c d_{3}} = \frac{... | \frac{(a+b+c)^{2}}{2 S} | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,644 |
Example 7 (36th IMO Problem) Let $a, b, c \in \mathbf{R}^{+}$, and satisfy $a b c=1$. Try to prove: $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+$ $\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$. (Example 12 in [1]) | Prove that if $abc=1$, then $\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}$
$$\begin{array}{l}
+\frac{1}{c^{3}(a+b)}=\frac{b^{2} c^{2}}{a(b+c)}+\frac{c^{2} a^{2}}{b(c+a)}+\frac{a^{2} b^{2}}{c(a+b)} \geqslant \\
\frac{(bc+ca+ab)^{2}}{2(ab+ac+bc)}=\frac{ab+ac+ba}{2} \geqslant \frac{3 \sqrt[3]{(abc)^{2}}}{2}=\frac{3}{2}
\end{... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,645 |
Example 8 Find the minimum value of the function $y=\frac{5}{2 \sin ^{2} x+3}+\frac{8}{5 \cos ^{2} x+6}$. (Example 2 in [4]) | Solve $y=\frac{5}{2 \sin ^{2} x+3}+\frac{8}{5 \cos ^{2} x+6}=\frac{5^{2}}{10 \sin ^{2} x+15}$ $+\frac{4^{2}}{10 \cos ^{2} x+12} \geqslant \frac{(5+4)^{2}}{10 \sin ^{2} x+15+10 \cos ^{2} x+12}=\frac{81}{37}$ when and only when $\frac{5}{10 \sin ^{2} x+15}=\frac{4}{10 \cos ^{2} x+12}$, i.e., $\sin ^{2} x=$ $\frac{5}{9}$,... | \frac{81}{37} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,646 |
Example 9 (Tsinghua University Independent Admission Examination Question 2 (1) in 2009) $x, y$ are real numbers, and $x+y=1$, prove that for any positive integer $n, x^{2 n}+y^{2 n} \geqslant \frac{1}{2^{2 n-1}}$. (Example 4 in Reference [5]) | Prove that when $x=0$ or $y=0$, the inequality obviously holds; when $x, y$ has one negative number, the inequality obviously holds;
When $x, y \in \mathbf{R}^{+}$, then $x^{2 n}+y^{2 n}=\frac{x^{2 n}}{1^{2 n-1}}+\frac{y^{2 n}}{1^{2 n-1}} \geqslant$ $\frac{(x+y)^{2 n}}{(1+1)^{2 n-1}}=\frac{1}{2^{2 n-1}}$. Equality hol... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,647 |
Example 10 Find the minimum value of the function $f(x, y)=\frac{3}{2 \cos ^{2} x \sin ^{2} x \cos ^{2} y}+$ $\frac{5}{4 \sin ^{2} y}\left(x, y \neq \frac{k \pi}{2}, k \in Z\right)$. (Example 3 in [4]) | $$\begin{aligned}
& \text { Solve } \frac{3}{2 \cos ^{2} x \sin ^{2} x \cos ^{2} y}+\frac{5}{4 \sin ^{2} y} \\
= & \frac{3\left(\sin ^{2} x+\cos ^{2} x\right)}{2 \cos ^{2} x \sin ^{2} x \cos ^{2} y}+\frac{5}{4 \sin ^{2} y} \\
= & \frac{3}{2 \cos ^{2} x \cos ^{2} y}+\frac{3}{2 \sin ^{2} x \cos ^{2} y}+\frac{5}{4 \sin ^{... | \frac{29+4 \sqrt{30}}{4} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,648 |
Prove: $\frac{x_{1}}{2-x_{1}}+\frac{x_{2}}{2-x_{2}}+\frac{x_{3}}{2-x_{3}} \geqslant \frac{3}{5}$. | Prove: It can be found that when $x_{1}=x_{2}=x_{3}=\frac{1}{3}$, the inequality achieves equality. Construct the function $f(x)=\frac{x}{2-x}, x \in (0,1)$, then $f^{\prime}(x)=\frac{2}{(2-x)^{2}}, f^{\prime}\left(\frac{1}{3}\right)=\frac{18}{25}$, $f\left(\frac{1}{3}\right)=\frac{1}{5}$. Therefore, the equation of th... | \frac{3}{5} | Inequalities | proof | Yes | Yes | inequalities | false | 736,649 |
Example 1 Let positive numbers $a, b, c$ satisfy $a b c=1$, prove:
$$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leqslant 1 .(2000$$ | Proof: Let $x=a, y=1, z=\frac{1}{b}=a c$, then $a=$ $\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$, so the original inequality $\Leftrightarrow\left(\frac{x}{y}-1+\right.$ $\left.\frac{z}{y}\right)\left(\frac{y}{z}-1+\frac{x}{z}\right)\left(\frac{z}{x}-1+\frac{y}{x}\right) \leqslant 1 \Leftrightarrow$ $\frac{(x-y+z)(y-z+x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,650 |
Example 2 Given that $x, y, z$ are non-negative real numbers satisfying $x+y+z=1$, prove that: $0 \leqslant x y+y z+z x-2 x y z \leqslant \frac{7}{27}$. | Proof: The proof of the first part of the inequality is omitted, and we only prove the latter part.
By the variant of Schur's inequality (3), we have
$$
\begin{array}{l}
x y + y z + z x - 2 x y z = (x + y + z)(x y + y z + z x) - 2 x y z \\
\leqslant \frac{(x + y + z)^{3}}{4} + \frac{9 x y z}{4} - 2 x y z = \frac{(x + ... | \frac{7}{27} | Inequalities | proof | Yes | Yes | inequalities | false | 736,651 |
Example 3 Let $x, y, z \in R^{+}$, prove: $\frac{x y}{z}+\frac{y z}{x}+$ $\frac{z x}{y}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}} \cdot(2000$ IMO China National | Proof: Let $\frac{x y}{z}=a^{2}, \frac{y z}{x}=b^{2}, \frac{z x}{y}=c^{2}$, then $x=$ $a c, y=a b, z=b c$, so, the original inequality $\Leftrightarrow\left(a^{2}+b^{2}+\right.$ $\left.c^{2}\right)^{3}>8\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)$. By the variant of Schur's inequality (1) and the AM-GM inequality,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,652 |
Example 4 Let $a, b, c \in R^{+}$, prove: $\sum(b+c-$ $a)(c+a-b) \leqslant \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c}) .(2001$ | Prove: Since $\Sigma(b+c-a)(c+a-b)=$ $2(a b+b c+c a)-\left(a^{2}+b^{2}+c^{2}\right)$. Therefore, by the variant (4) of Schur's inequality, we only need to prove $3 \sqrt[3]{a^{2} b^{2} c^{2}}$ $\leqslant \sqrt{a b c}(\sqrt{a}+\sqrt{b}+\sqrt{c})$. By the AM-GM inequality, we have $\sqrt{a}+\sqrt{b}+\sqrt{c} \geqslant 3 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,653 |
Example 5 Given $x, y, z \in R^{+}, x^{2}+y^{2}+z^{2}=$ 1 , find the minimum value of $\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{x^{2}+y^{2}-x y}+\frac{z^{5}}{x^{2}+y^{2}-x y}$. (2005 Jiangsu Province Mathematical Olympiad Winter Camp) | Solution: By the power mean inequality and a variant of Schur's inequality
(1) we get $\frac{x^{5}}{y^{2}+z^{2}-y z}+\frac{y^{5}}{x^{2}+y^{2}-x y}+$ $\frac{z^{5}}{x^{2}+y^{2}-x y}=\frac{x^{6}}{x y^{2}+x z^{2}-x y z}+$ $\frac{y^{6}}{y x^{2}+y z^{2}-x y z}+\frac{z^{6}}{z x^{2}+z y^{2}-x y z} \geqslant$ $\frac{\left(x^{3}... | \frac{\sqrt{3}}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,654 |
Example 2 If real numbers $x, y, z$ satisfy $x y z=1$, prove: $x^{2}$
$$+y^{2}+z^{2}+3 \geqslant 2(x y+y z+z x) .$$ | Prove that since $x^{2} y^{2} z^{2}=1$, there must exist 2 out of $x^{2}, y^{2}, z^{2}$ that are simultaneously not greater than 1, or simultaneously not less than 1. Without loss of generality, assume these are $x^{2}, y^{2}$. Then we have $\left(\frac{1}{x^{2}}-1\right)\left(\frac{1}{y^{2}}-1\right) \geqslant 0$.
Be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,656 |
Example 3 If $a, b, c$ are positive numbers, prove:
$$\frac{b^{2}}{a^{2}}+\frac{c^{2}}{b^{2}}+\frac{a^{2}}{c^{2}}+3 \geqslant 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) .$$ | Proof: Let $\frac{a}{b}=x, \frac{b}{c}=y, \frac{c}{a}=z$, then the original inequality is equivalent to, if positive numbers $x, y, z$ satisfy $x y z=1$, then the inequality $\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}+3 \geqslant 2(x+y+z)$ holds, which is $x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}+3 \geqslant 2(x+y+z)$.... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,657 |
$$\begin{array}{l}
\text { Example } 5 \text { Let } a, b, c>0 \text {, prove that: } \\
\left(a^{5}-a^{3}+3\right)\left(b^{5}-b^{3}+3\right)\left(c^{5}-c^{3}+3\right) \geqslant 3(a \\
+b+c)^{2}
\end{array}$$ | Prove that since $\left(a^{5}-a^{3}+3\right)-\left(a^{2}+2\right)=(a-1)^{2}(a+1)\left(a^{2}+a+1\right) \geqslant 0$,
then $a^{5}-a^{3}+3 \geqslant a^{2}+2$.
Similarly, we get two more inequalities, and multiplying the three inequalities, we have
$$\begin{array}{l}
\quad\left(a^{5}-a^{3}+3\right)\left(b^{5}-b^{3}+3\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,658 |
Example 6 Given $x, y, z \in \mathbf{R}_{+}$, prove:
$$\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}} \leqslant \frac{3 \sqrt{2}}{2} .$$ | Proof According to the Pigeonhole Principle: $\sqrt{\frac{x}{x+y}}, \sqrt{\frac{y}{y+z}}, \sqrt{\frac{z}{z+x}}$ must have at least two that are simultaneously not greater than the constant $\frac{\sqrt{2}}{2}$, or not less than the constant $\frac{\sqrt{2}}{2}$. Thus, without loss of generality, let's assume $\sqrt{\fr... | \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}} \leqslant \frac{3 \sqrt{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,659 |
Example 2 Given $a, b, c > 0, a^{2}+b^{2}+c^{2}=3$, prove: the algebraic expressions $a^{2} b^{2}+c^{2}, b^{2} c^{2}+a^{2}, c^{2} a^{2}+b^{2}$, at least one of them is not greater than 2. | Proof: Since $a^{2}-1$, $b^{2}-1$, and $c^{2}-1$ must have 2 that are not both greater than zero or not both less than zero, without loss of generality, let them be $a^{2}-1$ and $b^{2}-1$. Thus, we have $\left(a^{2}-1\right)\left(b^{2}-1\right) \leqslant 0$, which implies $a^{2}+b^{2} \geqslant a^{2} b^{2}+1$.
Therefo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,660 |
Example 4 Let $x, y, z > 0, x^{2} + y^{2} + z^{2} = 1$, prove that: $x y + y z + z x - \sqrt{3} x y z \leqslant \frac{2}{3}$. | Proof: Given $x, y, z > 0, x^{2} + y^{2} + z^{2} = 1$, we know that among $x, y, z$, there must be 2 that are both not greater than $\frac{1}{\sqrt{3}}$ or both not less than $\frac{1}{\sqrt{3}}$. Without loss of generality, let these be $x, y$, then we have
$$\left(x - \frac{1}{\sqrt{3}}\right)\left(y - \frac{1}{\sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,661 |
Example 6 Given $a, b, c > 0$, prove:
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 3(a+b+c)^{2}$$ | Proof: Since $a^{2}-1, b^{2}-1, c^{2}-1$ must have 2 that are not greater than zero or not less than zero, let's assume they are $a^{2}-1, b^{2}-1$, thus we have $\left(a^{2}-1\right)\left(b^{2}-1\right) \geqslant 0$, which means $a^{2} b^{2}+1 \geqslant a^{2}+b^{2}$,
which is $\left(a^{2}+2\right)\left(b^{2}+2\right) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,662 |
Example 7 (Mathematics Bulletin, Issue 9, 2010, Problem 1872) Prove: Among any 13 real numbers, there must exist two real numbers $x, y$, such that $y>\frac{x-0.3}{1+0.3 x}$. | Prove: For any 13 real numbers $a_{1}, a_{2}, \cdots, a_{13}$, there must exist 13 real numbers $\theta_{1}, \theta_{2}, \cdots, \theta_{13}$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan \theta_{k}=a_{k}(k=1,2, \cdots, 13)$.
Divide the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ into 12 su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,663 |
Example 1 (20th Iran Olympiad Problem) Let $a, b, c \in \mathbf{R}^{+}$ and $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $a+b+c \leqslant 3$.
| Prove briefly that from $a^{2}+b^{2} \geqslant 2 a b$, we get $4=a^{2}+b^{2}+c^{2}+$ $a b c \geqslant 2 a b+c^{2}+a b c=a b(2+c)+c^{2}$,
i.e., $4-c^{2} \geqslant a b(2+c)$, simplifying to
$$a b \leqslant 2-c$$
By the Pigeonhole Principle, among the three positive real numbers $a, b, c$, there must be two that are not... | a+b+c \leqslant 3 | Inequalities | proof | Yes | Yes | inequalities | false | 736,664 |
Example 2 (2010 National Competition Guangdong Preliminary Question) Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$, prove that: $9 a b c \leqslant a b+$ $b c+c a \leqslant \frac{1}{4}(1+9 a b c)$. | To prove $a b+b c+c a \leqslant \frac{1}{4}(1+9 a b c)$, it suffices to prove $4(a b+b c+c a) \leqslant 1+9 a b c$.
By the pigeonhole principle, among the three non-negative real numbers $a, b, c$, there must be two that are not greater than or not less than $\frac{1}{3}$. Without loss of generality, assume these are ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,665 |
Example 3 For any non-negative real numbers $a, b, c, \lambda$, prove: $\left(a^{2}+\right.$ $\lambda)\left(b^{2}+\lambda\right)\left(c^{2}+\lambda\right) \geqslant \frac{3 \lambda^{2}}{4}(a+b+c)^{2}$. | By the Pigeonhole Principle, among the three non-negative real numbers $a^{2}, b^{2}, c^{2}$, there must be two that are both not greater than or not less than $\frac{\lambda}{2}$.
Without loss of generality, let these be $a^{2}$ and $b^{2}$. Therefore, $\left(a^{2}-\frac{\lambda}{2}\right) \cdot\left(b^{2}-\frac{\lam... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,666 |
Example 4 (Problem from the American Mathematical Monthly) Let $x, y, z \in(0, 1)$, and $x^{2}+y^{2}+z^{2}=1$. Find the range of the function $f=x+y+z-xyz$. | On the one hand, from the known conditions, we have $x, y, z \in (0,1)$, $y > y^2$, $z > z^2$, and $x^2 + y^2 = 1 - z^2$.
Thus, $f = x + y + z - xyz$
$\geq x + y + z - x \cdot \frac{y^2 + z^2}{2}$
$= x + y + z - x \cdot \frac{1 - x^2}{2}$
$= y + z + x \cdot \frac{1 + x^2}{2}$
$\geq y + z + x \cdot \frac{2x}{2}$
$> y^2... | \left(1, \frac{8\sqrt{3}}{9}\right] | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,667 |
Example 5 Let $a, b, c \geqslant 0$, try to prove: $a^{2}+b^{2}+c^{2}+$ $2 a b c+1 \geqslant 2(a b+b c+c a)$ | By the Pigeonhole Principle, among three non-negative real numbers $a, b, c$, there must be two that are both not greater than or not less than 1. Without loss of generality, assume these are the non-negative real numbers $a$ and $b$, so we have $(a-1) \cdot (b-1) \geqslant 0$. Therefore,
$$\begin{array}{l}
a^{2}+b^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,668 |
Example 6 Let $a, b, c$ be non-negative real numbers satisfying $a+b+c=3$, then $\left(1+a^{3}\right)\left(1+b^{3}\right)\left(1+c^{3}\right) \geqslant 8$.
| Proof: By the Pigeonhole Principle, among the three non-negative real numbers $a, b, c$, there must be two that are both not greater than or not less than 1. Without loss of generality, let these be the non-negative real numbers $a$ and $b$. Therefore, we have $(a-1) \cdot (b-1) \geqslant 0$, which implies $a b \geqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,669 |
Example 7 Let $x, y, z$ be positive real numbers satisfying $x y z=1$, and $\lambda \geqslant 1$. Prove that: $\frac{1}{\sqrt{\lambda+x}}+\frac{1}{\sqrt{\lambda+y}}+\frac{1}{\sqrt{\lambda+z}} \leqslant \frac{3}{\sqrt{\lambda+1}}$. | Proof: By the Pigeonhole Principle, among the three positive real numbers $x, y, z$, there must be two that are both not greater than or not less than 1. Without loss of generality, let these be the positive real numbers $x$ and $y$. Therefore, we have
$$
\left(\frac{1}{\sqrt{\lambda+x}}-\frac{1}{\sqrt{\lambda+1}}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,670 |
Example 8 In $\triangle ABC$, prove that: $\cos A + \cos B + \cos C \leqslant \frac{3}{2}$, with equality holding if and only if $A = B = C$. | Proof: By the Pigeonhole Principle, among angles $A, B, C$, there must be two angles that are either both not greater than or both not less than $60^{\circ}$.
Without loss of generality, let these angles be $A$ and $B$. Therefore, we have $\left[\cos A - \cos 60^{\circ}\right] \cdot \left[\cos B - \cos 60^{\circ}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,671 |
Example 9 In $\triangle A B C$, prove that: $\sin ^{2} A+\sin ^{2} B+$ $\sin ^{2} C \leqslant \frac{9}{4}$, with equality if and only if $A=B=C$. | Proof: By the Pigeonhole Principle, among angles $A, B, C$, there must be two angles that are both not greater than or not less than $60^{\circ}$.
Without loss of generality, let these angles be $A$ and $B$. Therefore, we have
\[
\left[\sin ^{2} A-\sin ^{2} 60^{\circ}\right] \cdot \left[\sin ^{2} B-\sin ^{2} 60^{\cir... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,672 |
Example 2-202010 National Competition Guangdong Preliminary Question) Let non-negative real numbers $a, b, c$ satisfy $a+b+c=1$, prove that: $9 a b c \leqslant a b + b c + c a \leqslant \frac{1}{4}(1+9 a b c)$. | To prove $a b+b c+c a \leqslant \frac{1}{4}(1+9 a b c)$, it suffices to prove $4(a b+b c+c a) \leqslant 1+9 a b c$.
By the pigeonhole principle, among the three non-negative real numbers $a, b, c$, there must be two that are both not greater than or not less than $\frac{1}{3}$. Without loss of generality, let these be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,673 |
Example 5 Let $a, b, c \geqslant 0$, try to prove: $a^{2}+b^{2}+c^{2}+$ $2 a b c+1 \geqslant 2(a b+b c+c a)$. | By the Pigeonhole Principle, among three non-negative real numbers $a, b, c$, there must be two that are both not greater than or not less than 1. Without loss of generality, assume these are the non-negative real numbers $a$ and $b$, so we have $(a-1) \cdot (b-1) \geqslant 0$. Therefore,
$$\begin{array}{l}
a^{2}+b^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,674 |
Given positive real numbers $x, y, z$ satisfying $x y z \geqslant 1$, prove:
$$\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^{11} \text {. }$$ | Obviously, equation (1)
$$\begin{array}{l}
\Leftrightarrow \frac{x^{5}}{x^{5}+y^{2}+z^{2}}+\frac{y^{5}}{y^{5}+z^{2}+x^{2}}+\frac{z^{5}}{z^{5}+x^{2}+y^{2}} \\
\geqslant \frac{x^{2}}{x^{5}+y^{2}+z^{2}}+\frac{y^{2}}{y^{5}+z^{2}+x^{2}}+\frac{z^{2}}{z^{5}+x^{2}+y^{2}}
\end{array}$$
Lemma: For positive real numbers $x, y, z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,675 |
Proposition 1 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$ real numbers $a$ 、 $b$ satisfy $4 b \geqslant a>b \geqslant 0$ Prove:
$$\frac{x^{a}}{x^{a}+y^{b}+z^{b}}+\frac{y^{a}}{y^{a}+z^{b}+x^{b}}+\frac{z^{a}}{z^{a}+x^{b}+y^{b}} \geqslant 1$$ | Proof: From $x y z \geqslant 1, \frac{a-b}{3} \geqslant 0$ we know
$$x^{\frac{a-b}{3}} \geqslant \frac{1}{y^{\frac{a-b}{3}} z^{\frac{a-6}{3}}} .$$
Since $4 b \geqslant a>b \geqslant 0$ and by the rearrangement inequality, we have
$$\begin{array}{l}
y^{b}+z^{b} \geqslant y^{\frac{4 b-a}{3}} z^{\frac{a-b}{3}}+z^{\frac{4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,676 |
Proposition 2 Positive real numbers $x, y, z$ satisfy $x y z \geqslant 1$, real numbers $a, b$ satisfy $3 b \geqslant a > b \geqslant 0$ Prove:
$$\frac{x^{b}}{x^{a}+y^{b}+z^{b}}+\frac{y^{b}}{y^{a}+z^{b}+x^{b}}+\frac{z^{b}}{z^{a}+x^{b}+y^{b}} \leqslant 1$$ | Proof: It is known that $0 \leqslant 3 b-a<2 h$. By the lemma, we have $x^{3 b-a}+y^{3-a}+z^{3 b-a} \leqslant x^{2 b}+y^{2 b}+z^{2 b}$.
By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(x^{a}+y^{b}+z^{b}\right)\left(x^{2 b-a}+y^{b}+z^{b}\right) \geqslant\left(x^{b}+y^{b}+z^{b}\right)^{2} . \\
\text { T... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,677 |
Example 1. $P$ is a point inside an equilateral $\triangle A B C$, and the ratio of the sizes of $\angle A P B, \angle B P C, \angle C P A$ is $5: 6: 7$. Then the ratio of the sizes of the three interior angles of the triangle with sides $P A, P B, P C$ is ( ) (from smallest to largest).
(A) $2: 3: 4$
(B) $3: 4: 5$
(C)... | From the given, we know that $\angle A P B=100^{\circ}, \angle B P C=$ $120^{\circ}, \angle C P A=140^{\circ}$.
From the proof of the previous problem, we know that the three interior angles of the triangle with sides of lengths $P A, P B, P C$ are (note the comparison and observation of the angles at one point):
$$10... | A | Geometry | MCQ | Yes | Yes | inequalities | false | 736,679 |
Example 3. In an equilateral $\triangle ABC$, a point $P$ inside the triangle is connected to the three vertices with $PA=6, PB=8, PC=10$. The integer closest to the area of $\triangle ABC$ is ( ).
(A) 159
(B) 131
(C) 95
(D) 79
(E) 50 | From the previous example, we know that $a=6, b=8, c=10$, the key to the problem lies in finding $u, z, zw$.
$$48+\frac{\sqrt{3}}{4} \cdot \frac{290}{3} \leqslant S_{\triangle A B C} \leqslant 48+\frac{\sqrt{3}}{4} \cdot 100,$$
i.e., $76.9 \leqslant S_{\triangle A B E} \leqslant 91.3$. Therefore, the answer is D.
If it... | D | Geometry | MCQ | Yes | Yes | inequalities | false | 736,680 |
Example 4. Positive numbers $x, y, z$ satisfy the system of equations
$$\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25 \\
\frac{y^{2}}{3}+z^{2}=9 \\
x^{2}+x y+z^{2}=16
\end{array}\right.$$
Try to find: $x y+2 y z+3 z x$. | Notice that $9=3^{2}, 16=4^{2}, 25=5^{2}$, then we can form a right-angled triangle $\triangle ABC$ with side lengths $3, 4, 5$. Inside this triangle, find a point such that the angles formed by its connections to the three vertices are $90^{\circ}, 120^{\circ}, 150^{\circ}$ (as shown in the figure). Let
$$P B=x, P C=z... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,681 |
Example 5. As shown in the figure, $AB$ and $CD$ are two mutually perpendicular diameters of circle $O$, $P$ is a point on circle $O$, $PM$ is perpendicular to $OA$, $PN$ is perpendicular to $OD$, $M$ and $N$ are the feet of the perpendiculars, $OM=u, MP=z'$, and $u=p^{m}, z'=q^{n}$, where $p, q$ are both prime numbers... | Proof from the Pythagorean theorem we get
$u^{9}+v^{2}=r^{2}$.
Since $r$ is odd, $u$ and $v$ must be one odd and one even.
If $u$ is even, then we can set
$u=2 c d, v=c^{2}-d^{2}, r=c^{2}+d^{2}$.
Since $u=p=p^{m}$ is even and $p$ is a prime, then $p=2$.
Thus, we have $u=p^{\prime m}=2^{m}=2 \cdot d$.
Therefore, $c=2^{... | proof | Geometry | proof | Yes | Yes | inequalities | false | 736,683 |
Example 3. If $a, b, c$ are positive numbers, prove:
$$a^{2a} \cdot b^{2b} \cdot c^{2c} \geqslant a^{b+c} \cdot b^{c+a} \cdot c^{a+b}$$ | Given $a \geqslant b \geqslant c>0$, then
$$\lg a \geqslant \lg b \geqslant \lg c$$
By the rearrangement inequality, we have
$$\begin{array}{l}
a \lg a + b \lg b + c \lg c \geqslant b \lg a + c \lg b + a \lg c \\
a \lg a + b \lg b + c \lg c \geqslant c \lg a + a \lg b + b \lg c
\end{array}$$
Therefore, $2 a \lg a + 2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,686 |
Example 5. Let $a, b, c$ be the lengths of the three sides of a triangle. Prove:
$$\begin{aligned}
& a^{2}(b+c-a)+b^{2}(c+a-b) \\
+ & c^{2}(a+b-c) \leqslant 3 a b c .
\end{aligned}$$ | Assume without loss of generality that $a \geqslant b \geqslant c$,
then
$$\begin{array}{l}
b(c+a-b)-a(b+c-a) \\
=(a-b)(a+b-c) \geqslant 0, \\
c(a+b-c)-b(c+a-b) \\
=(b-c)(b+c-a) \geqslant 0,
\end{array}$$
Therefore, $a(b+c-a) \leqslant b(c+a-b)$
$$\leqslant c(a+b-c)$$
Since $a \geqslant b \geqslant c$,
by the rearran... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,688 |
Example 6. Let $a, b, c$ be the side lengths of a triangle. Prove:
$$\begin{array}{l}
a^{2} b(a-b)+b^{2} c(b-c) \\
+c^{2} a(c-a) \geqslant 0
\end{array}$$ | Given $a \geqslant b \geqslant c$, from the proof of Example 5, we have
$$\begin{array}{l}
a(b+c-a) \leqslant b(c+a-b) \\
\leqslant c(a+b-c) .
\end{array}$$
Since $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c}$,
by the rearrangement inequality, we get
$$\begin{array}{l}
\frac{a}{c}(b+c-a)+\frac{b}{a}(c+a-b) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,689 |
$$\begin{array}{c}
\text { Example 7. Let } x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}, \\
y_{1} \leqslant y_{2} \leqslant \cdots \leqslant y_{n}, \\
\text { and } z_{1}, \quad z_{2}, \cdots, z_{n} \text { be a permutation of } y_{1}, y_{2}, \cdots, y_{n}.
\end{array}$$
Prove:
$$\sum_{i=1}^{n}\left(x_{i}-y_... | Prove that by the sorting principle,
i.e., $-\sum_{i=1}^{n} 2 x_{i} y_{i} \leqslant-\sum_{i=1}^{n} 2 x_{i} z_{i}$.
But $\sum_{i=1}^{n}\left(x_{i}^{2}+y_{i}^{2}\right)=\sum_{i=1}^{n}\left(x_{i}^{2}+z_{i}^{2}\right)$,
Therefore, $\sum_{\substack{i=1 \\ n}}^{n}\left(x_{i}^{n}-2 x_{i} y_{i}+y_{i}^{2}\right)$
$$\begin{arra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,690 |
Example 8. Given $a_{1}, a_{2}, \cdots, a_{n}$ are pairwise distinct positive integers. Prove that:
$$\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \geqslant \sum_{k=1}^{n} \frac{1}{k}$$ | To prove that for $a_{1}, a_{2}, \cdots, a_{n}$ arranged in order of magnitude, let them be
$$a_{1}^{\prime}\frac{1}{2^{2}}>\cdots>\frac{1}{n^{2}},$$
By the rearrangement inequality, we have
$$\sum_{k^{2} 1}^{n} \frac{a_{k}}{k^{2}} \geqslant \sum_{k=1}^{n} \frac{a_{k}^{\prime}}{k^{2}}$$
Since $a_{1}^{\prime}, a_{2}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,691 |
Example 3. Let $a_{1}, a_{2}, \cdots, a_{n}$ be pairwise distinct positive integers. Prove that: $\sum_{k=1}^{n} \frac{a_{k}}{k^{2}} \geqslant \sum_{k=1}^{n} \frac{1}{k}$. | Here $a_{1}, a_{2}, \cdots, a_{n}$ are given positive integers. In the left sum $\frac{a_{1}}{1^{2}}+\frac{a_{2}}{2^{2}}+\cdots+\frac{a_{n}}{n^{2}}$, the terms cannot be permuted. If we use "without loss of generality" $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}$, it would be incorrect. We should use an ord... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,695 |
Example 4. Let $a, b, c$ be positive numbers.
Prove that $\frac{a^{3}}{b c}+\frac{b^{3}}{a c}+\frac{c^{3}}{a b} \geqslant a+b+c$.
(*) | Assume $a \leqslant b \leqslant c$. Then $\frac{1}{b c} \leqslant \frac{1}{a c} \leqslant \frac{1}{a b}$. The left side of (*) is the sum in order.
$$\begin{array}{l}
\text { Left side }=\frac{a}{b c} \cdot a^{2}+\frac{b}{a c} \cdot b^{2}+\frac{c}{a b} \cdot c^{2} \\
\geqslant \frac{a}{b c} \cdot b^{2}+\frac{b}{a c} \c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,696 |
Example 5. Let $a, b, c$ be positive numbers. Prove:
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{1}{2}(a+b+c) \text {. (*) }$$ | Proof: Let $a \leqslant b \leqslant c$. Then $a^{2} \leqslant b^{2} \leqslant c^{2}, \frac{1}{b+c} \leqslant$ $\frac{1}{c+a} \leqslant \frac{1}{a+b}$. It is known that the left side of (*) is an ordered sum, denoted as $S$. By the rearrangement inequality, we have
$$\begin{array}{l}
S \geqslant \frac{b^{2}}{b+c}+\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,697 |
Example 6. (1) If $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{n}, b_{1} \leqslant b_{2} \leqslant \cdots \leqslant$ $b_{n}$. Then we have
$$\begin{array}{l}
\left(a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}\right) \\
\geqslant \frac{1}{n}\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(b_{1}+b_{2}+\cdots\right. \\
\l... | Proof (1) The left side is the ordered sum, denoted as $S$, then
$$\begin{array}{l}
S=a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}, \\
S \geqslant a_{1} b_{2}+a_{2} b_{3}+\cdots+a_{n} b_{1} \\
S \geqslant a_{1} b_{3}+a_{2} b_{4}+\cdots+a_{n} b_{2} \\
\cdots \cdots \\
S \geqslant a_{1} b_{n}+a_{2} b_{1}+\cdots+a_{n} b_{n-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,698 |
Example 7. Let $a, b, c$ be positive real numbers, and $abc=1$. Prove that:
$$\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \frac{3}{2}$$ | Prove that let the left side be $S$, then
$$\begin{aligned}
S & =\frac{(a b c)^{2}}{a^{3}(b+c)}+\frac{(a b c)^{2}}{b^{3}(c+a)}+\frac{(a b c)^{2}}{c^{3}(a+b)} \\
& =\frac{b c}{a(b+c)} \cdot b c+\frac{a c}{b(c+a)} \cdot a c \\
& +\frac{a b}{c(a+b)} \cdot a b
\end{aligned}$$
Let $a \leqslant b \leqslant c$, then $a b \le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,699 |
Example 8. $a_{1}, a_{2}, \cdots, a_{n}$ are positive numbers. Prove:
$$\frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \geqslant \sqrt{a_{1} a_{2} \cdots a_{n}} .$$
(Arithmetic Mean Theorem) | To prove that if the proposition holds for $n-1$ positive numbers, it also holds for $n$ positive numbers, we need to show that:
$$\frac{a_{1}+a_{2}+\cdots+a_{n}}{n} \geqslant \sqrt[n]{a_{1} a_{2} \cdots a_{n}}$$
which is equivalent to:
$$\frac{a_{1}+a_{2}+\cdots+a_{n}}{\sqrt[4]{a_{1} a_{2} \cdots a_{n}}} \geqslant n.$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,700 |
Example 9. $n \geqslant 2, x_{1}, x_{2}, \cdots, x_{n}$ are positive numbers, and $x_{1}+x_{2}+\cdots+x_{n}=1$. Prove:
$$\begin{array}{l}
\frac{x_{1}}{\sqrt{1-x_{1}}}+\frac{x_{2}}{\sqrt{1-x_{2}}}+\cdots+\frac{x_{n}}{\sqrt{1-x_{n}}} \\
\geqslant \frac{\sqrt{x_{1}}+\sqrt{x_{2}}+\cdots+\sqrt{x_{n}}}{\sqrt{n-1}}
\end{array... | Proof: Let $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$. It is easy to see that the left side of (1) is an ordered sum, denoted as $S$. Then
$$\begin{array}{l}
S \geqslant \frac{x_{2}}{\sqrt{1-x_{1}}}+\frac{x_{3}}{\sqrt{1-x_{2}}}+\cdots+\frac{x_{1}}{\sqrt{1-x_{n}}} \\
S \geqslant \frac{x_{3}}{\sqrt{1-x_{1}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,701 |
Example 10. In $\triangle A B C$, prove that;
$$\frac{\pi}{3} p \leqslant a A+b B+c C<\frac{\pi}{2} p$$ | Prove that if $M=a A+b B+c C$ (M is the cyclic sum), then
$$\begin{array}{l}
M \geqslant a B+b C+c A \\
M \geqslant a C+b A+c B
\end{array}$$
Adding the three inequalities, we get
$$\begin{aligned}
3 M \geqslant a & (A+B+C)+b(A+B+C) \\
& +c(A+B+C) \\
& =\pi(a+b+c), \\
\therefore M \geqslant & \frac{\pi}{3} p . \\
\tex... | \frac{\pi}{3} p \leqslant M < \frac{\pi}{2} p | Inequalities | proof | Yes | Yes | inequalities | false | 736,702 |
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