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Example 17 Proof: For non-negative real numbers $a, b, c$, we have
$$\sum_{c \text { cc }} \sqrt{\frac{2 a(b+c)}{(2 b+c)(b+2 c)}} \geqslant 2$$ | Prove that by Hölder's inequality,
$$\left(\sum_{\text {cyc }} \sqrt{\frac{a(b+c)}{(2 b+c)(b+2 c)}}\right)^{2} \sum_{\text {cyc }} \frac{a^{2}(2 b+c)(b+2 c)}{b+c} \geqslant(a+b+c)^{3},$$
Therefore, it suffices to prove that
$$(a+b+c)^{3} \geqslant 2 \sum_{\text {cyc }} \frac{a^{2}(2 b+c)(b+2 c)}{b+c}$$
In fact,
$$(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,178 |
Example 25 Prove: For positive real numbers $a, b, c$, we have
$$\sum_{\text {cyc }} \frac{a^{5}}{a^{3}+b^{3}}+\sum_{\text {cyc }} \frac{a^{3}}{a+b} \geqslant \sum_{\text {cyc }} \frac{a^{4}}{a^{2}+b^{2}}+\frac{a^{2}+b^{2}+c^{2}}{2} .$$ | $$\begin{aligned}
& \sum_{\text {cyc }} \frac{a^{5}}{a^{3}+b^{3}}+\sum_{\text {cyc }} \frac{a^{3}}{a+b}-\sum_{\text {cyc }} \frac{a^{4}}{a^{2}+b^{2}}-\frac{a^{2}+b^{2}+c^{2}}{2} \\
= & \sum_{\text {cyc }}\left(\frac{a^{5}}{a^{3}+b^{3}}-\frac{a^{2}+b^{2}}{4}+\frac{a^{3}}{a+b}-\frac{a^{2}+b^{2}}{4}-\frac{a^{4}}{a^{2}+b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,180 |
Example 26 Prove: For positive real numbers $a, b, c$, we have
$$\frac{a^{5}}{a^{3}+b^{3}}+\frac{b^{5}}{b^{3}+c^{3}}+\frac{c^{5}}{c^{3}+a^{3}} \geqslant \frac{a^{2}+b^{2}+c^{2}}{2} .$$ | $$\begin{aligned}
& \sum_{\text {cyc }} \frac{a^{5}}{a^{3}+b^{3}} \geqslant \frac{a^{2}+b^{2}+c^{2}}{2} \Leftrightarrow \sum_{\text {cyc }}\left(\frac{a^{5}}{a^{3}+b^{3}}-\frac{a^{2}+b^{2}}{4}\right) \geqslant 0 \\
\Leftrightarrow & \sum_{\mathrm{cyc}}\left(\frac{(a-b)\left(3 a^{4}+3 a^{3} b+2 a^{2} b^{2}+a b^{3}+b^{4}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,181 |
Example 28 Prove: For any two non-zero non-negative real numbers $a, b, c$, we have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a^{2} b+b^{2} c+c^{2} a}{a b^{2}+b c^{2}+c a^{2}} \geqslant \frac{5}{2} .$$ | Prove $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a^{2} b+b^{2} c+c^{2} a}{a b^{2}+b c^{2}+c a^{2}} \geqslant \frac{5}{2}$
$$\begin{array}{l}
\Leftrightarrow \sum_{\text {cyc }}\left(\frac{a}{b+c}-\frac{1}{2}\right) \geqslant \frac{\sum_{\text {cyc }}\left(a^{2} c-a^{2} b\right)}{\sum_{\text {cyc }} a^{2} c} \\
\L... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,182 |
Example 29 Prove: For non-negative real numbers $a, b, c$, we have
$$a^{3}+b^{3}+c^{3}+12 a b c \leqslant \sum a^{2} \sqrt{a^{2}+24 b c} .$$ | Prove $a^{3}+b^{3}+c^{3}+12 a b c \leqslant \sum_{\mathrm{cyc}} a^{2} \sqrt{a^{2}+24 b c}$
$$\Leftrightarrow \sum_{\mathrm{cyc}}\left(a^{3} b^{3}+24 a^{2} b^{2} c^{2}\right) \leqslant \sum_{\mathrm{cyc}} a^{2} b^{2} \sqrt{\left(a^{2}+24 b c\right)\left(b^{2}+24 a c\right)} .$$
Since
$$\sqrt{\left(a^{2}+24 b c\right)\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,183 |
Example 32 Prove: For non-negative real numbers $a, b, c$, we have
$$\sum_{\text {cyc }} \frac{a^{3}+a b c}{b+c} \geqslant \sum_{\text {cyc }} \frac{a\left(b^{3}+c^{3}\right)}{a^{2}+b c}$$ | $$\begin{aligned}
& \sum_{\text {cyc }}\left(\left(a^{3}+a b c\right)(a+b)(a+c)\left(a^{2}+b c\right)\left(b^{2}+a c\right)\left(c^{2}+a b\right)\right. \\
& \left.-\left(a b^{3}+a c^{3}\right)\left(b^{2}+a c\right)\left(c^{2}+a b\right)(a+b)(a+c)(b+c)\right) \\
= & \sum_{\text {cyc }}\left(a^{5}+a^{4} b+a^{4} c+2 a^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,184 |
3. The three roots $\alpha, \beta, \gamma$ of the equation $x^{3}+a x^{2}+b x+c=0$ are all real numbers, and $a^{2}=2 b +2$. Prove: $|a-c| \leqslant 2$. | 3. Using Vieta's formulas for a cubic equation, and then through appropriate identity transformations to prove.
Keep the original text's line breaks and format, and output the translation result directly. | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,185 |
Example 34 Prove: For positive real numbers $x, y, z$, we have
$$\begin{aligned}
& \frac{(y+z)^{2}}{x^{2}+y z}+\frac{(z+x)^{2}}{y^{2}+z x}+\frac{(x+y)^{2}}{z^{2}+x y} \\
\geqslant & \frac{2}{3}\left(x^{2}+y z+y^{2}+z x+z^{2}+x y\right)\left(\frac{1}{x^{2}+y z}+\frac{1}{y^{2}+z x}+\frac{1}{z^{2}+x y}\right) \geqslant 6 ... | $$\begin{aligned}
& 3 \sum_{\text {cyc }}(y+z)^{2}\left(y^{2}+z x\right)\left(z^{2}+x y\right)-2 \sum_{\text {cyc }}\left(x^{2}+y z\right) \sum_{\text {cyc }}\left(y^{2}\right. \\
& +z x)\left(z^{2}+x y\right) \\
= & \left(x^{5}(y+z)+y^{5}(z+x)+z^{5}(x+y)\right)-\left(x^{4}\left(y^{2}+z^{2}\right)+y^{4}\left(z^{2}+x^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,186 |
Example 35 Prove: For positive real numbers $a, b, c$, we have
$$\begin{aligned}
& (8 a+5 b+5 c)^{3}+(5 a+8 b+5 c)^{3}+(5 a+5 b+8 c)^{3} \\
\geqslant & 1944(a+b+c)(a b+b c+c a) .
\end{aligned}$$ | $$\begin{aligned}
& \sum_{\text {cyc }}(8 a+5 b+5 c)^{3} \geqslant \frac{1}{3} \sum_{\text {cyc }}(8 a+5 b+5 c) \sum_{\text {cyc }}(8 a+5 b+5 c)^{2} \\
= & \sum_{\text {cyc }} 6 a \sum_{\text {cyc }}\left(114 a^{2}+210 a b\right) \geqslant \sum_{\text {cyc }} 6 a \sum_{\text {cyc }} 324 a b \\
= & 1944(a+b+c)(a b+a c+b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,187 |
Example 38 Prove: For non-negative real numbers $a, b, c$, we have
$$\sum_{\text {cyc }}\left(a^{2}-b c\right) \sqrt{a^{2}+4 b c} \geqslant 0 .$$ | Prove
$$\begin{aligned}
& \sum_{\text {cyc }}\left(a^{2}-b c\right) \sqrt{a^{2}+4 b c} \geqslant 0 \\
\Leftrightarrow & \sum_{\mathrm{cyc}} a^{2} \sqrt{a^{2}+4 b c} \geqslant \sum_{\mathrm{cyc}} b c \sqrt{a^{2}+4 b c} \\
\Leftrightarrow & \sum_{\text {cyc }}\left(a^{6}+4 a^{4} b c+2 a^{2} b^{2} \sqrt{\left(a^{2}+4 b c\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,188 |
Example 41 When $1 \leqslant a, b, c, d \leqslant 2$, prove:
$$\frac{4}{3} \leqslant \frac{a}{b+c d}+\frac{b}{c+d a}+\frac{c}{d+a b}+\frac{d}{a+b c} \leqslant 2$$ | Prove
$$\begin{aligned}
\sum_{\text {cyc }} \frac{a}{b+c d} & =\sum_{\text {cyc }} \frac{a^{2}}{a b+a c d} \\
& \geqslant \sum_{\text {cyc }} \frac{a^{2}}{a b+2 c d} \\
& \geqslant \frac{(a+b+c+d)^{2}}{3(a b+b c+c d+d a)} \\
& =\frac{(a+b+c+d)^{2}}{3(a+c)(b+d)} \geqslant \frac{4}{3}
\end{aligned}$$
On the other hand, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,189 |
Example 44 Prove: For non-negative real numbers $a, b, c$, we have
$$\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a} \geqslant \frac{a^{2}}{a^{2}+2 b c}+\frac{b^{2}}{b^{2}+2 c a}+\frac{c^{2}}{c^{2}+2 a b}$$ | Proof
$$\begin{aligned}
& \frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a} \geqslant \sum_{\text {cyc }} \frac{a^{2}}{a^{2}+2 b c} \\
\Leftrightarrow & \sum_{\text {cyc }}\left(\frac{a^{2}}{a b+a c+b c}-\frac{a^{2}}{a^{2}+2 b c}\right) \geqslant 0 \\
\Leftrightarrow & \sum_{\text {cyc }} \frac{a^{2}(a-b)(a-c)}{a^{2}+2 b c} \geqsl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,191 |
Example 48 Prove: For positive real numbers $a, b, c$, we have
$$\frac{a b}{c^{2}+c a}+\frac{b c}{a^{2}+a b}+\frac{c a}{b^{2}+b c} \geqslant \frac{a}{a+c}+\frac{b}{b+a}+\frac{c}{c+b}$$ | Prove that after eliminating the denominator and rearranging, we get
$$\sum_{\mathrm{cyc}}\left(a^{4} c^{2}-a^{3} c^{2} b\right)+\sum_{\mathrm{cyc}}\left(a^{3} b^{3}-a^{2} b^{2} c^{2}\right) \geqslant 0,$$
By the Arithmetic Mean-Geometric Mean Inequality, it is easy to see that the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,192 |
Example 50 Proof: For positive real numbers $x, y, z$, we have
$$\sum_{x y c} \sqrt{3 x(x+y)(x+z)} \leqslant \sqrt{4(x+y+z)^{3}}$$ | Prove
$$\begin{aligned}
& \sum_{\mathrm{cyc}} \sqrt{3 x(x+y)(x+z)} \leqslant \sqrt{4(x+y+z)^{3}} \\
\Leftrightarrow & 4(x+y+z)^{3} \geqslant \sum_{\mathrm{cyc}} 3 x(x+y)(x+z) \\
& +6 \sum_{\mathrm{cyc}}(x+y) \sqrt{x y(x+z)(y+z)}
\end{aligned}$$
Since
$$2 \sqrt{x y(x+z)(y+z)} \leqslant x(y+z)+y(x+z),$$
it suffices to ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,193 |
Example 52 Let $a, b, c \in [0,1]$. Prove:
$$\frac{a}{b^{3}+c^{3}+7}+\frac{b}{c^{3}+a^{3}+7}+\frac{c}{a^{3}+b^{3}+7} \leqslant \frac{1}{3} .$$ | Prove $\begin{aligned} \sum_{\mathrm{cyc}} \frac{a}{b^{3}+c^{3}+7} & =\sum_{\mathrm{cyc}} \frac{a}{b^{3}+2+c^{3}+2+3} \\ & \leqslant \sum_{\mathrm{cyc}} \frac{a}{3 b+3 c+3 a}=\frac{1}{3} .\end{aligned}$ | \frac{1}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 736,194 |
1. Prove: For positive real numbers $a, b, c$, we have
$$\frac{a}{a^{2}+b c}+\frac{b}{b^{2}+a c}+\frac{c}{c^{2}+b a} \geqslant \frac{3}{a+b+c}$$ | 1. Removing the denominators, expanding, and rearranging yields
$$(a-b)^{2}(a-c)^{2}(b-c)^{2}+\sum_{\mathrm{cyc}}\left(a^{3} b^{3}+a^{3} b^{2} c+a^{3} c^{2} b+a^{2} b^{2} c^{2}\right) \geqslant 0$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,195 |
5. Proof:
$$\left(a^{2}+b^{2}+c^{2}\right)\left(a^{4}+b^{4}+c^{4}\right) \geqslant(b-c)^{2}(c-a)^{2}(a-b)^{2} .$$ | 5. Prove the stronger inequality:
$$\begin{aligned}
\left(a^{2}+b^{2}+c^{2}\right)\left(a^{4}+b^{4}+c^{4}\right) & \geqslant(b-c)^{2}(c-a)^{2}(a-b)^{2} \\
& +\frac{74 a^{2} b^{2} c^{2}}{9}+\frac{7(a+b+c)^{6}}{6561}
\end{aligned}$$
The equality holds when $a=b=c$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,196 |
2. Prove: For positive real numbers $a, b, c$, we have
$$\sum_{\mathrm{cyc}} \frac{1}{5\left(a^{2}+b^{2}\right)-a b} \geqslant \frac{1}{a^{2}+b^{2}+c^{2}}$$ | $$\text { 2. } \begin{aligned}
\sum_{\mathrm{cyc}} \frac{1}{5 a^{2}+5 b^{2}-a b} & =\sum_{\mathrm{cyc}} \frac{(a+b+2 c)^{2}}{(a+b+2 c)^{2}\left(5 a^{2}+5 b^{2}-a b\right)} \\
& \geqslant \frac{16(a+b+c)^{2}}{\sum_{\mathrm{cyc}}(a+b+2 c)^{2}\left(5 a^{2}+5 b^{2}-a b\right)},
\end{aligned}$$
To prove the original inequa... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,197 |
3. Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$\frac{\sqrt{a^{2}+a b c}}{a b+c}+\frac{\sqrt{b^{2}+a b c}}{b c+a}+\frac{\sqrt{c^{2}+a b c}}{c a+b} \leqslant \frac{1}{2 \sqrt{a b c}} .$$ | $$\text { 3. } \begin{array}{l}
\sum_{\text {cyc }} \frac{\sqrt{a^{2}+a b c}}{a b+c}=\sum_{\mathrm{cyc}} \frac{\sqrt{a(a+b)(a+c)}}{(a+c)(b+c)}=\sum_{\text {cyc }} a \sqrt{\frac{a+b}{a(a+c)(b+c)^{2}}} \\
\leqslant \sqrt{\sum_{\mathrm{cyc}} \frac{a+b}{(a+c)(b+c)^{2}}}, \\
(a+b)^{2}(a+c)^{2}(b+c)^{2}(a+b+c) \geqslant 4 a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,198 |
4. Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \geqslant \frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c} .$$ | 4. Let $c=\min \{a, b, c\}$, then
$$\begin{aligned}
& 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \geqslant \frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c} \\
\Leftrightarrow & 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\right) \geqslant \sum_{\text {cyc }}\left(\frac{2 a}{b+c}-1\right) \\
\Leftrightarrow & 2\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,199 |
5. Using only the Arithmetic Mean-Geometric Mean Inequality, prove:
$$\sin A+\sin B+\sin C \leqslant \frac{3 \sqrt{3}}{2}$$ | 5. $\begin{aligned} & \sin A+\sin B+\sin C=\frac{2}{\sqrt{3}}\left(\sin A \cdot \frac{\sqrt{3}}{2}+\sin B \cdot \frac{\sqrt{3}}{2}\right) \\ & +\sqrt{3}\left(\frac{\sin A}{\sqrt{3}} \cdot \cos B+\frac{\sin B}{\sqrt{3}} \cdot \cos A\right) \\ \leqslant & \frac{1}{\sqrt{3}}\left(\left(\sin ^{2} A+\frac{3}{4}\right)+\left... | \frac{3 \sqrt{3}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,200 |
6. Let non-negative real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Prove
$$\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{y}{1-z^{2}} \geqslant \frac{3 \sqrt{3}}{2}$$ | $$\text { 6. } \begin{array}{l}
\sum_{\text {cyc }} \frac{x}{1-x^{2}} \geqslant \frac{3 \sqrt{3}}{2} \\
\Leftrightarrow \sum_{\text {cyc }}\left(\frac{x}{1-x^{2}}-\frac{\sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}\left(x^{2}-\frac{1}{3}\right)\right) \geqslant 0 \\
\Leftrightarrow \sum_{\text {cyc }} \frac{x\left(3 \sqrt{3} x^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,201 |
7. Let positive real numbers $a, b, c$ satisfy $a+b+c=3$. Prove:
$$\sum_{c=1} \frac{a}{3 a^{2}+a b c+27} \leqslant \frac{3}{31}$$ | 7. It is easy to obtain
$$3 a b c \geqslant 4(a b+b c+c a)-9$$
Therefore, to prove the original inequality, it is only necessary to prove
$$\sum_{\mathrm{cyc}} \frac{3 a}{9 a^{2}+4(a b+b c+c a)+72} \leqslant \frac{3}{31}$$
That is,
$$\begin{array}{l}
\sum_{\text {cyc }}\left(1-\frac{31 a(a+b+c)}{9 a^{2}+4(a b+b c+c a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,202 |
8. Let non-negative real numbers $a, b, c$ satisfy $abc=1$. Prove:
$$18\left(\frac{1}{a^{3}+1}+\frac{1}{b^{3}+1}+\frac{1}{c^{3}+1}\right) \leqslant(a+b+c)^{3}$$ | $$\begin{array}{l}
\text { 8. } 18\left(\frac{1}{a^{3}+1}+\frac{1}{b^{3}+1}+\frac{1}{c^{3}+1}\right) \leqslant(a+b+c)^{3} \\
\Leftrightarrow(a+b+c)^{3} \geqslant 18 a^{2} b^{2} c^{2} \sum_{\text {cyc }} \frac{1}{a^{3}+a b c} \\
\Leftrightarrow(a+b+c)^{3} \geqslant 18 a b c \sum_{\text {cyc }} \frac{b c}{a^{2}+b c} \\
\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,203 |
9. Let positive real numbers $a, b, c$ satisfy $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=1$. Prove: $\left(a^{2}+b^{2}+c^{2}\right)^{2}+a b c\left(\sqrt{a^{2}+b^{2}+c^{2}}\right)^{3} \geqslant 4$. | 9.
$$\begin{array}{l}
\left(a^{2}+b^{2}+c^{2}\right)^{2}+a b c\left(\sqrt{a^{2}+b^{2}+c^{2}}\right)^{3} \\
=\left(a^{2}+b^{2}+c^{2}\right)^{2}+a b c \sqrt{a^{2}+b^{2}+c^{2}} \cdot \sqrt{\left(a^{2}+b^{2}+c^{2}\right)^{2}} \\
\geqslant\left(a^{2}+b^{2}+c^{2}\right)^{2}+a b c \sqrt{3\left(a^{2}+b^{2}+c^{2}\right)} \\
\ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,204 |
6. Prove: For non-negative real numbers $x, y, z$, we have
$$\left(x^{2}+y^{2}+z^{2}+y z+z x+x y\right)^{2} \geqslant 4(x+y+z)\left(x^{2} y+y^{2} z+z^{2} x\right) \text {. }$$ | 6. $\begin{array}{l}\left(x^{2}+y^{2}+z^{2}+yz+zx+xy\right)^{2}-4(x+y+z)\left(x^{2}y+y^{2}z+z^{2}x\right) \\ =(z+x)^{2}(x-y)(x-z)+(x+y)^{2}(y-z)(y-x) \\ \quad+(y+z)^{2}(z-x)(z-y) \geqslant 0 .\end{array}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,205 |
7. Prove: For real numbers $a, b, c, x, y, z$, when $|x|+|z| \geqslant|y|$, we have $x^{2}(a-b)(a-c)+y^{2}(b-c)(b-a)+z^{2}(c-a)(c-b) \geqslant 0$. | 7. $\begin{array}{l} x^{2}(a-b)(a-c)+y^{2}(b-c)(b-a)+z^{2}(c-a)(c-b) \\ \quad=(|x| a-(|z|+|x|) b+|z| c)^{2} \\ +\left((|z|+|x|)^{2}-y^{2}\right)(a-b)(b-c) \geqslant 0 .\end{array}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,206 |
10. Let positive real numbers $x, y, z$ satisfy $x+y+z=1$. Prove:
(1) $\frac{x}{\sqrt{\frac{1}{y}-1}}+\frac{y}{\sqrt{\frac{1}{z}-1}}+\frac{z}{\sqrt{\frac{1}{x}-1}} \leqslant \frac{3 \sqrt{3}}{4} \sqrt{(1-x)(1-y)(1-z)}$;
(2) $\frac{x}{\sqrt[3]{\frac{1}{y}-1}}+\frac{y}{\sqrt[3]{\frac{1}{z}-1}}+\frac{z}{\sqrt[3]{\frac{1}{... | $$\begin{array}{l}
10 . \\
\text { (1) }\left(7 x^{3}(y+z)+x^{2}\left(9 y^{2}+16 y z+9 z^{2}\right)+11 x y z(y+z)\right. \\
\left.+2 y^{2} z^{2}\right)^{2}-48 x(x+y)(x+z)(x+y+z)(y z+z x+x y)^{2} \\
=\frac{x^{6}(y-z)^{2}\left(y^{2}+6 y z+z^{2}\right)}{(y+z)^{2}} \\
+\frac{2 x^{5}(y-z)^{2}\left(3 z^{2}+8 y z+3 y^{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,209 |
11. Let $x, y, z>0, x^{2}+y^{2}+z^{2}=1$. Prove:
$$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}-\frac{2\left(x^{3}+y^{3}+z^{3}\right)}{x y z} \geqslant 3 .$$ | 11.
$$\begin{aligned}
& \frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}-\frac{2\left(x^{3}+y^{3}+z^{3}\right)}{x y z}-3 \\
= & \sum_{\text {cyc }}\left(\frac{x^{2}+y^{2}+z^{2}}{x^{2}}-\frac{2 x^{2}}{y z}\right)-3 \\
= & \sum_{\text {cyc }} x^{2}\left(\frac{1}{y^{2}}-\frac{2}{y z}+\frac{1}{z^{2}}\right) \\
= & \sum_{\te... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,210 |
12. Let positive real numbers $a, b, c$ satisfy $a+b+c=1$. Prove:
$$\left(\frac{a}{1+a}\right)^{2}+\left(\frac{b}{1+b}\right)^{2}+\left(\frac{c}{1+c}\right)^{2} \geqslant \frac{3}{16}$$ | 12. Since
$$\sum_{\text {cyc }}\left(\frac{a^{2}}{(1+a)^{2}}-\frac{1}{16}-\frac{9}{32} \cdot\left(a-\frac{1}{3}\right)\right) \geqslant 0$$
i.e.,
$$\sum_{\mathrm{cyc}} \frac{(3 a-1)^{2}(1-a)}{(1+a)^{2}} \geqslant 0$$
we can deduce that the original inequality holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,211 |
13. Let non-negative real numbers $a, b, c, d$ satisfy $a+b+c+d=4$. Prove:
$$a b(a+b+2 c)+b c(b+c+2 d)+c d(c+d+2 a)+d a(d+a+2 b) \leqslant 16 .$$ | 13. $\begin{aligned} & (a+b+c+d)^{3}-8(b c d+c d a+d a b+a b c)-4\left(a^{2}(b+d)\right. \\ & \left.+b^{2}(c+a)+c^{2}(d+b)+d^{2}(a+c)\right) \\ = & (a+b+c+d)(a-b+c-d)^{2} \geqslant 0\end{aligned}$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,212 |
14. Prove: For positive numbers $x, y, z$, we have
$$\frac{x}{\sqrt{x^{2}+2 y z}}+\frac{y}{\sqrt{y^{2}+2 z x}}+\frac{z}{\sqrt{z^{2}+2 x y}}<2$$ | 14. Since
$$\begin{aligned}
& \left(x^{2}+2 y z\right)\left(x(y+z)+y^{2}+z^{2}\right)^{2}-\left(x^{2}(y+z)+y^{2}(z+x)\right. \\
& \left.+z^{2}(x+y)\right)^{2} \\
= & 2 x(y+z)\left(y^{2}+z^{2}\right)+2 y^{4}-y^{3} z+2 y^{2} z^{2}-y z^{3}+2 z^{4} \\
= & 2 x(y+z)\left(y^{2}+z^{2}\right)+(y-z)^{2}\left(y^{2}+y z+z^{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,213 |
15. For any real number $x$, prove:
$$5 x^{4}+x^{2}+2>5 x$$ | 15. $5 x^{4}+x^{2}+2-5 x=5\left(x^{2}-\frac{1}{3}\right)^{2}+\frac{13}{3}\left(x-\frac{15}{26}\right)^{2}+\frac{1}{468}>0$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,214 |
16. Real numbers $x, y, z$ are not equal to 1 and satisfy $x y z=1$. Prove:
$$\frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}} \geqslant 1$$ | 16. $\begin{aligned} & \frac{x^{2}}{(x-1)^{2}}+\frac{y^{2}}{(y-1)^{2}}+\frac{z^{2}}{(z-1)^{2}}-1 \\ \equiv & \frac{a^{6}}{\left(a^{3}-a b c\right)^{2}}+\frac{b^{6}}{\left(b^{3}-a b c\right)^{2}}+\frac{c^{6}}{\left(c^{3}-a b c\right)^{2}}-1 \\ = & \frac{(b c+c a+a b)^{2}\left(b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}-a^{2} b ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,215 |
17. Let $a_{n}=\sum_{k=1}^{n} \frac{1}{k(n+1-k)}$. Prove that for any positive integer $n \geqslant 2$, $a_{n+1}<a_{n}$. | 17. From the known, we have
$$a_{n}=\sum_{k=1}^{n} \frac{1}{k(n+1-k)}=\frac{2}{n+1} \sum_{k=1}^{n} \frac{1}{k},$$
In fact,
$$\frac{1}{k(n+1-k)}=\frac{1}{n+1}\left(\frac{1}{k}+\frac{1}{n+1-k}\right),$$
Thus, for any positive integer \( n \geqslant 2 \), we have
$$\begin{aligned}
\frac{1}{2}\left(a_{n}-a_{n+1}\right) &... | proof | Algebra | proof | Yes | Yes | inequalities | false | 736,216 |
18. If $x, y, z$ are all positive real numbers, and $x^{2}+y^{2}+z^{2}=1$. Prove:
(a) $\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z} \geqslant \sqrt{3}$;
(b) $\frac{y^{2} z}{x^{2}}+\frac{x^{2} z}{y^{2}}+\frac{x^{2} y}{z^{2}} \geqslant \sqrt{3}$;
(c) $\frac{y^{2} z^{3}}{x^{4}}+\frac{x^{2} z^{3}}{y^{4}}+\frac{x^{2} y^{3}}{z^... | 18. (a) From $\left(\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}\right)^{2}=\sum_{\mathrm{cyc}}\left(\frac{x z}{y}-\frac{x y}{z}\right)^{2}+3 \sum_{\mathrm{cyc}} x^{2} \geqslant 3$, we know the original inequality holds, with equality if and only if $x=y=z=\frac{\sqrt{3}}{3}$.
(b) Squaring the left side of the original in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,217 |
19. Let $x, y$ be any real numbers. Prove:
$$3(x+y+1)^{2}+1 \geqslant 3 x y .$$ | 19. $3(x+y+1)^{2}+1-3 x y=\frac{1}{4}(3 x+3 y+4)^{2}+\frac{3}{4}(x-y)^{2} \geqslant 0$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,218 |
20. Let $x, y, z \in \mathbf{R}^{+}$. Prove:
$$\frac{2\left(x^{3}+y^{3}+z^{3}\right)}{x y z}+\frac{9(x+y+z)^{2}}{x^{2}+y^{2}+z^{2}} \geqslant 33 .$$ | 20. Since
$$\begin{aligned}
& 2\left(\sum_{\mathrm{cyc}} x^{3}\right)\left(\sum_{\mathrm{cyc}} x^{2}\right)+9 x y z\left(\sum_{\mathrm{cyc}} x\right)^{2}-33 x y z\left(\sum_{\mathrm{cyc}} x^{2}\right) \\
= & \left(\sum_{\mathrm{cyc}}(y-z)^{2}\right)\left(\sum_{\mathrm{cyc}} x^{3}+\sum_{\mathrm{cyc}} x^{2}(y+z)-9 x y z\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,219 |
21. Let \(a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant 0, B_{k}=\sum_{i=1}^{k} B_{i}\) (with the convention \(B_{0}=0\)), and
$$B^{\prime} \leqslant B_{k} \leqslant B \quad (k=1,2, \cdots, n)$$
Prove the Abel inequality:
$$a_{1} B^{\prime} \leqslant \sum_{i=1}^{n} a_{i} b_{i} \leqslant a_{1} B .$$ | 21. $\sum_{i=1}^{n} a_{i} b_{i}=\sum_{i=1}^{n} a_{i}\left(B_{i}-B_{i-1}\right)=\sum_{i=1}^{n-1}\left(a_{i}-a_{i-1}\right) B_{i}+a_{n} B_{n}$,
Therefore,
$$\sum_{i=1}^{n} a_{i} b_{i} \leqslant B\left(\sum_{i=1}^{n-1}\left(a_{i}-a_{i+1}\right)+a_{n}\right)=a_{1} B$$
Similarly, the other half of the original inequality ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,220 |
22. When $x, y, z \in [1,2]$, prove the following inequality holds, and specify the conditions under which equality holds:
$$\left(\sum_{\text{cyc}} x\right)\left(\sum_{\text{cyc}} \frac{1}{x}\right) \geqslant 6\left(\sum_{\text{cyc}} \frac{x}{y+z}\right).$$ | 22. First, we have
$$\begin{array}{c}
\left(\sum_{\mathrm{cyc}} x\right)\left(\sum_{\mathrm{cyc}} \frac{1}{x}\right)-9=\sum_{\mathrm{cyc}} \frac{(x-y)^{2}}{x y} \\
6\left(\sum_{\mathrm{cyc}} \frac{x}{y+z}\right)-9=\sum_{\mathrm{cyc}} \frac{3(x-y)^{2}}{(x+z)(y+z)}
\end{array}$$
From equations (7) and (8), to prove the ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,221 |
23. Find the largest real number $m$ such that the inequality
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+m \leqslant \frac{1+x}{1+y}+\frac{1+y}{1+z}+\frac{1+z}{1+x}$$
holds for any positive real numbers $x, y, z$ satisfying $x y z=x+y+z+2$. | 23. First guess the value of $m$ to be $\frac{3}{2}$, which is achieved if and only if $x=y=z=2$.
Let $x=\frac{1}{a}-1, y=\frac{1}{b}-1, z=\frac{1}{c}-1$, then we have $\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1$, which means $a+b+c=1$. Thus, th... | \frac{3}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,222 |
25. Starting from the following lemma, prove the arithmetic mean-geometric mean inequality.
Lemma: If for all $\nu, a_{\nu-1} \leqslant a_{\nu}, b_{\nu-1} \leqslant b_{\nu}, a_{\nu} \leqslant b_{\nu}$, then when $a_{i}$ and $b_{i}$ are swapped, $\sum a_{\nu} \sum b_{\nu}$ does not decrease, and except for the cases wh... | 25. We can rewrite the arithmetic mean-geometric mean inequality as
$$\begin{array}{r}
\left(x_{1}+x_{1}+\cdots+x_{1}\right) \cdot\left(x_{2}+x_{2}+\cdots+x_{2}\right) \cdot \cdots \cdot\left(x_{n}+x_{n}+\cdots+x_{n}\right) \\
\leqslant\left(x_{1}+x_{2}+\cdots+x_{n}\right) \cdot\left(x_{1}+x_{2}+\cdots+x_{n}\right) \cd... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,224 |
26. Proof:
$$x_{1} x_{2} \cdots x_{n}=1, x_{i} \geqslant 0$$
implies $x_{1}+x_{2}+\cdots+x_{n} \geqslant n$. | 26. Assuming the conclusion holds for $n$, and
$$x_{1} x_{2} \cdots x_{n} x_{n+1}=1$$
Without loss of generality, assume $x_{1} \geqslant 1, x_{2} \leqslant 1$, then we have $\left(x_{1}-1\right)\left(x_{2}-1\right) \leqslant 0$, or
$$x_{1} x_{2}+1 \leqslant x_{1}+x_{2}$$
Therefore, applying the assumption to $n$ qua... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,225 |
Example 6 Prove the Cauchy inequality:
$$\left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}^{2}\right) \geqslant\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} .$$ | $$\begin{aligned}
& \left(\sum_{i=1}^{n} a_{i}^{2}\right)\left(\sum_{i=1}^{n} b_{i}^{2}\right)-\left(\sum_{i=1}^{n} a_{i} b_{i}\right)^{2} \\
= & \sum_{i=1}^{n} \sum_{j=1}^{n} a_{i}^{2} b_{j}^{2}-\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i} b_{i} a_{j} b_{j} \\
= & \frac{1}{2} \sum_{i=1}^{n} \sum_{j=1}^{n}\left(a_{i}^{2} b_{j}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,226 |
27. The logarithmic mean of two positive numbers $x$ and $y$ is defined as:
$$\begin{array}{c}
L(x, y)=\frac{x-y}{\ln x-\ln y}, x \neq y \\
L(x, x)=x
\end{array}$$
Prove: $L(x, y) \leqslant \frac{x+y}{2}$. | 27. Use the method of limits to prove the original statement. Construct a sequence of means involving $x$ and $y$:
$$I_{r}=I_{r}(x, y)=\frac{x+x^{\frac{r-1}{r}} y^{\frac{1}{r}}+\cdots+y}{r+1},$$
where $r=1,2,3, \cdots$. Specifically,
$I_{1}=\frac{x+y}{2}$ is the arithmetic mean,
$I_{2}=\frac{x+\sqrt{x y}+y}{3}$ is the... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,227 |
Example 1 For any real numbers $a, b, c$, prove:
$$\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geqslant 9(a b+b c+c a) .$$ | Proof: Let $a=\sqrt{2} \tan A, b=\sqrt{2} \tan B, c=\sqrt{2} \tan C$, where $A, B, C \in\left(0, \frac{\pi}{2}\right)$. Using $1+\tan ^{2} \theta=\frac{1}{\cos ^{2} \theta}$, equation (1) can be written as
$$\begin{aligned}
\frac{4}{9} \geqslant & \cos A \cos B \cos C(\cos A \sin B \sin C+\sin A \cos B \sin C \\
& +\si... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,228 |
Example 3 Non-negative real numbers $a, b, c$ satisfy $a^{2}+b^{2}+c^{2}+a b c=4$. Prove:
$$0 \leqslant a b+b c+c a-a b c \leqslant 2$$ | Prove that if $a=\min \{a, b, c\} \leqslant 1$, otherwise $a^{2}+b^{2}+c^{2}+a b c>4$. Then
$$a b+b c+c a-a b c \geqslant(1-a) b c \geqslant 0$$
On the other hand, let $a=2 p, b=2 q, c=2 r$, we get $p^{2}+q^{2}+r^{2}+2 p q r=1$. Thus, we can set $a=2 \cos A, b=2 \cos B, c=2 \cos C$, where $A, B, C \in\left[0, \frac{\p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,230 |
Example 7 (Neuberg-Pedoe Inequality) Let the side lengths of $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ be $a_{1}, a_{2}, a_{3}$ and $b_{1}, b_{2}, b_{3}$, respectively, and their areas be denoted as $S_{1}$ and $S_{2}$. Prove:
$$\begin{aligned}
& a_{1}^{2}\left(b_{2}^{2}+b_{3}^{2}-b_{1}^{2}\right)... | Prove that by slightly transforming equation (13), we can obtain its equivalent form:
$$16 S_{1} S_{2} \leqslant\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\right)\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)-2\left(a_{1}^{2} b_{1}^{2}+a_{2}^{2} b_{2}^{2}+a_{3}^{2} b_{3}^{2}\right) .$$
Rearranging terms and applying the Cauchy-Sc... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,231 |
Example 8 Real numbers $x, y, z>1$, and satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove:
$$\sqrt{x+y+z} \geqslant \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} .$$ | Proof Let $a=\sqrt{x-1}, b=\sqrt{y-1}, c=\sqrt{z-1}$, then
i.e. $\square$
$$\begin{array}{c}
\frac{1}{1+a^{2}}+\frac{1}{1+b^{2}}+\frac{1}{1+c^{2}}=2 \\
a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a^{2} b^{2} c^{2}=1
\end{array}$$
Thus, equation (25) can be rewritten as
$$\sqrt{a^{2}+b^{2}+c^{2}+3} \geqslant a+b+c$$
i.e. $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,232 |
Example 11 For the three sides $a, b, c$ of a triangle, prove:
$$\left(\sum_{\mathrm{cyc}} a\right)\left(\sum_{\mathrm{cyc}} \frac{1}{a}\right) \geqslant 6\left(\sum_{\mathrm{cyc}} \frac{a}{b+c}\right) .$$ | Proof: Let $a=y+z, b=z+x, c=x+y$, then we obtain the equivalent inequality for non-negative numbers $x, y, z$:
$$\left(\sum_{\text {cyc }} x\right)\left(\sum_{\text {cyc }} \frac{1}{y+z}\right) \geqslant 3\left(\sum_{\text {cyc }} \frac{y+z}{2 x+y+z}\right)$$
Clearing the denominators, subtracting the right side from ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,233 |
Example 12 Real numbers $x, y, z$ satisfy $x y z=8$. Prove:
$$\frac{2}{2+x^{2}}+\frac{2}{2+y^{2}}+\frac{2}{2+z^{2}} \geqslant 1 .$$ | Proof: Let $x^{2}=u^{3}, y^{2}=v^{3}, z^{2}=w^{3}$, then equation (36) can be rewritten as
$$\frac{v w}{v w+2 u^{2}}+\frac{w u}{w u+2 v^{2}}+\frac{u v}{u v+2 w^{2}} \geqslant 1,$$
where $u, v, w$ are positive real numbers, and satisfy $u v w=4$.
Since
$$\begin{aligned}
& \frac{v w}{v w+2 u^{2}}+\frac{w u}{w u+2 v^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,234 |
Example 13 Positive real numbers $x, y, z$ satisfy $x+y+z+2=x y z$. Prove:
$$x+y+z+6 \geqslant 2(\sqrt{y z}+\sqrt{z x}+\sqrt{x y}) .$$ | Proof: Let $x=\frac{v+w}{u}, y=\frac{w+u}{v}, z=\frac{u+v}{w}$, then equation (38) can be rewritten as
$$\sum_{\mathrm{cyc}} \frac{v+w}{u}+6 \geqslant 2 \sum_{\mathrm{cyc}} \sqrt{\frac{(w+u)(u+v)}{v w}}$$
where $u, v, w$ are positive real numbers.
Since $(2 v w+w u+u v)^{2}-4 v w(w+u)(u+v)=u^{2}(v-w)^{2} \geqslant 0$,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,235 |
Example 1 Given $x, y, z>0$, and $y z+z x+x y=1$. Prove:
$$\sum_{\text {cyc }} \frac{1+y^{2} z^{2}}{(y+z)^{2}} \geqslant \frac{5}{2}$$ | Proof: By the symmetry of equation (1), without loss of generality, let $x \leqslant y \leqslant z$, and $y=x+s, z=x+s+t$, where $s, t \geqslant 0$. Then,
$$\begin{aligned}
& 2 \sum_{\text {vec }}(z+x)^{2}(x+y)^{2}\left((y z+z x+x y)^{2}+y^{2} z^{2}\right) \\
& -5(y+z)^{2}(z+x)^{2}(x+y)^{2}(y z+z x+x y) \\
= & 32\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,236 |
Example 8 Let $\triangle A_{1} A_{2} A_{3}$ and $\triangle B_{1} B_{2} B_{3}$ have side lengths $a_{1}, a_{2}, a_{3}$ and $b_{1}, b_{2}, b_{3}$, and areas $S_{1}, S_{2}$, respectively. Also, let
$$H=a_{1}^{2}\left(-b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)+a_{2}^{2}\left(b_{1}^{2}-b_{2}^{2}+b_{3}^{3}\right)+a_{3}^{2}\left(b... | Prove that for the area of a triangle, we have Heron's formula:
$$\begin{array}{l}
S_{1}^{2}=2 a_{1}^{2} a_{2}^{2}+2 a_{2}^{2} a_{3}^{2}+2 a_{3}^{2} a_{1}^{2}-a_{1}^{4}-a_{2}^{4}-a_{3}^{4}, \\
S_{2}^{2}=2 b_{1}^{2} b_{2}^{2}+2 b_{2}^{2} b_{3}^{2}+2 b_{3}^{2} b_{1}^{2}-b_{1}^{4}-b_{2}^{4}-b_{3}^{4} .
\end{array}$$
Let
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,237 |
Example 7 Non-negative real numbers $x, y, z$ satisfy $x+y+z=3$. Prove: the inequality
$$\frac{x}{r y^{2}+1}+\frac{y}{r z^{2}+1}+\frac{z}{r x^{2}+1} \geqslant \frac{3}{r+1}$$
holds if and only if $r=0 \mathrm{~V} r \geqslant \frac{2}{\sqrt{3}}-1=0.1547$. The equality condition is $r=\frac{2}{\sqrt{3}}-1, x=0, y=3-\sqrt... | Proof: Without loss of generality, let $x=\min \{x, y, z\}$, then
$$\begin{aligned}
& F(x, y, z)=F(x, x+s, x+t) \\
= & 648 r(r+1)\left(s^{2}-s t+t^{2}\right) x^{4}+27\left(3\left(7 s^{3}+s^{2} t+10 s t^{2}+7 t^{3}\right) r^{2}\right. \\
& \left.+2\left(19 s^{3}-9 s^{2} t+19 t^{3}\right) r+s^{3}+3 s^{2} t-6 s t^{2}+t^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,238 |
For example, let $A, B, C$ be the interior angles of $\triangle ABC$. Prove that for any $x, y, z$, we have
$$x^{2}+y^{2}+z^{2}-2 x y \cos C-2 y z \cos A-2 z x \cos B \geqslant 0 .$$ | Prove that $x^{2}+y^{2}+z^{2}-2 x y \cos C-2 y z \cos A-2 z x \cos B$
$$=(x-y \cos C-z \cos B)^{2}+(y \sin C-z \sin B)^{2} .$$
Equation (17) immediately yields Equation (16). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,239 |
Example 7 Let $a, b, c \geqslant 0, \sum_{c y c} a=1$. Find the maximum value of $\sum_{c y c} \sqrt{a^{2}+b c}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | When $b=c=\frac{1}{2}, a=0$; or $c=a=\frac{1}{2}, b=0$; or $a=b=\frac{1}{2}, c=0$,
$$\sum_{\mathrm{cyc}} \sqrt{a^{2}+b c}=\frac{3}{2} \sum_{\mathrm{cyc}} a$$
Now we prove that the maximum value of $\sum_{\mathrm{cyc}} \sqrt{a^{2}+b c}$ is $\frac{3}{2} \sum_{\mathrm{cyc}} a=\frac{3}{2}$.
Assume without loss of generali... | \frac{3}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,240 |
Example 13 Let non-negative real numbers $a, b, c, d$ satisfy $a b c d=1$. Prove:
$$\frac{1}{2 a^{2}+a+1}+\frac{1}{2 b^{2}+b+1}+\frac{1}{2 c^{2}+c+1}+\frac{1}{2 d^{2}+d+1} \geqslant 1$$ | Prove that by successively rewriting $a, b, c, d$ as $a^{4}, b^{4}, c^{4}, d^{4}$, equation (26) becomes
$$\sum_{c y c} \frac{1}{2 a^{8}+a^{5} b c d+a^{2} b^{2} c^{2} d^{2}} \geqslant \frac{1}{a^{2} b^{2} c^{2} d^{2}}$$
Using Muirhead's theorem from Section 3.2, we get
$$\begin{array}{l}
\sum_{\text {cyc }}\left(\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,242 |
Example 14 For non-negative real numbers $a, b, c, d$, prove:
$$\sum_{\mathrm{cyc}} \frac{a}{\sqrt{b^{2}+c^{2}+d^{2}}} \geqslant 2$$ | Prove that because
$$a \sqrt{b^{2}+c^{2}+d^{2}} \leqslant \frac{a^{2}+b^{2}+c^{2}+d^{2}}{2},$$
so
$$\begin{aligned}
\sum_{c y c} \frac{a}{\sqrt{b^{2}+c^{2}+d^{2}}} & =\sum_{\mathrm{vcc}} \frac{a^{2}}{a \sqrt{b^{2}+c^{2}+d^{2}}} \\
& \geqslant \sum_{j c} \frac{2 a^{2}}{a^{2}+b^{2}+c^{2}+d^{2}}=2 .
\end{aligned}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,243 |
Example 16 Non-negative real numbers $a, b, c$ satisfy $a+b+c \neq 0$. Prove:
$$\begin{array}{l}
\frac{(b+c-3 a)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(c+a-3 b)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(a+b-3 c)^{2}}{2 c^{2}+(a+b)^{2}} \geqslant \frac{1}{2} \\
\frac{(b+c-a)^{2}}{a^{2}+(b+c)^{2}}+\frac{(c+a-b)^{2}}{b^{2}+(c+a)^{2}}+\frac{... | Prove that because
$$\begin{aligned}
& 2\left(a^{2}+b^{2}+c^{2}\right)(b+c-3 a)^{2}-\left(2 a^{2}+(b+c)^{2}\right)\left(9 a^{2}-4 b^{2}\right. \\
& \left.-4 c^{2}+12 b c-6 c a-6 a b\right) \\
= & (b-c)^{2}\left(19 a^{2}-6 a(b+c)+6(b+c)^{2}\right) \geqslant 0,
\end{aligned}$$
so
$$\sum_{\text {cyc }} \frac{(b+c-3 a)^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,244 |
Example 19 Positive real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Prove:
$$\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}-\frac{2}{x y z} \geqslant 9-6 \sqrt{3},$$
with equality if and only if $x=y=z=\frac{\sqrt{3}}{3}$. | Proof: Let $f(x, y, z)=\frac{1}{x^{2}}+\frac{1}{y^{2}}+\frac{1}{z^{2}}-\frac{2}{x y z}$, and assume without loss of generality that $x=\max \{x, y, z\} \in\left[\frac{\sqrt{3}}{3}, 1\right]$, then
$$\begin{aligned}
& f(x, y, z)-f\left(x, \sqrt{\frac{y^{2}+z^{2}}{2}}, \sqrt{\frac{y^{2}+z^{2}}{2}}\right) \\
= & \frac{\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,245 |
Example 20 For a natural number $n \geqslant 2$, and real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying $a_{1}+a_{2}+\cdots+a_{n}=$ 1. Prove:
$$\sum_{i=1}^{n} \frac{1}{1+a_{i}^{2}} \leqslant \frac{n^{3}}{n^{2}+1}$$
Equality holds if and only if $a_{1}=a_{2}=\cdots=a_{n}=\frac{1}{n}$. | Prove that for non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$,
$$\sum_{i=1}^{n} \frac{\left(\sum_{k=1}^{n} a_{k}\right)^{2}}{\left(\sum_{k=1}^{n} a_{k}\right)^{2}+a_{i}^{2}} \leqslant \sum_{i=1}^{n} \frac{\left(\sum_{k=1}^{n}\left|a_{k}\right|\right)^{2}}{\left(\sum_{k=1}^{n}\left|a_{k}\right|\right)^{2}+a_{i}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,246 |
Example 21 Let non-negative real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$. Prove:
$$\frac{x}{\sqrt{1+y z}}+\frac{y}{\sqrt{1+z x}}+\frac{z}{\sqrt{1+x y}} \leqslant \frac{3}{2}$$ | $$\begin{aligned}
& 9(10 x+7 y+7 z)^{2}\left(x^{2}+y^{2}+z^{2}+y z\right)-16\left(5 x^{2}+5 y^{2}+5 z^{2}\right. \\
& +7 y z+7 z x+7 x y)^{2} \\
= & (2 x-y-z)^{2}\left(125 x^{2}+160 x(y+z)+5 y^{2}+157 y z+5 z^{2}\right) \\
& +36(y-z)^{2}\left(7 x^{2}+y^{2}+3 y z+z^{2}\right) \geqslant 0 \\
\Rightarrow & \sum_{\text {cy... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 736,247 |
Proposition 1 If $0<\alpha<\beta \leqslant \frac{\pi}{4}$, then we have
$$\frac{\cot \beta}{\cot \alpha}<\frac{\csc \beta}{\csc \alpha}<\frac{\cos \beta}{\cos \alpha}<\frac{\sec \beta}{\sec \alpha}<\frac{\sin \beta}{\sin \alpha}<\frac{\beta}{\alpha}<\frac{\tan \beta}{\tan \alpha}$$ | Prove the following proof uses the conclusion: if $0<x<\frac{\pi}{2}$, then $\sin x<x<\tan x$.
Let $x \in\left(0, \frac{\pi}{4}\right], 0<\alpha<\beta<\frac{\pi}{4}$, and let $f_{1}(x)=\frac{\cot x}{\csc x} = \cos x$. Clearly, $f_{1}(x)$ is a decreasing function on $\left(0, \frac{\pi}{4}\right]$, then
$$\frac{\cot \b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,249 |
Proposition 2 If $\frac{\pi}{4}<\alpha<\beta \leqslant x_{0}$ (where $x_{0}$ is the solution of $\cot x=x$ in the interval $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$), then we have
$$\frac{\cot \beta}{\cot \alpha}<\frac{\cos \beta}{\cos \alpha}<\frac{\csc \beta}{\csc \alpha}<\frac{\sin \beta}{\sin \alpha}<\frac{\sec \... | Proof: Let $x \in\left(\frac{\pi}{4}, x_{0}\right], \frac{\pi}{4}\cot x_{0}=x_{0} \geqslant x$, we get $f_{8}^{\prime}(x)=\cos x-x \sin x=$ $\sin x(\cot x-x)>0$,
so $f_{8}(x)$ is an increasing function on $\left(\frac{\pi}{4}, x_{0}\right]$, then $\frac{\alpha}{\sec \alpha}<$ $\frac{\beta}{\sec \beta} \Leftrightarrow ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,250 |
Proposition 3 If $x_{0}<\alpha<\beta<\frac{\pi}{2}\left(x_{0}\right.$ is the solution of $\cot x=x$ in the interval $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$), then we have
$$\frac{\cot \beta}{\cot \alpha}<\frac{\cos \beta}{\cos \alpha}<\frac{\csc \beta}{\csc \alpha}<\frac{\sin \beta}{\sin \alpha}<\frac{\beta}{\alpha... | Proof: Let $x \in (x_0, \frac{\pi}{2}), x_0 < \alpha < \beta < \frac{\pi}{2}$ (where $x_0$ is the solution to $\cot x = x$ in the interval $(\frac{\pi}{4}, \frac{\pi}{2})$).
Let $f_9(x) = \frac{x}{\sec x} = x \cos x \left(x \in (x_0, \frac{\pi}{2})\right)$,
$\because \cot x < \cot x_0 = x_0 < x$, then $f_9'(x) = \cos ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,251 |
Example 1 Given $x>0$, find the minimum value of $f(x)=2 x+\frac{1}{x^{2}}$. | Analyze $f(x)=2 x+\frac{1}{x^{2}}=x+x+\frac{1}{x^{2}} \geqslant 3 \sqrt[3]{x \cdot x \cdot \frac{1}{x^{2}}}=$ 3, when and only when $x=\frac{1}{x^{2}}$, i.e., $x=1$, $f(x)$ has the minimum value 3. | 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,252 |
Example 2 Given $x<\frac{5}{4}$, find the maximum value of the function $f(x)=4 x-2+$ $\frac{1}{4 x-5}$. | ® Given $4 x-50$, so
$$\begin{aligned}
f(x)= & 4 x-2+\frac{1}{4 x-5}=-\left(5-4 x+\frac{1}{5-4 x}\right)+3 \leqslant \\
& -2 \sqrt{(5-4 x) \frac{1}{5-4 x}}+3=-2+3=1
\end{aligned}$$
The equality holds if and only if $5-4 x=\frac{1}{5-4 x}$, i.e., $x=1$. | 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,253 |
Example 3 Find the range of $y=\frac{x^{2}+7 x+10}{x+1}(x \neq-1)$ | This problem seems unable to use the AM-GM inequality, but we can factor the numerator to form $(x+1)$, and then separate it.
$y=\frac{x^{2}+7 x+10}{x+1}=\frac{(x+1)^{2}+5(x+1)+4}{x+1}=$
$$(x+1)+\frac{4}{x+1}+5$$
When $x+1>0$, i.e., $x>-1$, $y \geqslant 2 \sqrt{(x+1) \frac{4}{x+1}}+5=$ 9 (the equality holds if and onl... | 9 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,254 |
Example 4 Let $x \geqslant 0, y \geqslant 0, x^{2}+\frac{y^{2}}{2}=1$, then the maximum value of $x \sqrt{1+y^{2}}$ is $\qquad$ | Let $\left\{\begin{array}{l}x=\cos \theta, \\ y=\sqrt{2} \sin \theta\end{array}\left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)\right.$.
Then
$$\begin{array}{c}
x \sqrt{1+y^{2}}=\cos \theta \sqrt{1+2 \sin ^{2} \theta}= \\
\sqrt{2 \cos ^{2} \theta\left(1+2 \sin ^{2} \theta\right) \cdot \frac{1}{2}} \leqslant \\
\... | \frac{3 \sqrt{2}}{4} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,255 |
Example 5 Given $a>0, b>0, a+2b=1$, find the minimum value of $t=\frac{1}{a}+\frac{1}{b}$. | Let's multiply $\frac{1}{a}+\frac{1}{b}$ by 1, and replace 1 with $a+2b$.
$$\begin{array}{c}
\left(\frac{1}{a}+\frac{1}{b}\right) \cdot 1=\left(\frac{1}{a}+\frac{1}{b}\right) \cdot(a+2b)=1+\frac{2b}{a}+\frac{a}{b}+2= \\
3+\frac{2b}{a}+\frac{a}{b} \geqslant 3+2 \sqrt{\frac{2b}{a} \cdot \frac{a}{b}}=3+2 \sqrt{2}
\end{arr... | 3+2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,256 |
For all positive real numbers $a, b, c$, prove that
$$\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 c a}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geqslant 1 .$$
Reference [1] strengthens inequality (1) to:
For all positive real numbers $a, b, c$, prove that
$$\begin{array}{l}
\frac{a}{\sqrt{a^{2}+2(b+c)^{2}}}+\frac{b}{\sq... | Proof: Applying the generalization of the Cauchy-Schwarz inequality, we get
$$\begin{array}{l}
(a+b+c)^{3}=\left(\sum \sqrt[3]{\frac{a}{\sqrt{a^{2}+\lambda(b+c)^{2}}}}\right. \\
\cdot \sqrt[3]{\left.\frac{a}{\sqrt{a^{2}+\lambda(b+c)^{2}}} \cdot \sqrt[3]{a\left[a^{2}+\lambda(b+c)^{2}\right]}\right)^{3}} \\
\quad \leqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,260 |
Question 2 Let $x_{k}>0(k=1,2, \cdots, n, n \geqslant 2)$, prove:
$$\frac{x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}}{n x_{1} x_{2} \cdots x_{n}}+\frac{n \sqrt[n]{x_{1} x_{2} \cdots x_{n}}}{x_{1}+x_{2}+\cdots+x_{n}} \geqslant 2 .$$ | Prove that from the 2-element arithmetic-geometric mean inequality, we get
$$\begin{array}{l}
\frac{x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}}{n x_{1} x_{2} \cdots x_{n}}+\frac{n \sqrt[n]{x_{1} x_{2} \cdots x_{n}}}{x_{1}+x_{2}+\cdots+x_{n}} \geqslant 2 \sqrt{\frac{x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}}{n x_{1} x_{2} \cdots x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,262 |
$$\begin{array}{l}
\text { Form } 1^{\circ} \text { Let } x, y, z > 0, k_{1}, k_{2} \geqslant 0, k = k_{1} + k_{2} \text {, } \\
\text { then } p = \frac{k y^{2} - k_{1} z^{2} - k_{2} x^{2}}{z + x} + \frac{k z^{2} - k_{1} x^{2} - k_{2} y^{2}}{x + y} + \\
\frac{k x^{2} - k_{1} y^{2} - k_{2} z^{2}}{y + z} \geqslant 0
\en... | Let $z+x=s_{1}, x+y=s_{2}, y+z=s_{3}$, then
$$\begin{array}{l}
y-x=s_{3}-s_{1}, z-y=s_{1}-s_{2}, x-z=s_{2}-s_{3} . \\
p=\frac{k_{1}\left(y^{2}-z^{2}\right)+k_{2}\left(y^{2}-x^{2}\right)}{z+x} \\
+\frac{k_{1}\left(z^{2}-x^{2}\right)+k_{2}\left(z^{2}-y^{2}\right)}{x+y} \\
+\frac{k_{1}\left(x^{2}-y^{2}\right)+k_{2}\left(x... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,263 |
Let $x, y, z > 0, k \geqslant 1$, then
$$\begin{aligned}
I= & \frac{(k y+z)\left(y^{2}-x^{2}\right)}{z+x}+\frac{(k z+x)\left(z^{2}-y^{2}\right)}{x+y} \\
& +\frac{(k x+y)\left(x^{2}-z^{2}\right)}{y+z} \geqslant 0
\end{aligned}$$ | $$\text { Prove } \begin{aligned}
I= & \frac{[(k-1) y+(y+z)]\left(y^{2}-x^{2}\right)}{z+x} \\
& +\frac{[(k-1) z+(z+x)]\left(z^{2}-y^{2}\right)}{x+y} \\
& +\frac{[(k-1) x+(x+y)]\left(x^{2}-z^{2}\right)}{y+z} \\
= & (k-1)\left[\frac{y\left(y^{2}-x^{2}\right)}{z+x}\right. \\
& \left.+\frac{z\left(z^{2}-y^{2}\right)}{x+y}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,264 |
$$\begin{aligned}
Q= & \frac{(a y+b x+b z)\left(y^{2}-x^{2}\right)}{z+x} \\
& +\frac{(a z+b x+b y)\left(z^{2}-y^{2}\right)}{x+y} \\
& +\frac{(a x+b y+b z)\left(x^{2}-z^{2}\right)}{y+z} \geqslant 0
\end{aligned}$$
In the form $3^{\circ}$, let $x, y, z > 0, a \geqslant 0, b \in \mathrm{R}$, then | $$\text { Proof: } \begin{aligned}
Q= & a\left[\frac{y\left(y^{2}-x^{2}\right)}{z+x}+\frac{z\left(z^{2}-y^{2}\right)}{x+y}\right. \\
& \left.+\frac{x\left(x^{2}-z^{2}\right)}{y+z}\right]+b\left[\left(y^{2}-x^{2}\right)+\left(z^{2}-y^{2}\right)+\right. \\
& \left.\left(x^{2}-z^{2}\right)\right] \\
= & a\left[\frac{y\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,265 |
Theorem 1 Let $a, b, c$ be the lengths of the sides of a triangle, $1 \leqslant \lambda \leqslant \min : \frac{b+c}{a} \cdot \frac{c+a}{b} \cdot \frac{a+b}{c}$; then
$$\begin{array}{l}
\frac{c}{b+c-\lambda a}+\frac{d}{c+a-\lambda b}+\frac{b}{a+b-\lambda c} \geqslant \frac{3}{2-\lambda} \\
\frac{b}{b+c-\lambda a}+\frac{... | Proof: Let $x=b+c-\lambda a, y=c+a-\lambda b, z=a+b-\cdot$ $\lambda c{ }^{\circ}$
Then we get $a=\frac{v+z+(\lambda-1) x}{(2-\lambda)(1+\lambda)}, b=\frac{z+x+(\lambda-1) y}{(2-\lambda)(1+\lambda)}$, $c=\frac{x+y+(\lambda-1) z}{(2-\lambda)(1+\lambda)}$
(5) Left side $=\frac{1}{(2-\lambda)(1+\lambda)}\left[3+\left(\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,266 |
Let $x, y, z \in \mathrm{R}^{+}$, prove:
$$\frac{y^{2}-x^{2}}{z+x}+\frac{z^{2}-y^{2}}{x+y}+\frac{x^{2}-z^{2}}{y+z} \geqslant 0$$ | $$\begin{array}{l}
\text{Given } x, y, z \in \mathrm{R}^{+}, \\
\therefore \quad \frac{y^{2}-x^{2}}{z+x}+\frac{z^{2}-y^{2}}{x+y}+\frac{x^{2}-z^{2}}{y+z} \\
=\frac{y^{2}-x^{2}}{z+x}+(x-y)+\frac{z^{2}-y^{2}}{x+y}+(y-z)+ \\
\frac{x^{2}-z^{2}}{y+z}+(z-x) \\
=\frac{(y-x)(y-z)}{z+x}+\frac{(z-y)(z-x)}{x+y}+ \\
\frac{(x-z)(x-y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,267 |
Example 1 Given that $x, y, z$ are positive real numbers, and $x y + y z + z x = 1$, prove:
$$(x+y)(y+z)(z+x) \geq \frac{8}{9}(x+y+z)$$ | Proof: By the AM-GM inequality for three variables, we have
$$x+y+z=(x+y+z) \cdot(x y+y z+z x) \geq 3 \sqrt[3]{x y z} \cdot 3 \sqrt[3]{(x y z)^{2}}=9 x y z$$
$$\text { i.e., } 9 x y z \leq \frac{1}{9}(x+y+z) \text { , }$$
Thus, by the identity (*), we get
$$\begin{aligned}
(x+y)(y+z)(z+x) & =(x+y+z)(x y+y z+z x)-x y z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,268 |
Example 2 (2006 Turkey National Team Selection Exam) Given positive numbers $x, y, z$ satisfying $x y+y z+z x=1$, prove:
$$\frac{27}{4}(x+y)(y+z)(z+x) \geq(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x})^{2} \geq 6 \sqrt{3}$$ | To prove: In fact, in the proof of Example 1, it has already been proven that
$$x+y+z \geq 9xyz$$
Similarly, it is easy to prove that
$$x+y+z \geq \sqrt{3}$$
From Example 1 and the above two inequalities, it is easy to prove this problem. In fact,
on the one hand, we first prove:
$$\frac{27}{4}(x+y)(y+z)(z+x) \geq (\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,269 |
Example 3 (1992 Poland---Austria Mathematical Olympiad) Let $a, b, c$ be positive real numbers, prove the inequality:
$$2 \sqrt{a b+b c+c a} \leq \sqrt{3} \cdot \sqrt[3]{(b+c)(c+a)(a+b)}$$ | Proof: By the 3-variable mean inequality, it is easy to get
$$(a+b+c)(a b+b c+c a) \geq 3 a b c$$
Thus, we have $(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)-a b c \geq 8 a b c$, i.e., $a b c \leq \frac{1}{8}(a+b)(b+c)(c+a)$,
Therefore, $\quad(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)-a b c$
$$\begin{array}{c}
\geq(a+b+c)(a b+b c+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,270 |
Example 4 (2005 Romanian Mathematical Olympiad) Let $a, b, c$ be positive real numbers, and $(a+b)(b+c)(c+a)=1$, prove that: $a b+b c+c a \leq \frac{3}{4}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove: Since $(a b c)^{2}=a b \cdot b c \cdot c a \leq\left(\frac{a b+b c+c a}{3}\right)^{3}$,
thus $a b c \leq\left(\frac{a b+b c+c a}{3}\right)^{\frac{3}{2}}$.
$$(a+b+c)(a b+b c+c a)=(a+b)(b+c)(c+a)+a b c \leq 1+\left(\frac{a b+b c+c a}{3}\right)^{\frac{3}{2}}$$
and $(a+b+c)(a b+b c+c a) \geq \sqrt{3(a b+b c+c a)} \... | null | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 736,271 |
Example 5 (2004 Chinese National Team Training Problem) Let $a, b, c$ be positive real numbers, prove that:
$$\frac{a+b+c}{3} \geq \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}} \geq \frac{\sqrt{a b}+\sqrt{b c}+\sqrt{c a}}{3}$$ | To prove: First, we prove the left inequality. By the AM-GM inequality for three variables, we have
$$\begin{aligned}
3 \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}} & =3 \sqrt[3]{\frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+a}{2}} \\
& \leq \frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}=a+b+c
\end{aligned}$$
Thus, we have $\frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,272 |
Example 6 (31st IMO Preliminary Question) Let $a, b, c$ be positive real numbers. Prove that:
$$\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geq(a b+b c+c a)^{3}$$ | Proof: It is easy to prove:
$$\begin{array}{l}
a^{2}+a b+b^{2} \geq \frac{3}{4}(a+b)^{2} \\
b^{2}+b c+c^{2} \geq \frac{3}{4}(b+c)^{2} \\
c^{2}+c a+a^{2} \geq \frac{3}{4}(c+a)^{2}
\end{array}$$
Therefore, to prove the original inequality, it suffices to prove
$$27(a+b)^{2}(b+c)^{2}(c+a)^{2} \geq 64(a b+b c+c a)^{3}$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,273 |
Example 7 (2001 Korean Mathematical Olympiad) For positive real numbers $a, b, c$, prove:
$$\sqrt{\left(a^{2} b+b^{2} c+c^{2} a\right)\left(a b^{2}+b c^{2}+c a^{2}\right)} \geq 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)}$$ | Proof: Let $x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}$, then $x y z=1$, and $x, y, z$ are positive real numbers. Multiplying both sides of the original inequality by $\frac{1}{a b c}$, we transform it to
$$\sqrt{\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right)} \geq 1+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,274 |
If $a_{1}, a_{2}, a_{3}, a_{4} \in \mathrm{R}^{+}$, prove that
$$\begin{array}{l}
\frac{a_{1}^{3}}{a_{2}+a_{3}+a_{4}}+\frac{a_{2}^{3}}{a_{1}+a_{3}+a_{4}}+\frac{a_{3}^{3}}{a_{1}+a_{2}+a_{4}}+ \\
\frac{a_{4}{ }^{3}}{a_{1}+a_{2}+a_{3}} \geqslant \frac{\left(a_{1}+a_{2}+a_{3}+a_{4}\right)^{2}}{12} . \text { (1) }
\end{arra... | $$\begin{array}{l}
\text { Proof: } \because \frac{a^{2}}{b} \geqslant 2 a-b,\left(a, b \in \mathbb{R}^{+}\right) \\
\therefore \frac{\left(3 a_{1}\right)^{2}}{a_{2}+a_{3}+a_{4}} \geqslant 2\left(3 a_{1}\right)-\left(a_{2}+a_{3}+a_{4}\right) \\
=6 a_{1}-\left(a_{2}+a_{3}+a_{4}\right), \\
\text { i.e., } \frac{a_{1}^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,275 |
Promotion 1 If $a_{i} \in \mathrm{R}^{+}(i=1,2,3, \cdots, n)$, $S=\sum_{i=1}^{n} a_{i}$, and $2 \leqslant n \in \mathrm{N}$, prove that $\sum_{i=1}^{n} \frac{a_{i}^{3}}{S-a_{i}} \geqslant \frac{1}{n-1} \sum_{i=1}^{n} a_{i}^{2}$. | $$\begin{array}{l}
\left(S-a_{i}\right)^{2} \leqslant(n-1)\left[\sum_{i=1}^{n} a_{i}^{2}-a_{i}^{2}\right] \\
\because \frac{a_{i}^{3}}{S-a_{i}}+\frac{a_{i}^{3}}{S-a_{i}}+\frac{\left(S-a_{i}\right)^{2}}{(n-1)^{3}} \\
\geqslant 3 \sqrt[3]{\left(\frac{a_{i}^{3}}{S-a_{i}}\right)^{2} \frac{\left(S-a_{i}\right)^{2}}{(n-1)^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,276 |
Promotion 2 If $a_{i} \in \mathrm{R}^{+}(i=1,2,3, \cdots, n)$, $2 \leqslant n \in \mathrm{N}$, and $S=\sum_{i=1}^{n} a_{i}, m \in \mathrm{N}$, prove: $\sum_{i=1}^{n} \frac{a_{i}^{m}}{S-a_{i}} \geqslant \frac{1}{n-1} \sum_{i=1}^{n} a_{i}^{m-1}$ | To prove Generalization 2, we need the following lemma:
If \(a_{i} \in \mathrm{R}^{+} (i=1,2,3, \cdots, n), m \in \mathrm{N}\), then \(\left(\sum_{i=1}^{n} a_{i}\right)^{m} \leqslant n^{m-1} \sum_{i=1}^{n} a_{i}^{m}\). Below is the proof of Generalization 2.
Proof: By the Cauchy-Schwarz inequality, we have
\[
\begin{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,277 |
Question 1 Given $a, b, c>0$, and $a b c=1$, prove:
$$\frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2} \leqslant 1$$ | Prove that since $a^{2}+2=\left(a^{2}+1\right)+1 \geqslant 2 a+1$,
thus $a^{2}+2 \geqslant 2 a+1$.
Similarly, $b^{2}+2 \geqslant 2 b+1, c^{2}+2 \geqslant 2 c+1$.
Therefore, $\frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2} \leqslant \frac{a}{2 a+1}+\frac{b}{2 b+1}+$ $\frac{c}{2 c+1}$,
Thus, to prove $\frac{a}{a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,279 |
Question 2 Given $a, b, c>0$, and $a b c=1$, prove:
$$\frac{1}{2 a+1}+\frac{1}{2 b+1}+\frac{1}{2 c+1} \geqslant 1$$ | To prove that given $a, b, c > 0$, and $abc = 1$, we can set $a = \frac{yz}{x^2}, b = \frac{zx}{y^2}, c = \frac{xy}{z^2} (x, y, z > 0)$. Using a variant of the Cauchy-Schwarz inequality, we get:
First, we need to prove that for any $a, b, c$ satisfying the condition, there exist $x, y, z$ such that this holds, which m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,280 |
Theorem 3 Let $a, b, c$ be distinct real numbers, and $\lambda$ be a real number, then
$$\begin{array}{l}
\left(\frac{c}{a-b}+\lambda\right)^{2}+\left(\frac{a}{b-c}+\lambda\right)^{2}+\left(\frac{b}{c-a}+\lambda\right)^{2} \\
\geqslant 2\left(\lambda^{2}+1\right) .
\end{array}$$ | Prove that for $x=\frac{c}{a-b}, y=\frac{a}{b-c}, z=\frac{b}{c-a}$, we have
$$\begin{array}{l}
(x-1)(y-1)(z-1) \\
=\frac{(b+c-a)(c+a-b)(a+b-c)}{(a-b)(b-c)(c-a)} \\
=(x+1)(y+1)(z+1)
\end{array}$$
Thus, $x y+y z+z x=-1$, and we get
$$\begin{array}{l}
\left(\frac{c}{a-b}+\lambda\right)^{2}+\left(\frac{a}{b-c}+\lambda\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,287 |
A sphere with a radius of 1 can move freely in all directions inside a regular tetrahedron container with inner wall edge length of $4 \sqrt{6}$. The area of the container's inner wall that the sphere can never touch is ${ }^{-}$ $\qquad$ | Solution 1: Let the radius of the small sphere $O$ be 1, and the edge length of the inner tetrahedron $ABCD$ be $4\sqrt{6}$. The points of contact between the small sphere $O$ and the faces of the tetrahedron $ABCD$ form four congruent small equilateral triangles (as shown in Figure 1, two small equilateral triangles a... | not found | Geometry | math-word-problem | Yes | Yes | inequalities | false | 736,288 |
Problem 1 Let real numbers $a, b, c \geqslant 1$, and satisfy
$$a b c+2 a^{2}+2 b^{2}+2 c^{2}+c a-b c-4 a+4 b-c=28$$
Find the maximum value of $a+b+c$. | From the transformation of the conditional equation, we get
$$\begin{aligned}
& \left(\frac{a-1}{2}\right)^{2}+\left(\frac{b+1}{2}\right)^{2}+\left(\frac{c}{2}\right)^{2}+\left(\frac{a-1}{2}\right)\left(\frac{b+1}{2}\right) \frac{c}{2} \\
= & 4
\end{aligned}$$
Let \( x=\frac{a-1}{2}, y=\frac{b+1}{2}, z=\frac{c}{2} \),... | 6 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 736,290 |
Given $x, y, z$ are positive real numbers, $x^{2}+y^{2}+z^{2}$ $+x y z=4$, prove: $x+y+z \leqslant 3$. | Prove that among the positive real numbers $x, y, z$, there must be 2 that are simultaneously not less than 1, or not greater than 1. Without loss of generality, let these be $y, z$, then we have
$$(y-1)(z-1) \geqslant 0$$
which means $y z \geqslant y+z-1$.
Since $4-x^{2}=y^{2}+z^{2}+x y z$
$$\geqslant 2 y z+x y z=y z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,291 |
Conclusion 1 Given $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}^{+}$, prove:
$$\begin{array}{l}
\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\cdots+\frac{x_{n}^{2}}{x_{1}} \geqslant x_{1}+x_{2}+\cdots \\
+x_{n}+\frac{4\left(x_{1}-x_{2}\right)^{2}}{x_{1}+x_{2}+\cdots+x_{n}}
\end{array}$$ | Given that $x^{2}+y^{2}-2 x y=(x-y)^{2}$, we have
$$\frac{x^{2}}{y}=2 x-y+\frac{(x-y)^{2}}{y}$$
Therefore,
$$\begin{array}{l}
\frac{x_{1}^{2}}{x_{2}}=2 x_{1}-x_{2}+\frac{\left(x_{1}-x_{2}\right)^{2}}{x_{2}} \\
\frac{x_{2}^{2}}{x_{3}}=2 x_{2}-x_{3}+\frac{\left(x_{2}-x_{3}\right)^{2}}{x_{3}} \\
\cdots \cdots \\
\frac{x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 736,292 |
Question 2 Let $a, b, c \in \mathbf{R}_{+}$, and $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $a b c \leqslant 1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Prove: Using the AM-GM inequality for three variables, we get
$$4=a^{2}+b^{2}+c^{2}+a b c \geqslant 3 \sqrt[3]{(a b c)^{2}}+a b c,$$
Let $x=\sqrt[3]{a b c}$, then we have $x^{3}+3 x^{2}-4 \leqslant 0$, i.e., $(x-1)(x+2)^{2} \leqslant 0$.
Noting that $x>0$, we get $x \leqslant 1$.
Therefore, $a b c \leqslant 1$. | null | Inequalities | proof | Yes | Yes | inequalities | false | 736,294 |
Question 3 Let $a, b, c \in \mathbf{R}_{+}$, and $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $a b+b c+c a \leqslant 3$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: From problem 1, we know $a+b+c \leqslant 3$, squaring both sides gives
$$a^{2}+b^{2}+c^{2}+2(a b+b c+c a) \leqslant 9,$$
Noting the common inequality $a b+b c+c a \leqslant a^{2}+b^{2}+c^{2}$, we get $3(a b+b c+c a) \leqslant 9$,
Thus, $a b+b c+c a \leqslant 3$. | null | Inequalities | proof | Yes | Yes | inequalities | false | 736,295 |
Question 4 Let $a, b, c \in \mathbf{R}_{+}$, and $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $a^{2}+b^{2}+c^{2} \geqslant 3$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: For the conditional equation, using the conclusion from problem 2, $a b c \leqslant 1$, we can get $a^{2}+b^{2}+c^{2}+1 \geqslant 4$, therefore, we have $a^{2}+b^{2}+c^{2} \geqslant 3$. | null | Inequalities | proof | Yes | Yes | inequalities | false | 736,296 |
Question 6 Let $a, b, c \in \mathbf{R}_{+}$, and $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $0<a b+b c+c a-a b c \leqslant 2$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | Proof: From the conditional equation $a^{2}+b^{2}+c^{2}+a b c=4$, we know that one of $a, b, c$ must be no greater than 1. Without loss of generality, let $c \leqslant 1$.
On one hand, from $a b+b c+c a-a b c=a b(1-c)+b c+c a \geqslant b c+c a > 0$,
we get $a b+b c+c a-a b c > 0$.
On the other hand, for the condition... | null | Inequalities | proof | Yes | Yes | inequalities | false | 736,298 |
Question 7 Let $a, b, c \in \mathbf{R}_{+}$, and $a^{2}+b^{2}+c^{2}+a b c=4$, prove: $\frac{a}{a^{2}+4}+\frac{b}{b^{2}+4}+\frac{c}{c^{2}+4} \leqslant \frac{3}{5}$. | Prove: First, prove $\frac{t}{t^{2}+4} \leqslant \frac{3 t}{25}+\frac{2}{25}(0<t<2)$, which can be proven using the 5-term AM-GM inequality. In fact,
$$\begin{array}{l}
\frac{t}{t^{2}+4}=\frac{t}{t^{2}+1+1+1+1} \leqslant \frac{t}{5 \sqrt[5]{t^{2}}} \\
=\frac{1}{25} \cdot 5 \sqrt[5]{t \cdot t \cdot t \cdot 1 \cdot 1} \\... | \frac{a}{a^{2}+4}+\frac{b}{b^{2}+4}+\frac{c}{c^{2}+4} \leqslant \frac{3}{5} | Inequalities | proof | Yes | Yes | inequalities | false | 736,299 |
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