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Example $6.22 a, b, c \in \mathbf{R}$, prove that
$$2(1+a b c)+\sqrt{2\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)} \geqslant(1+a)(1+b)(1+c)$$ | $$\begin{array}{l}
\sqrt{2\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)}= \\
\sqrt{\left[(1+a)^{2}+(1-a)^{2}\right]\left[(b+c)^{2}+(1-b c)^{2}\right]} \geqslant \\
(1+a)(b+c)+(1-a)(b c-1)= \\
(1+a)(1+b)(1+c)-2(1+a b c)
\end{array}$$
Rearranging terms yields the desired inequality! | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,809 |
Example $6.23$ a, b, c > 0, prove that
$$\sqrt[3]{\frac{a+b}{a+c}}+\sqrt[3]{\frac{b+c}{b+a}}+\sqrt[3]{\frac{c+a}{c+b}} \leqslant \frac{a+b+c}{\sqrt[3]{a b c}}$$ | Prove that from Cauchy's generalization,
$$\begin{array}{r}
\left(\sqrt[3]{\frac{a+b}{a+c}}+\sqrt[3]{\frac{b+c}{b+a}}+\sqrt[3]{\frac{c+a}{c+b}}\right)^{3} \leqslant 6(a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) \leqslant \\
3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
\end{array}$$
It suff... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,810 |
Example 6.24 (Li Li) Cyclic quadrilateral \(ABCD\) (counterclockwise), let \(AB = b\), \(BC = a\), \(AC = c\), \(AD = d\), \(CD = e\), \(BD = f\), prove that
\[ a \sqrt{d^2 + x} + b \sqrt{e^2 + x} = c \sqrt{f^2 + x} \]
has only the real solution \(x = 0\). | Proof (Han Jingjun) By Ptolemy's theorem, we know $a d + b e = c f$. Clearly, $x = 0$ is a solution.
Next, we prove that when $x \neq 0$, the equation has no solution.
$$\begin{array}{l}
a \sqrt{d^{2} + x} + b \sqrt{e^{2} + x} = c \sqrt{f^{2} + x} \Leftrightarrow \\
a\left(\sqrt{d^{2} + x} - d\right) + b\left(\sqrt{e^{... | proof | Geometry | proof | Yes | Yes | inequalities | false | 731,811 |
Example $6.25$ Given $a, b, c \geqslant 0$ and $a b c=1$, prove that
$$\frac{1+a+a b}{a(a+b)^{4}}+\frac{1+b+b c}{b(b+c)^{4}}+\frac{1+c+c a}{c(c+a)^{4}} \geqslant \frac{81}{16(a b+b c+c a)^{2}}$$ | Prove that when $abc=1$, we have
$$\sum \frac{a}{1+a+ab}=1$$
By the Cauchy-Schwarz inequality and the Iran 1996 inequality, we have
$$\begin{aligned}
\sum \frac{1+a+ab}{a(a+b)^{4}}= & \left(\sum \frac{1+a+ab}{a(a+b)^{4}}\right)\left(\sum \frac{a}{1+a+ab}\right) \geqslant \\
& \left(\sum \frac{1}{(a+b)^{2}}\right)^{2} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,812 |
Example 6.26 (2008 IMO) $a, b, c \in \mathbf{R}$, prove that
$$\left(\frac{a}{a-b}\right)^{2}+\left(\frac{b}{b-c}\right)^{2}+\left(\frac{c}{c-a}\right)^{2} \geqslant 1$$ | Prove that by Cauchy's inequality we have
and there is also the identity
$$\begin{array}{l}
{\left[\sum_{c y c}\left(\frac{a}{a-b}\right)^{2}\right]\left[\sum_{c y c}(a-b)^{2}(a-c)^{2}\right] \geqslant} \\
\left(\sum_{c y c}|a||a-c|\right)^{2} \geqslant\left(\sum_{c y c} a^{2}-\sum_{c y c} a b\right)^{2}
\end{array}$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,813 |
Example 6.27 (2002 Vietnam Mathematical Olympiad) $a, b, c \in \mathbf{R}$, and $a^{2}+b^{2}+c^{2}=9$, prove that
$$2(a+b+c)-abc \text{ is constant }$$ | Proof: Without loss of generality, assume $a \leqslant b \leqslant c \Rightarrow 9 \geqslant \frac{3}{2}\left(a^{2}+b^{2}\right) \geqslant 3 a b \Leftrightarrow 3 \geqslant a b$. Thus, we have
$$\begin{array}{l}
2(a+b+c)-a b c=2(a+b)+c(2-a b) \leq \\
\sqrt{\left(a^{2}+b^{2}+2 a b+c^{2}\right)\left(8-4 a b+a^{2} b^{2}\r... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,814 |
Example $1.18 \ a \geqslant 1 \geqslant b>0$, prove that
$$a^{2 b}+b^{2 a} \leqslant 2$$ | Prove that by Bernoulli's inequality we have
$$\begin{array}{c}
a^{b} \leqslant 1+b(a-1)=1+b-b^{2} \\
b^{a}=b \cdot b^{a-1} \leqslant b[1+(a-1)(b-1)]=b\left[1-(b-1)^{2}\right]=b^{2}(2-b)
\end{array}$$
Since
$$a^{2 b}+b^{2 a} \leqslant\left(1+b-b^{2}\right)^{2}+b^{4}(2-b)^{2}$$
It suffices to prove
$$\begin{array}{l}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,815 |
$$x+y+z \leqslant x y z+2$$
Example 6.28 Non-negative real numbers $a, b, c$ satisfy $a+b+c=2$, prove
$$\sqrt{a+b-2 a b}+\sqrt{b+c-2 b c}+\sqrt{c+a-2 c a} \geqslant 2$$ | Proof First, it is easy to see that the original inequality is equivalent to
$$\begin{array}{l}
\sum \sqrt{(a+b)(a+b+c)-4 a b} \geqslant 2 \sqrt{2} \Leftrightarrow \\
\sum \sqrt{c(a+b)+(a-b)^{2}} \geqslant 2 \sqrt{2}
\end{array}$$
Squaring and using the Cauchy inequality, it suffices to prove
$$\sum a^{2}+\sum \sqrt{b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,816 |
Example $6.29 a, b, c>0$, prove that
$$\sum_{c y c} \frac{1}{\sqrt{a^{2}+b c}} \leqslant \sum_{c y c} \frac{\sqrt{2}}{a+b}$$ | Prove that using the Cauchy-Schwarz inequality we have
$$\begin{aligned}
\left(\sum \frac{1}{\sqrt{a^{2}+b c}}\right)^{2} \leqslant & \left(\sum \frac{1}{(a+b)(a+c)}\right)\left(\sum \frac{(a+b)(a+c)}{a^{2}+b c}\right)= \\
& \frac{2 \sum a}{(a+b)(b+c)(c+a)}\left(\sum \frac{a(b+c)}{a^{2}+b c}+3\right)
\end{aligned}$$
T... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,817 |
Example 6.32 Given $a, b, c > 0, abc = 1$, prove that
$$\sqrt{3a^2 + 4} + \sqrt{3b^2 + 4} + \sqrt{3c^2 + 4} \leq \sqrt{7}(a + b + c)$$ | First, we transform the original expression into a homogeneous form. For this, let $a=x^{3}, b=y^{3}, c=z^{3}$, then $x y z=1$. The desired inequality can be rewritten as:
$$\begin{array}{l}
\sum_{cyc} \sqrt{3 x^{6}+4 x^{2} y^{2} z^{2}} \leqslant \sqrt{7}\left(x^{3}+y^{3}+z^{3}\right) \Leftrightarrow \\
\sum_{cyc} x \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,820 |
Example $6.33 a, b, c \geqslant 0, k \geqslant-2$, prove that
$$\sqrt{\frac{a^{2}}{a^{2}+k a b+b^{2}}}+\sqrt{\frac{b^{2}}{b^{2}+k b c+c^{2}}}+\sqrt{\frac{c^{2}}{c^{2}+k c a+a^{2}}} \geqslant \min \left\{1, \frac{3}{\sqrt{k+2}}\right\}$$ | Proof: Let $x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}$, and $x y z=1$. We need to prove
$$\sum \frac{1}{\sqrt{x^{2}+k x+1}} \geqslant \min \left\{1, \frac{3}{\sqrt{k+2}}\right\}$$
When $k \geqslant 7$, since $x, y, z>0$ and $x y z=1$, there exist $m, n, p>0$ such that $x=\frac{n^{2} p^{2}}{m^{4}}, y=\frac{p^{2} m^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,821 |
Example 6.34 (2006 China National Training Team) Let $x_{1}, x_{2}, \cdots, x_{n} \geqslant 0$, and $\sum_{i=1}^{n} x_{i}=1$, prove that
$$\sum_{i=1}^{n} \sqrt{x_{i}} \sum_{i=1}^{n} \frac{1}{\sqrt{1+x_{i}}} \leqslant \frac{n^{2}}{\sqrt{n+1}}$$ | Prove that by Cauchy's inequality,
$$\begin{array}{l}
\sum_{i=1}^{n} \sqrt{x_{i}} \sum_{i=1}^{n} \frac{1}{\sqrt{1+x_{i}}}=\sum_{i=1}^{n} \sqrt{x_{i}}\left(\sum_{i=1}^{n} \sqrt{1+x_{i}}-\sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1+x_{i}}}\right) \leqslant \\
\sum_{i=1}^{n} \sqrt{x_{i}}\left(\sum_{i=1}^{n} \sqrt{1+x_{i}}-\frac{\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,822 |
Let $x_{1}, x_{2}, \cdots, x_{n} \geqslant 0, 0 \leqslant \lambda \leqslant n$ and $\sum_{i=1}^{n} x_{i}=1$, prove that
$$\sum_{i=1}^{n} \sqrt{x_{i}} \sum_{i=1}^{n} \frac{1}{\sqrt{1+\lambda x_{i}}} \leqslant \frac{n^{2}}{\sqrt{n+\lambda}}$$ | Prove that by Cauchy's inequality,
$$\begin{array}{l}
\left(\sum_{i=1}^{n} \sqrt{x_{i}}\right)^{2} \leqslant \sum_{i=1}^{n}\left(1+\lambda x_{i}\right) \sum_{i=1}^{n} \frac{x_{i}}{1+\lambda x_{i}}= \\
\quad \frac{n+\lambda}{\lambda} \sum_{i=1}^{n} \frac{\lambda x_{i}}{1+\lambda x_{i}}=\frac{n+\lambda}{\lambda}\left(n-\... | \frac{n^{2}}{\sqrt{n+\lambda}} | Inequalities | proof | Yes | Yes | inequalities | false | 731,823 |
Example 1.19 (Han Jingjun) $x_{i} \geqslant 0, i=1,2, \cdots, n . \sum_{i=1}^{n} x_{i}=1$, find
$$\max \left\{x_{1}+x_{2}+\cdots+x_{j}, x_{2}+x_{3}+\cdots+x_{j+1}, \cdots, x_{n-j+1}+x_{n-j+2} \cdots+x_{n}\right\}$$
the minimum value. | Notice
$$\begin{array}{l}
\max \left\{x_{1}+x_{2}+\cdots+x_{j}, x_{2}+x_{3}+\cdots+x_{j+1}, \cdots, x_{n-j+1}+x_{n-j+2}+\cdots+x_{n}\right\} \geqslant \\
\max \left\{x_{n-j+1}+x_{n-j+2}+\cdots+x_{n}, x_{1}+x_{2}+\cdots+x_{j}, x_{j+1}+x_{j+2}+\cdots+\right. \\
\left.x_{2 j}, \cdots, x_{\left[\frac{n}{j}\right]+1-j}+x_{\... | \frac{1}{\left[\frac{n+j-1}{j}\right]} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,826 |
Example $6.37$ a, b, c are non-negative real numbers, at most one of which is 0, prove that
$$\frac{1}{(a+2 b)^{2}}+\frac{1}{(b+2 c)^{2}}+\frac{1}{(c+2 a)^{2}} \geqslant \frac{1}{a b+b c+c a}$$ | Prove We discuss in two cases.
(1) If $4(a b+b c+c a) \geqslant a^{2}+b^{2}+c^{2}$, then by the Cauchy-Schwarz inequality we have
$$\left(\sum \frac{1}{(a+2 b)^{\frac{1}{2}}}\right)\left(\sum(a+2 b)^{2}(a+2 c)^{2}\right) \geqslant 9\left(\sum a\right)^{2}$$
Thus, it suffices to prove
$$\begin{array}{l}
9\left(\sum a\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,827 |
Example 6.38 (2007 China Western Mathematical Olympiad) Let $a, b, c$ be real numbers, satisfying $a+b+c=3$, prove that
$$\frac{1}{5 a^{2}-4 a+11}+\frac{1}{5 b^{2}-4 b+11}+\frac{1}{5 c^{2}-4 c+11} \leqslant \frac{1}{4}$$ | Proof: This problem has been introduced in local inequalities, here we provide a proof based on the Cauchy-Schwarz inequality. Clearly, there exist \(a, b\) such that
$$(a-1)(b-1) \geqslant 0 \Rightarrow a^{2}+b^{2} \leqslant 1+(a+b-1)^{2}=c^{2}-4 c+5$$
The original inequality is equivalent to
$$\left(5-\frac{51}{5 a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,828 |
Example 6.39 Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{100}$ satisfy $a_{1}^{2}+a_{2}^{2}+\cdots+a_{100}^{2}=1$, prove that $a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\cdots+a_{100}^{2} a_{1}<\frac{12}{25}$ | Prove that for $S=\sum_{k=1}^{100} a_{k}^{2} a_{k+1}$, where $a_{101}=a_{1}$, by Cauchy's inequality and the AM-GM inequality, we have
$$\begin{aligned}
(3 S)^{2}= & \left(\sum_{k=1}^{100} a_{k+1}\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\right)\right)^{2} \leqslant\left(\sum_{k=1}^{100} a_{k+1}^{2}\right)\left(\sum_{k=1}^{100}... | S \leqslant \frac{\sqrt{2}}{3}<\frac{12}{25} | Inequalities | proof | Yes | Yes | inequalities | false | 731,829 |
Example 6. $40 a, b, c$ are non-negative, and not all are 0, prove that
$$\frac{a}{a+b+7 c}+\frac{b}{b+c+7 a}+\frac{c}{c+a+7 b}+\frac{2}{3} \cdot \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \leqslant 1$$ | To prove
$$\frac{a}{a+b+c}-\frac{a}{a+b+7 c}=\frac{6 c a}{(a+b+c)(a+b+7 c)}$$
Thus, we only need to prove
$$\sum \frac{c a}{a+b+7 c} \geqslant \frac{(a+b+c)(a b+b c+c a)}{9\left(a^{2}+b^{2}+c^{2}\right)}$$
If $a=b=0$ or $b=c=0$ or $c=a=0$, the inequality becomes an equality.
For $a+b>0, b+c>0, c+a>0$, by the Cauchy-S... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,830 |
Example $6.41 a, b, c, d \geqslant 0$, not all three are 0 at the same time, prove that
$$\sqrt{\frac{a}{a+b+c}}+\sqrt{\frac{b}{b+c+d}}+\sqrt{\frac{c}{c+d+a}}+\sqrt{\frac{d}{d+a+b}} \leqslant \frac{4}{\sqrt{3}}$$ | To prove that this problem is cyclic symmetric, we hope to eliminate the square root so that one of the terms in the product is symmetric, which facilitates calculation. By the Cauchy inequality, we have
$$\begin{array}{l}
\left(\sum_{c x c} \sqrt{\frac{a}{a+b+c}}\right)^{2} \leqslant\left(\sum_{v x}(a+b+d)(a+c+d)\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,831 |
Example 6.42 (Ma Tengyu, Huang Chendi) $x_{i}>0, i=1,2, \cdots, n$, and satisfy $\sum_{i=1}^{n} x_{i}=1$, prove that
$$\sum_{i=1}^{n} \sqrt{x_{i}^{2}+x_{i+1}^{2}} \leqslant 2-\frac{1}{\frac{\sqrt{2}}{2}+\sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i+1}}}$$ | $$\begin{array}{l}
\Leftrightarrow \sum_{i=1}^{n}\left(x_{i}+x_{i+1}-\sqrt{x_{i}^{2}+x_{i+1}^{2}}\right) \geqslant \frac{1}{\frac{\sqrt{2}}{2}+\sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i+1}}} \Leftrightarrow \\
\sum_{i=1}^{n} \frac{x_{i}^{2}}{\frac{x_{i}^{2}}{x_{i+1}}+x_{i}+\frac{x_{i}}{x_{i+1}} \sqrt{x_{i}^{2}+x_{i+1}^{2}}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,832 |
Example 6.43 Let $a, b, c$ be non-negative real numbers, and not two of them are zero at the same time, prove that
$$\frac{a^{3}}{a^{2}+b^{2}}+\frac{b^{3}}{b^{2}+c^{2}}+\frac{c^{3}}{c^{2}+a^{2}} \geqslant \frac{\sqrt{3\left(a^{2}+b^{2}+c^{2}\right)}}{2}$$ | Proof Considering the inequality has only the denominator in odd power, we consider using the Cauchy inequality to convert it to even power, after which a substitution can be made to reduce the degree.
By the generalized Cauchy inequality, we have
$$\begin{array}{l}
\left(\sum_{c r c} \frac{a^{3}}{a^{2}+b^{2}}\right)^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,833 |
Example 6.44 (Jack Garfunkel, Crux, 2007 China National Training Team) $a, b, c \geqslant 0$, prove that
$$\sum_{\vartheta x} \frac{a}{\sqrt{a+b}} \leqslant \frac{5}{4} \sqrt{a+b+c}$$ | Proof of this problem is quite challenging, with equality holding when $a=3, b=1, c=0$ and its cyclic permutations, which adds significant difficulty to the proof.
By the Cauchy-Schwarz inequality, we have
$$\left(\sum \frac{a}{\sqrt{a+b}}\right)^{2} \leqslant \sum a(x a+y b+z c) \sum \frac{a}{(a+b)(x a+y b+z c)}$$
w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,834 |
Example $6.45 a, b, c \geqslant 0$, not two of them are 0 at the same time, satisfying $a+b+c=1$, prove that $a \sqrt{4 b^{2}+c^{2}}+b \sqrt{4 c^{2}+a^{2}}+c \sqrt{4 a^{2}+b^{2}} \leqslant \frac{3}{4}$ | Prove that by Cauchy's inequality, we have
$$\left(\sum a \sqrt{4 b^{2}+c^{2}}\right)^{2} \leqslant \sum a(x a+y b+z c) \sum \frac{a\left(4 b^{2}+c^{2}\right)}{(2 b+c)(x a+y b+z c)}$$
where \(x, y, z\) are non-negative and are to be determined.
Following the previous problem, we can obtain
$$\frac{(x a+y b+z c)^{2}}{4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,835 |
Example $6.46 a, b, c \geqslant 0$, prove that
$$\begin{array}{l}
\sqrt{a^{2}-a+1}+\sqrt{b^{2}-b+1}+\sqrt{c^{2}-c+1} \geqslant \\
\sqrt{(a+b+c)^{2}-3(a+b+c)+9}
\end{array}$$ | $$\begin{array}{l}
\frac{1}{\sqrt{2}}\left(\sqrt{2 a^{2}-2 a+2}+\sqrt{2 b^{2}-2 b+2}+\sqrt{2 c^{2}-2 c+2}\right)= \\
\frac{1}{\sqrt{2}}\left(\sqrt{a^{2}+(a-1)^{2}+1}+\sqrt{b^{2}+(b-1)^{2}+1}+\sqrt{c^{2}+(c-1)^{2}+1}\right) \geqslant \\
\frac{1}{\sqrt{2}} \sqrt{(a+b+c)^{2}+(a+b+c-3)^{2}+3^{2}}= \\
\frac{1}{\sqrt{2}} \sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,836 |
Example $1.20 x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}_{+}$, prove that
$$\sum_{i=1}^{n} \frac{x_{1}}{\sum_{j \neq i} x_{j}} \sum_{1 \leqslant i<j<n} x_{i} x_{j} \leqslant \frac{n}{2} \sum_{i=1}^{n} x_{i}^{2}$$ | Proof: Let $S=\sum_{i=1}^{n} x_{i}$, then the inequality to be proved $\Leftrightarrow$
$$\begin{array}{l}
\left(\sum_{i=1}^{n} \frac{x_{i}}{S-x_{i}}\right)\left(S^{2}-\sum_{i=1}^{n} x_{i}^{2}\right) \leqslant n \sum_{i=1}^{n} x_{i}^{2} \Leftrightarrow \\
S^{2} \sum_{i=1}^{n} \frac{x_{i}}{S-x_{i}} \leqslant \sum_{i=1}^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,837 |
Example $6.48 a, b, c>0, c \geqslant b \geqslant a$ and $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, prove that $a b^{2} c^{3} \geqslant 1$ | Proof (Wang Ye, selected for the 2008 National Training Team) From the conditions, we have
$$\begin{aligned}
( & \left.\sum a b\right)^{2} \geqslant 3 \sum a \cdot a b c=3 \sum a b \Rightarrow a b+b c+c a \geqslant 3 \\
a b^{2} c^{3}= & a^{2} b^{2} c^{2} \cdot \frac{c}{a} \geqslant \\
& \frac{1}{3} a^{2} b^{2} c^{2}\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,839 |
Example 6.49 (2007 Serbia Mathematical Olympiad) $x, y, z>0, x+y+z=1$, prove that
$$\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}} \geqslant \frac{1}{7}$$ | Assume without loss of generality that $x \geqslant y \geqslant z$, it is easy to see that
$$\begin{array}{l}
\frac{x^{k+1}}{x^{k+1}+y^{k}+z^{k}} \geqslant \frac{y^{k+1}}{y^{k+1}+z^{k}+x^{k}} \geqslant \frac{z^{k+1}}{z^{k+1}+x^{k}+y^{k}} \\
z^{k+1}+x^{k}+y^{k} \geqslant y^{k+1}+z^{k}+x^{k} \geqslant x^{k+1}+y^{k}+z^{k}... | \frac{1}{7} | Inequalities | proof | Yes | Yes | inequalities | false | 731,840 |
Example 6.50 (Crux1988, Walther Janous) Let $a, b, c \geqslant 0$, prove that
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}} \geqslant \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$$ | Proof: Let $a=x^{2}, b=y^{2}, c=z^{2}$.
$$\begin{array}{l}
4\left(\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}+\frac{y^{2}}{\sqrt{y^{2}+z^{2}}}+\frac{z^{2}}{\sqrt{z^{2}+x^{2}}}\right)^{2} \geqslant 2(x+y+z)^{2} \Leftrightarrow \\
\sum \frac{4 x^{4}}{x^{2}+y^{2}}+\sum \frac{8 x^{2} y^{2}}{\sqrt{\left(x^{2}+y^{2}\right)\left(y^{2}+z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,841 |
Example 6.51
Let $a, b, c>0$, prove that
$$\sum \sqrt{\frac{a^{3}}{a^{2}+a b+b^{2}}} \geqslant \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{3}}$$ | Prove that after squaring both sides, the original inequality is equivalent to
$$\begin{array}{l}
\sum \frac{a^{3}}{a^{2}+a b+b^{2}}+2 \sum \sqrt{\frac{a^{3} b^{3}}{\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)}} \geqslant \\
\frac{1}{3}\left(\sum a+2 \sum \sqrt{a b}\right)
\end{array}$$
By the rearrangemen... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,842 |
Example $6.52 a, b, c \geqslant 1$, prove that
$$\frac{(a+b)^{2 c-1}}{(a+b+1)^{2 c}}+\frac{(b+c)^{2 a-1}}{(b+c+1)^{2 a}}+\frac{(c+a)^{2 b-1}}{(c+a+1)^{2 b}} \leqslant \frac{2}{a+b+c}$$ | From Bernoulli's inequality, we have
$$\begin{aligned}
\left(\frac{a+b+1}{a+b}\right)^{c}= & \left(1+\frac{1}{a+b}\right)^{c} \geqslant 1+\frac{c}{a+b}=\frac{a+b+c}{a+b} \Rightarrow \\
& \left(\frac{a+b+1}{a+b}\right)^{2 c} \geqslant \frac{(a+b+c)^{2}}{(a+b)^{2}}
\end{aligned}$$
Therefore,
$$\begin{aligned}
\sum \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,843 |
Example $6.53 \quad 1>a \geqslant b \geqslant c>0$, prove that
$$a^{b}+b^{c}+c^{a} \geqslant \frac{3}{2}$$ | We first prove a lemma.
Lemma If $1>a, b>0$, then $a^{b} \geqslant \frac{a}{a+b}$, which is obvious when $a \geqslant 1$.
If $1>a, b>0$, by Bernoulli's inequality we have
$$a^{b}=\frac{a}{a^{1-b}} \geqslant \frac{a}{1-(1-b)(1-a)}=\frac{a}{a+b-a b} \geqslant \frac{a}{a+b}$$
Thus the lemma is proved.
Returning to the or... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,844 |
Example 6.54 Let $a, b, c>0$, prove that $a^{b+c}+b^{c+a}+c^{a+b} \geqslant 1$. | Proof We divide the proof into two parts. First, we prove that if at least one of $a, b, c$ is greater than 1, then the inequality holds. Without loss of generality, assume $a>1$. We first prove a lemma.
Lemma
$$x^{y}>\frac{x}{x+y}(x>0,0<y<1)$$
Let $f(x)=x^{y}+y x^{y+1}$, then $f^{\prime}(x)=y x^{y-1}(x+y-1)$.
It is ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,845 |
Example 6.55 Given that $a_{i}(i=1,2, \cdots, n)$ are positive numbers, $S=\sum_{i=1}^{n} a_{i}$, prove that
$$\sum_{i=1}^{n}\left(S-a_{i}\right)^{a_{i}}>n-1$$ | Prove that if there exists $a_{i} \geqslant 1$, then the desired inequality obviously holds.
Below, assume $0 < a_{i} < 1$, for $i=1,2, \cdots, n$.
We have
$$
a_{i} > \frac{S-a_{i}}{S}, \quad i=1,2, \cdots, n
$$
Adding the above $n$ inequalities, the original proposition is immediately proved.
If $S-a_{i} < 1$, then
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,846 |
Example 6.56 Let $a_{i} \geqslant-1(i=1,2, \cdots, n), \sum_{i=1}^{n} a_{i} \geqslant 0$, prove that
$$\prod_{i=1}^{n}\left(a_{i}+1\right) \geqslant 1-\frac{n}{4}\left(\sum_{i=1}^{n} a_{i}^{2}\right)$$ | Prove that by the generalized Bernoulli inequality and the AM - CM inequality, we have
$$\begin{aligned}
\prod_{i=1}^{n}\left(a_{i}+1\right)= & \prod_{a_{i} \geqslant 0}\left(a_{i}+1\right) \prod_{a_{j} \leqslant 0}\left(a_{j}+1\right) \geqslant \\
& \left(1+\sum_{a_{i} \geqslant 0} a_{i}\right)\left(1+\sum_{a_{j} \leq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,847 |
Example 1.3 $a, b$ are positive real numbers, satisfying $a+b=2$, prove that
$$a^{\frac{2}{a}}+b^{\frac{2}{b}} \leqslant 2$$ | Proof: Without loss of generality, let $a \geqslant 1 \geqslant b$. By Bernoulli's inequality,
$$\frac{1}{\left(\frac{1}{a}\right)^{\frac{2}{a}}} \geqslant \frac{1}{1+\frac{2}{a}\left(\frac{1}{a}-1\right)}=\frac{a^{2}}{a^{2}-2 a+2}$$
Similarly $\square$
$$b^{\frac{2}{b}} \leqslant \frac{b^{2}}{b^{2}-2 b+2}=\frac{(a-2)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,848 |
Example $1.21 a, b, c, d \geqslant 0, a+b+c+d=a b c+b c d+c d a+d a b$. Prove that $\sqrt{\frac{a^{2}+1}{2}}+\sqrt{\frac{b^{2}+1}{2}}+\sqrt{\frac{c^{2}+1}{2}}+\sqrt{\frac{d^{2}+1}{2}} \leqslant a+b+c+d$ | Note that
$$(a+b)(a+c)(a+d)=\sum a \cdot a^{2}+\sum_{9 c} a b c=\left(a^{2}+1\right) \sum a$$
Therefore, it suffices to prove
$$\sum \sqrt{(a+b)(a+c)(a+d)} \leqslant \sqrt{2\left(\sum a\right)^{3}}$$
The above inequality can be obtained by using Cauchy. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,849 |
Example 6.58 (Han Jingjun) $a_{1}, a_{2}, \cdots, a_{n}(n \geqslant 2)$ satisfy $a_{1}=0,\left(2 a-a_{i}\right) a_{i+1}=1(i=$ $1,2, \cdots, n-1), a_{n}=2 a$. Prove that
$$\frac{2 n^{2}}{(n+1)^{2}}<a_{n}<\frac{2 n^{2}+4 n-6}{(n+1)^{2}}$$ | Prove that from $\left(2 a-a_{1}\right)=1,\left(2 a-a_{2}\right) a_{3}=1, \cdots,\left(2 a-a_{n-1}\right) a_{n}=1$, we get
$$4 a^{2} \cdot a^{n-2} \geqslant\left(2 a-a_{1}\right) a_{2}\left(2 a-a_{2}\right) a_{3} \cdots\left(2 a-a_{n-1}\right) a_{n}=1 \Rightarrow a \geqslant 4^{\frac{-1}{n}}$$
Thus, by Bernoulli's ine... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,851 |
Example 7.1 (2008 Chinese National Team Training Problem) $0 \leqslant a, b \leqslant 1$, prove that
$$a^{a}+b^{b} \geqslant a^{b}+b^{a}$$ | Proof (Han Jingjun) Without loss of generality, assume $a \geqslant b$, and consider the function
$$f(x)=x^{y a}-x^{{ }^{b}}, a \geqslant x \geqslant b, 1 \geqslant y a-y b \geqslant 0$$
Taking the derivative with respect to $x$ yields
$$\left(x^{\gamma(a-b)}\right)^{\prime}=\frac{y(a-b)}{x} \cdot x^{\gamma(a-b)}>0 \R... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,852 |
For example, $7.2 a, b, c>0$, and satisfy $7\left(a^{2}+b^{2}+c^{2}\right)=11(a b+b c+c a)$, prove that
(1) $\frac{51}{28} \leqslant \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqslant 2$;
(2) $\frac{35}{3} \leqslant(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leqslant \frac{135}{7}$. | Proof: Let $a+b=1, x=ab$, then $x=\frac{7c^2-11c+7}{25}$, and $x \leqslant \frac{1}{4}$, so we have $c \in \left[\frac{1}{14}, \frac{3}{2}\right]$. Using this, we have
$$\begin{aligned}
\sum \frac{a}{b+c}= & (a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3= \\
& (1+c)\left(\frac{a+b+2c}{c^2+c(a+b)+ab}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,853 |
For example, let $7.3 x, y, z$ be non-negative real numbers, and satisfy $x y+y z+x z=1$, prove
$$\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}} \geqslant 2+\frac{1}{\sqrt{2}}$$ | Without loss of generality, let $x=\max (x, y, z)$, and set $a=y+z>0$. Clearly, $a x=1- y z \leqslant 1$. Consider the function
$$\begin{aligned}
f(x)= & \frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}}= \\
& \frac{1}{\sqrt{y+z}}+\sqrt{\frac{2 x+y+z+2 \sqrt{x^{2}+1}}{x^{2}+1}}= \\
& \frac{1}{\sqrt{a}}+\sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,854 |
Example $7.4 n \geqslant 2, a_{k}, b_{k}>0, k=1,2, \cdots, n . S=\sum_{k=1}^{n} a_{k}, T=b_{1} b_{2} \cdots b_{n}$, prove that $\frac{1}{n-1} \sum_{i=1}^{n}\left(1-\frac{a_{i}}{S}\right) b_{i} \geqslant\left(\frac{T}{S} \sum_{j=1}^{n} \frac{a_{j}}{b_{j}}\right)^{\frac{1}{n-1}}$ | Proof: Let $x_{i}=\frac{a_{i}}{S}, y_{i}=\frac{b_{i}}{\sqrt[n]{T}}$, then
$$\prod_{i=1}^{n} y_{i}=1 \Leftrightarrow \frac{\sqrt[n]{T}}{n-1} \sum_{i=1}^{n}\left(1-x_{i}\right) y_{i} \geqslant\left(\sqrt[n]{T^{n-1}} \sum_{j=1}^{n} \frac{x_{j}}{y_{j}}\right)^{\frac{1}{n-1}}$$
If $n=2$, then $x_{1}=\left(1-x_{2}\right), y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,855 |
Prove that
$$\left(\sum_{i=1}^{n} a_{i} \sum_{j \neq i} b_{j}\right)^{n-1} \geqslant(n-1)^{n-1}\left(\sum_{i=1}^{n} a_{i}\right)^{n-2}\left(\sum_{i=1}^{n} a_{i} \prod_{j \neq i} b_{j}\right)$$ | Prove that when $n=2$, the inequality becomes an equality.
When $n \geqslant 3$, we assume $b_{1}=\max \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$ and $b_{n}=\min \left\{b_{1}, b_{2}, \cdots, b_{n}\right\}$. By the AM - GM inequality, we have
$$(n-1)^{n-1} \sqrt{\left(\sum_{i=1}^{n} a_{i}\right)^{n-2}\left(\sum_{i=1}^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,856 |
Example 7.5 (2006 China National Team Training Problem) Given $a \geqslant b \geqslant c \geqslant d>0$, prove that
$$\left(1+\frac{c}{a+b}\right)\left(1+\frac{d}{b+c}\right)\left(1+\frac{a}{c+d}\right)\left(1+\frac{b}{d+a}\right) \geqslant\left(\frac{3}{2}\right)$$ | Considering the equality holds for $a=b=c=d$, we try to reduce dimensions, bringing $a$ closer to $b$ and $c$ closer to $d$.
$$\Leftrightarrow \frac{(a+b+c)(a+c+d)(a+b+d)}{(a+b)(a+d)} \cdot \frac{b+c+d}{(b+c)(c+d)} \geqslant\left(\frac{3}{2}\right)^{4}$$
Fix $b, c, d$, and let
$$\begin{array}{l}
f(a)=\frac{(a+b+c)(a+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,857 |
Example 7.6 (Michael Rozenberg) $a, b, c, d, e$ are positive numbers satisfying
$$a b c+a b d+a b e+a c d+a c e+a d e+b c d+b c e+b d e+c d e=10$$
Prove that
$$1+\frac{13}{a+b+c+d+e} \geqslant \frac{36}{a b+b c+c d+d a+a c+b d+a e+b e+c e+d e}$$ | Prove that if $a+b+c+d+e=A, a b+b c+c d+d a+a c+b d+a e+b e+c e+d e=B$.
Consider the function
$$\begin{aligned}
f(x)= & (x-a)(x-b)(x-c)(x-d)(x-e)= \\
& x^{5}-A x^{4}+B x^{3}-10 x^{2}+x \sum a b c d-a b c d e
\end{aligned}$$
Clearly, $f(x)=0$ has 5 positive real roots $a, b, c, d, e$, so $f''(x)=20 x^{3}-12 A x^{2}+6 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,858 |
Example $7.7 a, b, c, d \geqslant 0$, no two are zero at the same time, and $a+b+c+d=1$, prove that $E(a, b, c, d)=\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+d}}+\frac{d}{\sqrt{d+a}} \leqslant \frac{3}{2}$ | Suppose $(a, b, c, d)$ is an extremum point of $E$.
If $a, b, c, d$ are not all zero, then by Fermat's theorem, $f(t)=E(a, b+t, c-t, d)$ must have $f^{\prime}(0)=0$. Note that
$$f^{\prime}(0)=\frac{-c-2 d}{2(c+d)^{\frac{3}{2}}}+\frac{1}{\sqrt{b+c}}-\frac{a}{2(a+b)^{\frac{3}{2}}}$$
Thus,
$$\frac{-c-2 d}{2(c+d)^{\frac{3... | \frac{3}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,859 |
Example $1.22 a, b, c, x, y, z \geqslant 0, a+b+c=x+y+z$, prove that
$$a x(a+x)+b y(b+y)+c z(c+z) \geqslant 3(a b c+x y z)$$ | Proof: By Cauchy's inequality, we have
$$\begin{array}{l}
a^{2} x+b^{2} y+c^{2} z \geqslant \frac{(a+b+c)^{2}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \\
a x^{2}+b y^{2}+c z^{2} \geqslant \frac{(x+y+z)^{2}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}
\end{array}$$
Therefore, it suffices to prove
$$\frac{(a+b+c)^{2}}{\frac{1}{x}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,860 |
Example $7.9 x_{i}>0, m=\min \left\{x_{i}\right\}, M=\max \left\{x_{i}\right\}, i=1,2,3, \cdots, n$, prove $\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} \frac{1}{x_{i}} \leqslant n^{2}+\left[\frac{n^{2}}{4}\right]\left(\sqrt{\frac{m}{M}}-\sqrt{\frac{M}{m}}\right)^{2}$ | Prove that if we consider the left side of the inequality as a function of $x_{1}$, then
$$\sum_{i=1}^{n} x_{i} \sum_{i=1}^{n} \frac{1}{x_{i}}=a x_{1}+\frac{b}{x_{1}}+c=f\left(x_{1}\right)$$
$f^{\prime \prime}\left(x_{i}\right)=\frac{2 b}{x^{3}}>0$, which is a convex function, so $f\left(x_{1}\right)$ must achieve its ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,863 |
Example $7.10$ For $a, b, c \geqslant 0$ satisfying $a \leqslant 1 \leqslant b \leqslant c$, then
(1) If $a+b+c=3$, we have
$$a^{2} b+b^{2} c+c^{2} a \geqslant a b c+2$$
(2) If $a b+b c+c a=3$, we have
$$a^{2} b+b^{2} c+c^{2} a \geqslant 3$$ | Proof (1) From the condition, we have $c \geqslant b \geqslant \frac{a+c}{2}$.
By the AM - GM inequality, we have
$$b^{2} \geqslant b(a+c)-\frac{(a+c)^{2}}{4}$$
Thus, we only need to prove
$$f(b)=a^{2} b+\left[b(a+c)-\frac{(a+c)^{2}}{4}\right] c+c^{2} a-a b c-\frac{2}{27}(a+b+c)^{3} \geqslant 0$$
It is easy to see th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,864 |
Example 7.11 Let $x, y, z$ be positive real numbers satisfying
$$\max \{x, y\}<z \leqslant 1, 2 \sqrt{3} x z \leqslant \sqrt{3} x+z, \sqrt{2} y+z \leqslant 2$$
Prove that
$$P=3 x^{2}+2 y^{2}+5 z^{2} \leqslant 7$$ | We prove
$$x^{2}+z^{2} \leqslant \frac{4}{3}, y^{2}+z^{2} \leqslant \frac{3}{2}$$
In fact, if $z \leqslant \frac{\sqrt{3}+1}{2 \sqrt{3}}0$
Thus $f(z)$ is convex, so
$$f(z) \leqslant \max \left\{f(1), f\left(\frac{\sqrt{3}+1}{2 \sqrt{3}}\right)\right\}=\max \left\{\frac{4}{3}, \frac{2+\sqrt{3}}{3}\right\}=\frac{4}{3}$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,865 |
Example $7.12 a_{i}>0, i=1,2, \cdots, n$, prove that
$$\prod_{i=1}^{n} a_{i}^{a_{i}} \geqslant\left(a_{1} a_{2} \cdots a_{n}\right)^{\frac{1}{2}\left(a_{1}+a_{2}+\cdots+a_{n}\right)}$$ | Prove that for the function $f(x)=x \ln x, f^{\prime \prime}(x)=\frac{1}{x}>0$, hence $f(x)$ is a convex function. Using Jensen's inequality, we have
$$\begin{array}{l}
\frac{f\left(a_{1}\right)+f\left(a_{2}\right)+\cdots+f\left(a_{n}\right)}{n} \geqslant f\left(\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}\right) \Leftrightarrow... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,867 |
Example 7.13 (2004 China Western Mathematical Competition) $a, b, c>0$, prove that
$$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} \leqslant \frac{3 \sqrt{2}}{2}$$ | Proof: Let $a+b+c=1$, then
$$\begin{aligned}
S= & \sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}= \\
& (a+c) \sqrt{\frac{a}{(a+b)(a+c)^{2}}}+(b+a) \sqrt{\frac{b}{(b+c)(b+a)^{2}}}+ \\
& (c+b) \sqrt{\frac{b}{(c+a)(c+b)^{2}}}
\end{aligned}$$
Since $\sqrt{x}$ is a concave function, by Jensen's inequality w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,868 |
Example 7.14 If $x \geqslant y \geqslant 1$, prove
$$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+1}}+\frac{1}{\sqrt{x+1}} \geqslant \frac{y}{\sqrt{x+y}}+\frac{x}{\sqrt{x+1}}+\frac{1}{\sqrt{y+1}}$$ | It is not hard to observe that the equality holds when $y=1$ or $x=y$. Therefore, we set $x=y+a, y=1+b, b \geqslant 0$. The inequality can be rewritten as follows:
$$\begin{array}{l}
\frac{x-y}{\sqrt{x+y}}+\frac{y-1}{\sqrt{y+1}}+\frac{1-x}{\sqrt{1+x}} \geqslant 0 \Leftrightarrow \\
\frac{a}{\sqrt{2+a+2 b}}+\frac{b}{\sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,869 |
Theorem $7.4 x_{1}, x_{2}, \cdots, x_{n}$ are $n$ real numbers, satisfying:
(1) $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$
(2) $x_{1}, x_{2}, \cdots, x_{n} \in[a, b]$
(3) $x_{1}+x_{2}+\cdots+x_{n}=C(C$ is a constant).
$f$ is a function defined on $[a, b]$, if $f$ is convex (concave) on $[a, c]$, and conca... | Prove the case for taking the minimum value (the case for the maximum value can be similarly proved).
We use mathematical induction to prove.
If there do not exist $x_{1}, x_{2}, \cdots, x_{n}$ or only $x_{1} \in[a, c]$, then the theorem is obviously correct.
This is because $x_{2}, x_{3}, \cdots, x_{n} \in[c, b]$, so
... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,870 |
Example 1.23 Positive numbers $a, b, c$ satisfy $a+b+c=1$, prove
$$(a-b c)(b-c a)(c-a b) \leqslant 8(a b c)^{2}$$ | It is easy to see that we only need to consider the case where $a-b c, b-c a, c-a b$ are all positive. Consider the local inequality
$$\begin{array}{c}
(a-b c)(b-c a) \leqslant 4 a^{2} b^{2} \Leftrightarrow a b\left(1+c^{2}\right) \leqslant 4 a^{2} b^{2}+c\left(a^{2}+b^{2}\right) \Leftrightarrow \\
a b(1-c)^{2} \leqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,871 |
Example 7.15 Let $\triangle A B C$ be an acute triangle, prove that
$$\sum \frac{\cos ^{2} A}{\cos A+1} \geqslant \frac{1}{2}$$ | Let $f(x)=\frac{\cos ^{2} x}{\cos x+1}$, it is easy to prove that $f$ satisfies the conditions of the theorem. Therefore, we only need to prove
$$\frac{\cos ^{2} A}{\cos A+1}+\frac{\cos ^{2} B}{\cos B+1} \geqslant \frac{1}{2}\left(A+B=\frac{\pi}{2}\right)$$
or
$$\sum \frac{\cos ^{2} A}{\cos A+1} \geqslant \frac{1}{2}(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,873 |
Example $7.16 a, b, c \geqslant 0$
$$\sqrt{1+\frac{48 a}{b+c}}+\sqrt{1+\frac{48 b}{c+a}}+\sqrt{1+\frac{48 c}{a+b}} \geqslant 15$$ | Let $x \doteq \frac{a}{a+b+c}, y, z$ be similarly defined, and note that
$$\sqrt{1+\frac{48 a}{b+c}}=\sqrt{\frac{48}{1-x}-47}$$
Let $f(t)=\sqrt{\frac{48}{1-t}-47}$. It is easy to prove that $f$ satisfies the above theorem. Without loss of generality, assume $x \leqslant y \leqslant z$. Thus, we only need to prove
$$f(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,874 |
Theorem $7.6 \ x_{1}, x_{2}, \cdots, x_{n}$ are $n$ real numbers, satisfying:
(1) $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$
(2) $x_{1}, x_{2}, \cdots, x_{n} \in(-\infty,+\infty)$;
(3) $x_{1}+x_{2}+\cdots+x_{n}=C$ ( $C$ is a constant).
$f$ is a function on $(-\infty,+\infty)$, if $f$ is convex (concave) o... | Prove the case for minimum value (the case for maximum value can be proved similarly).
Assume without loss of generality that $x_{1}, x_{2}, \cdots, x_{i} \in(-\infty, c]$.
Since $f$ is convex on $(-\infty, c]$, we have
$$\begin{array}{l}
f\left(x_{1}\right)+f\left(x_{2}\right)+\cdots+f\left(x_{i}\right) \geqslant(i-1)... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,875 |
Example $7.17 x, y, z \geqslant 0, x y z=1$, find the maximum value of the following expression
$$\frac{1}{(1+x)^{k}}+\frac{1}{(1+y)^{k}}+\frac{1}{(1+z)^{k}}$$ | Let $f(t)=\frac{1}{\left(1+\mathrm{e}^{t}\right)^{k}}$, then
$$f^{\prime \prime}(t)=\frac{\mathrm{e}^{x}\left(k(k+1) \mathrm{e}^{x}-k\right)}{\left(1+\mathrm{e}^{x}\right)^{k+2}}$$
Assume $x \leqslant y \leqslant z$, by the theorem above, we only need to consider the case $y=z$. | not found | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,877 |
Example $7.18$ Given $a, b, c \geqslant 0, a+b+c=1$, find the maximum value of the following expression.
$$F(a, b, c)=\sqrt{\frac{1-a}{1+a}}+\sqrt{\frac{1-b}{1+b}}+\sqrt{\frac{1-c}{1+c}}$$ | The maximum value is $F(0.5,0.5,0)$.
Notice that
$$f^{\prime \prime}=\frac{1-2 x}{\sqrt{(1+x)^{5}(1-x)^{3}}}$$
Assume without loss of generality that $a \leqslant b \leqslant c$.
If $a+b \leqslant \frac{1}{2}$, we have
$$f(a)+f(b) \leqslant f(0)+f(a+b)$$
If $a+b>\frac{1}{2}$, we have
$$f(a)+f(b) \leqslant f\left(a+b-... | F(0.5,0.5,0) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,878 |
Example 1.24 Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers. Prove that
$$\left(1+x_{1}\right)\left(1+x_{1}+x_{2}\right) \cdots\left(1+x_{1}+x_{2}+\cdots+x_{n}\right) \geqslant \sqrt{(n+1)^{n+1} x_{1} x_{2} \cdots x_{n}}$$ | Prove that
$$s=\frac{x_{1} x_{2} x_{3} \cdots x_{n}}{\left(1+x_{1}\right)^{2}\left(1+x_{1}+x_{2}\right)^{2} \cdots\left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{2}}$$
The original inequality is equivalent to $s \leqslant\left(\frac{1}{n+1}\right)^{n+1}$, let
$$\begin{array}{c}
y_{1}=\frac{x_{1}}{1+x_{1}}, y_{2}=\frac{x_{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,882 |
Example $7.20 \quad a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n} \geqslant 0, b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{n} \geqslant 0$. If $\sum_{i=1}^{j} \frac{a_{i}}{b_{i}} \geqslant j(n \geqslant$ $j \geqslant 1$ ), then we have
$$\left(a_{1}, a_{2}, \cdots, a_{n}\right)>\left(b_{1}, b_{2}, \cdo... | To prove in fact there is a stronger proposition:
$$\begin{array}{c}
\left(a_{1}, a_{2}, \cdots, a_{n}\right)>\left(b_{1}, b_{2}, \cdots, b_{n}\right) \\
\sum_{i=1}^{n-1} a_{i}-\sum_{i=1}^{n-1} b_{i} \geqslant\left(\sum_{i=1}^{n-1} \frac{a_{i}}{b_{i}}-n+1\right) b_{n}
\end{array}$$
We use mathematical induction to pro... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,883 |
$\begin{array}{r}\text { Example } 722 \text { (2006 China National Team Training Problem) } a \geqslant b \geqslant c \geqslant d>0 \text {, prove that } \\ \left(1+\frac{c}{a+b}\right)\left(1+\frac{d}{b+c}\right)\left(1+\frac{a}{c+d}\right)\left(1+\frac{b}{d+a}\right) \geqslant\left(\frac{3}{2}\right)^{4}\end{array}$ | Prove that the following inequality is equivalent to:
$$\begin{array}{l}
\ln \frac{a+b+c}{3}+\ln \frac{b+c+d}{3}+\ln \frac{c+d+a}{3}+\ln \frac{d+a+b}{3} \geqslant \\
\ln \frac{a+b}{2}+\ln \frac{b+c}{2}+\ln \frac{c+d}{2}+\ln \frac{d+a}{2}
\end{array}$$
If \(a+d \geqslant b+c\), note that \(f(x)=\ln x\) is a concave fun... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,885 |
Example $7.23$ Let $a, b, c, d$ be non-negative real numbers satisfying $a+b+c+d=1$. Prove that
$$\sqrt{a+b+c^{2}}+\sqrt{b+c+d^{2}}+\sqrt{c+d+a^{2}}+\sqrt{d+a+b^{2}} \geqslant 3$$ | To prove the desired inequality is equivalent to
$$\begin{aligned}
\Leftrightarrow & \sum \sqrt{(a+b)(a+b+c+d)+c^{2}} \geqslant 3(a+b+c+d) \Leftrightarrow \\
& \sum \sqrt{P_{2}+Q_{1}} \geqslant \sum \sqrt{P_{1}+Q_{1}}
\end{aligned}$$
where \( P_{1}=b^{2}+c^{2}+d^{2}+b c+c d+d b, Q_{1}=b c+c d+d b \) etc. Note that \( ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,886 |
Theorem $7.11 f\left(x_{1}, x_{2}, \cdots, x_{n}\right): \mathbf{R}_{+}^{n} \rightarrow \mathbf{R}$, continuous on $\mathbf{R}_{+}^{n}$. $[f]:=\mathbf{R}_{+}^{n} \rightarrow \mathbf{R}_{+}$, then the inequality $f\left(x_{1}, x_{2}, \cdots, x_{n}\right) \geqslant 0$ holds if and only if $\exists i$ such that $x_{i}=0$ ... | Proof by contradiction, if $x_{i}$ are not all 0. For the numbers $\left(x_{1}^{1}, x_{2}^{1}, \cdots, x_{n}^{1}\right)$ on $\mathbf{R}_{+}^{n}$, we let $a_{1}=\left(x_{1}^{1}+x_{2}^{1}+\cdots+x_{n}^{1}\right)$.
Since $[f] \geqslant 0$, there exists $x_{i}^{1}$ such that $f^{\prime}\left(x_{i}\right)>0$, or $f^{\prime... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,888 |
Example $7.25 P: \mathbf{R}_{+}^{3} \rightarrow \mathbf{R}$, is a ternary cyclic symmetric homogeneous polynomial of degree 3, the necessary and sufficient condition for $P \geqslant 0$ to hold is
$$P(1,1,1) \geqslant 0, P(a, b, 0) \geqslant 0, \forall a, b \geqslant 0$$ | To prove the necessity is obvious, we now prove the sufficiency. A cyclic symmetric inequality of degree 3 in three variables can be written as
$$\begin{aligned}
P(a, b, c)= & m\left(a^{3}+b^{3}+c^{3}\right)+n\left(a^{2} b+b^{2} c+c^{2} a\right)+ \\
& p\left(a b^{2}+b c^{2}+c a^{2}\right)+3 q a b c \geqslant 0
\end{ali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,889 |
Example $7.26 a, b, c \geqslant 0$, then
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}} \leqslant \frac{5}{4} \sqrt{a+b+c}$$ | Proof First, we prove the following lemma:
Lemma If $x, y, z$ are the lengths of the sides of an acute triangle, then
(1) $x+y+z \geqslant \sqrt{2\left(x^{2}+y^{2}+z^{2}\right)}$;
(2) $x y z \geqslant 4(x-y)(y-z)(x-z)$.
Proof of the lemma (1) is obvious, in fact, it holds for any triangle.
For (2), note that it suffic... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,890 |
Example 7.27 (Vasile's Inequality) \(a, b, c \in \mathbf{R}\), prove that
\[
\left(a^{2}+b^{2}+c^{2}\right)^{2} \geqslant 3\left(a^{3} b+b^{3} c+c^{3} a\right)
\] | Proof: Let $F_{0}=\left(a^{2}+b^{2}+c^{2}\right)^{2}-\left(a^{3} b+b^{3} c+c^{3} a\right)$, then we have
$$\begin{aligned}
& F_{1}=2\left(a^{2}+b^{2}+c^{2}\right)(2 a+2 b+2 c)-3 \sum_{\text {ccc }}\left(3 a^{2} b+a^{3}\right) \\
F_{2}= & 2(2 a+2 b+2 c)^{2}+2\left(a^{2}+b^{2}+c^{2}\right) 6-3 \sum_{\gamma c}\left(6 a b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,892 |
Example $1.25 a, b, c, d>0$, prove that
$$\frac{c}{a}(8 b+c)+\frac{d}{b}(8 c+d)+\frac{a}{c}(8 d+a)+\frac{b}{d}(8 a+b) \geqslant 9(a+b+c+d)$$ | Proof: By the AM - GM inequality we have
$$\begin{array}{l}
\frac{b c}{a}+\frac{d a}{c} \geqslant 2 \sqrt{b d} \\
\frac{c d}{b}+\frac{a b}{d} \geqslant 2 \sqrt{a c}
\end{array}$$
Thus, it suffices to prove
$$\begin{array}{l}
\frac{c^{2}}{a}+\frac{a^{2}}{c}+16 \sqrt{a c}+16 \sqrt{b d} \geqslant 9(a+b+c+d) \Leftrightarr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,893 |
Example $7.28 a, b, c \geqslant 0$, prove that
$$\left(a^{2}+b^{2}+c^{2}\right)^{2} \geqslant 4(a-b)(b-c)(c-a)(a+b+c)$$ | Proof: Let $f(a, b, c)=\left(a^{2}+b^{2}+c^{2}\right)^{2}-4(a-b)(b-c)(c-a)(a+b+c)$. Also, let $g(t)=f(a+t, b+t, c+t)$, where $t \geqslant 0$.
Then
$$\begin{aligned}
g(t)= & {\left[3 t^{2}+2 t(a+b+c)+a^{2}+b^{2}+c^{2}\right]^{2}-} \\
& 4(a-b)(b-c)(c-a)(3 t+a+b+c)
\end{aligned}$$
We have
$$\begin{aligned}
g^{\prime}(t)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,894 |
For example, if $a, b, c \in \mathbf{R}$, then
$$a^{4}+b^{4}+c^{4}+a^{3} b+b^{3} c+c^{3} a \geqslant 2\left(a b^{3}+b c^{3}+c a^{3}\right)$$ | Proof: Let
$$f(a, b, c)=a^{4}+b^{4}+c^{4}+a b^{3}+b c^{3}+c a^{3}-2\left(a^{3} b+b^{3} c+c^{3} a\right)$$
Then
$$\begin{array}{l}
f(a+t, b+t, c+t)=6\left(\sum a^{2}-\sum a b\right) t^{2}+3\left(\sum a^{3}+\sum_{\text {cr }} a^{2} b-2 \sum_{c y c} a b^{2}\right) t+ \\
a^{4}+b^{4}+c^{4}+a b^{3}+b c^{3}+c a^{3}-2\left(a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,895 |
Example $7.30 a, b, c, d \geqslant 0$, prove that
$$(a+b+c+d)^{6} \geqslant 1728(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)$$ | Prove (Han Jingjun) Without loss of generality, assume $a \geqslant b \geqslant c \geqslant d$, it is easy to see that
$$\begin{array}{l}
(a+b+c+d)^{6}-1728(a-b)(a-c)(a-d)(b-c)(b-d)(c-d) \geqslant \\
(a-d+b-d+c-d+d-d)^{6}- \\
1728(a-b)(a-c)(a-d)(b-c)(b-d)(c-d)
\end{array}$$
Thus, we only need to prove the case when $d... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,896 |
Example $8.1$ Let $a, b, c$ be the lengths of the three sides of a triangle. Prove that
$$\begin{array}{c}
8 a^{2} b^{2} c^{2} \geqslant \\
(a+b)(b+c)(c+a)(a+b-c)(b+c-a)(c+a-b)
\end{array}$$ | Prove that by making the substitution $x=b+c-a, y=c+a-b, z=a+b-c$, the original inequality is equivalent to
$$\prod(x+y)^{2} \geqslant x y z(x+2 z+y)(y+2 x+z)(z+2 y+x)$$
Then, by making the substitution $x=Y Z, y=Z X, z=X Y$, it can be transformed into
$$\begin{array}{l}
\prod(X+Z)^{2} \geqslant \prod(Y Z+2 Z X+X Y) \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,897 |
For example, $8.2$ given $a, b, c>0$ and $a+b+c=abc$, prove that
$$\sum \sqrt{\left(1+a^{2}\right)\left(1+b^{2}\right)}-\sqrt{\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)} \geqslant 4$$ | Given the conditions, we can let $a=\cot \frac{A}{2}, b=\cot \frac{B}{2}, c=\cot \frac{C}{2}$ and satisfy $A+B+C=\pi$. Using the following two identities:
$$\begin{array}{c}
1+\cot ^{2} x=\csc ^{2} x \\
\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2}=4 \sin \frac{A+B}{4} \sin \frac{B+C}{4} \sin \frac{C+A}{4}+1
\end{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,898 |
Example 8.4 Let positive numbers $x, y, z$ satisfy $x y + y z + z x = 1$, prove that
$$\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x} \geqslant \frac{5}{2}$$ | Prove the inequality is equivalent to in $\triangle A B C$,
$$\begin{array}{c}
\frac{\cos \left(\frac{B}{2}\right) \cos \left(\frac{C}{2}\right)}{\cos \left(\frac{A}{2}\right)}+\frac{\cos \left(\frac{B}{2}\right) \cos \left(\frac{A}{2}\right)}{\cos \left(\frac{C}{2}\right)}+\frac{\cos \left(\frac{A}{2}\right) \cos \lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,900 |
Example $8.5 a b c=1, a, b, c>0$, prove that
$$(a+b)(b+c)(c+a) \geqslant 4(a+b+c-1)$$ | Proof. For convenience, we write the original inequality as
$$(x+y)(y+z)(z+x) \geqslant 4(x+y+z-1)$$
where $xyz=1$. Making the substitution $x=s-a, y=s-b, z=s-c$, then $a, b, c$ are the side lengths of a triangle, and $s$ is the semi-perimeter of the triangle. Thus, the inequality becomes
$$abc \geqslant 4(s-1)$$
Sin... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,901 |
Example 8.6 (2007 China National Training Team) $u, v, w>0$, satisfying $u+v+w+\sqrt{u v w}=4$, prove that
$$\sqrt{\frac{u v}{w}}+\sqrt{\frac{v w}{u}}+\sqrt{\frac{w u}{v}} \geqslant u+v+w$$ | Proof: Let $a^{2}=\frac{u}{4}, b^{2}=\frac{v}{4}, c^{2}=\frac{w}{4}$, then the problem is transformed to
$$a, b, c>0, a^{2}+b^{2}+c^{2}+a b c=1$$
Prove that
$$\frac{b c}{a}+\frac{c a}{b}+\frac{a b}{c} \geqslant 2\left(a^{2}+b^{2}+c^{2}\right)$$
Let $a=\cos A, b=\cos B, c=\cos C$, where $\triangle A B C$ is an acute t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,902 |
Proof: Let $\dot{u}=\frac{4 y z}{(x+y)(x+z)}, v=\frac{4 x z}{(x+y)(y+z)}, w=\frac{4 x y}{(x+z)(y+z)}$, where $x, y, z$ are positive real numbers. | Equivalent to proving
$$\begin{array}{l}
\sqrt{\frac{u v}{w}}+\sqrt{\frac{v w}{u}}+\sqrt{\frac{w u}{v}} \geqslant u+v+w \Leftrightarrow \\
u v+u w+v w \geqslant(u+v+w) \sqrt{u v w} \Leftrightarrow \\
\sum_{c y c} \frac{16 z^{2} x y}{(x+y)^{2}(x+z)(y+z)} \geqslant \sum_{c x c} \frac{4 x y}{(x+z)(y+z)} \cdot \frac{8 x y ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,903 |
Example 1.26 For positive numbers $a, b, c$, prove
$$\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}+\frac{3(a b+b c+c a)}{(a+b+c)^{2}} \geqslant 4$$ | $$\begin{array}{l}
b=\max \{a, b, c\} \Leftrightarrow \\
\left(\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}-3\right)+\left(\frac{3(a b+b c+c a)}{(a+b+c)^{2}}-1\right) \geqslant 0 \Leftrightarrow \\
\left(\frac{(a-c)^{2}}{(a+b)(b+c)}+\frac{(b-a)(b-c)}{(a+b)(c+a)}\right)- \\
\frac{(a-c)^{2}+(b-a)(b-c)}{(a+b+c)^{2}} \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,904 |
Example 8.9 (G - B Inequality) In $\triangle ABC$, we have
$$\tan ^{2} \frac{A}{2}+\tan ^{2} \frac{B}{2}+\tan ^{2} \frac{C}{2} \geqslant 2-8 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$ | Prove that from the identities in a triangle
$$r=4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}, s=4 R \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$$
we have
$$\begin{array}{c}
\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}=\frac{r}{4 R} \\
\sum \tan \frac{A}{2}=\frac{1+\sin \frac{A}{2} \sin \frac{B}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,907 |
Example 8.10 (Han Jingjun) $a, b, c, x, y, z>0$, prove that
$$\frac{x+a}{a c x y}+\frac{y+b}{b a y z}+\frac{z+c}{c b z x} \geqslant \frac{3(a+x)(b+y)(c+z)}{(a b c+x y z)^{2}}$$ | Prove that after combining the left side of the inequality, it is equivalent to
$$\frac{(a+x)(b+y)(c+z)-a b c-x y z}{a b c x y z} \geqslant \frac{3(a+x)(b+y)(c+z)}{(a b c+x y z)^{2}}$$
Let $(a+x)(b+y)(c+z)=m, a b c=p, x y z=q$.
$$\begin{array}{l}
\Leftrightarrow \frac{m-p-q}{p q} \geqslant \frac{3 m}{(p+q)^{2}} \Leftr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,908 |
Example 8.11 Let $a, b, c \geqslant 0$ and not all zero, prove that
$$\frac{a}{b\left(a^{2}+2 b^{2}\right)}+\frac{b}{c\left(b^{2}+2 c^{2}\right)}+\frac{c}{a\left(c^{2}+2 a^{2}\right)} \geqslant \frac{3}{a b+b c+c a}$$ | Prove that the original inequality is quite challenging. Try the inverse substitution, let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, then the original inequality is equivalent to
$$\frac{x^{2}}{y\left(2 z^{2}+x^{2}\right)}+\frac{y^{2}}{z\left(2 x^{2}+y^{2}\right)}+\frac{z^{2}}{x\left(2 y^{2}+z^{2}\right)} \geqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,909 |
Example 8.13 (1997 Belarus Mathematical Olympiad) $a, b, c>0$, prove that
$$\sum_{c y c} \frac{a}{b} \geqslant \sum \frac{a+b}{a+c}$$ | Proof (Han Jingjun) Let $\frac{a}{b}=x, \frac{b}{c}=y, \frac{c}{a}=z$, then $x y z=1, x, y, z>0 \Leftrightarrow$
$$\begin{array}{l}
\sum_{c x} \frac{x}{y+1} \geqslant \sum \frac{1}{y+1} \Leftrightarrow \\
\sum_{c x} \frac{1}{y z(y+1)} \geqslant \sum \frac{1}{y+1} \Leftrightarrow \\
\sum_{c c} x(z+1)(x+1) \geqslant \sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,911 |
Example 8. Let $a, b, c$ be positive numbers, prove that
$$\sum_{c y c} \frac{(a+b) a}{(b+c)(2 a+b+c)} \geqslant \frac{3}{4}$$ | Proof: Let $x=a+b, y=b+c, z=c+a$, then the original inequality is equivalent to
$$\frac{x(x-y+z)}{y(z+x)}+\frac{y(y-z+x)}{z(x+y)}+\frac{z(z-x+y)}{x(y+z)} \geqslant \frac{3}{2}$$
Notice that
$$\sum \frac{x(x-y+z)}{y(z+x)}=\sum\left(\frac{x}{y}-\frac{z+x}{z+y}\right)+\sum \frac{x}{y+z}$$
By Cauchy's inequality, we have... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,912 |
Example $8.15$ a, b, c > 0, prove that
$$\sqrt{\frac{a^{3}}{a^{3}+(b+c)^{3}}}+\sqrt{\frac{b^{3}}{b^{3}+(c+a)^{3}}}+\sqrt{\frac{c^{3}}{c^{3}+(a+b)^{3}}} \geqslant 1$$ | Prove the substitution
$$x=\frac{b+c}{a}, y=\frac{c+a}{b}, z=\frac{a+b}{c}$$
We have
$$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=1 \Leftrightarrow 2+x+y+z=x y z$$
At this point, we need to prove
$$\frac{1}{\sqrt{1+x^{3}}}+\frac{1}{\sqrt{1+y^{3}}}+\frac{1}{\sqrt{1+z^{3}}} \geqslant 1$$
Note that for all \( u \geqslan... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,913 |
Example 1.27 Given a cubic equation $x^{3} + a x^{2} + b x + c = 0 (a, b, c \in \mathbf{R})$ with three roots $\alpha$, $\beta$, $\gamma$ whose magnitudes are all no greater than 1, find
$$\frac{1+|a|+|b|+|c|}{|\alpha|+|\beta|+|\gamma|}$$
the minimum value. | Given the problem, let's assume $1 \geqslant|\alpha| \geqslant|\beta| \geqslant|\gamma|, \beta=s \alpha, \gamma=t \alpha$, then $1 \geqslant|s| \geqslant|t|$, and let
then
$$\begin{aligned}
\theta= & |\alpha| \leqslant 1, u=\frac{1+|a|+|b|+|c|}{|\alpha|+|\beta|+|\gamma|} \\
u= & \frac{1+|a|+|b|+|c|}{|\alpha|+|\beta|+|... | \frac{\sqrt[3]{2}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,915 |
Example 8.17 (2009 Vietnam) Determine the minimum value of $k$ such that the following inequality holds for all positive real numbers $a, b, c$
$$\left(k+\frac{a}{b+c}\right)\left(k+\frac{b}{c+a}\right)\left(k+\frac{c}{a+b}\right) \geqslant\left(k+\frac{1}{2}\right)^{3}$$ | Prove that by setting $a=b=1, c=0$, we can obtain $k \geqslant \frac{\sqrt{5}-1}{4}$.
Let $m=2k, x=\frac{2a}{b+c}, y=\frac{2b}{c+a}, z=\frac{2c}{a+b}$. It is not hard to verify that $xy+yz+zx+xyz=4$.
First, we need to prove that under this condition, we have $x+y+z \geqslant xy+yz+zx$.
Clearly, if $x+y+z > 4$, then it ... | \frac{\sqrt{5}-1}{4} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,916 |
Example 9. $1 a, b, c>0, a b+b c+c a=1$, prove that
$$\frac{1}{a+b}+\frac{1}{c+b}+\frac{1}{a+c} \geqslant \frac{5}{2}$$ | Prove that multiplying both sides by $(a+b)(b+c)(c+a)$, and noting that $(a+b)(a+c)=a^{2}+1$ and $(a+b)(b+c)(c+a)=a+b+c-abc$, is equivalent to proving
$$\begin{array}{c}
2\left(a^{2}+b^{2}+c^{2}\right)+6+5abc \geqslant 5(a+b+c) \\
(a+b+c-2)^{2} \geqslant 0 \Rightarrow 2(a+b+c)^{2} \geqslant 8(a+b+c)-8
\end{array}$$
Su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,919 |
Example $9.2 a b c=1, a, b, c>0$, prove that
$$(a+b)(b+c)(c+a) \geqslant 4(a+b+c-1)$$ | Prove that since $abc=1$, at least one of $a, b, c$ is not less than 1. Without loss of generality, assume $a \geqslant 1$. Since
$$\begin{aligned}
(a+b)(b+c)(c+a)= & (b+c)\left(a^{2}+a b+b c+c a\right) \geq \\
& (b+c)\left(a^{2}+3 \sqrt[3]{(a b c)^{2}}\right)= \\
& (b+c)\left(a^{2}+3\right)
\end{aligned}$$
It suffice... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,920 |
Example $9.4$ When $a+b+c=3, a, b, c \geqslant 0$, we have
$$a^{2} b+b^{2} c+c^{2} a+a b c \leqslant 4$$ | Proof: Let $b=\min \{a, b, c\}$, then
$$a^{2} b+b^{2} c+c^{2} a+a b c=b(a+c)^{2}-c(a-b)(b-c) \leqslant b(a+c)^{2} \leqslant 4$$ | 4 | Inequalities | proof | Yes | Yes | inequalities | false | 731,922 |
Example 9.5 (1980 USA Mathematical Olympiad) Let $0 \leqslant a, b, c \leqslant 1$, prove that
$$\frac{a}{b+c+1}+\frac{b}{c+a+1}+\frac{c}{a+b+1}+(1-a)(1-b)(1-c) \leqslant 1$$ | Prove that due to the symmetry of the inequality about $a, b, c$, without loss of generality, assume $0 \leqslant a \leqslant b \leqslant c \leqslant 1$, then
$$\begin{aligned}
\text { LHS }= & \sum \frac{a}{b+c+1}+(1-a)(1-b)(1-c) \leqslant \\
& \frac{a+b+c}{a+b+1}+(1-a)(1-b)(1-c)= \\
& 1-\frac{1-c}{a+b+1}[1-(1+a+b)(1-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,923 |
Example 9.7 (1991 Polish Mathematical Olympiad) \(x^{2}+y^{2}+z^{2}=2\), prove that
\[x+y+z \leqslant x y z+2\] | Assume $x \leqslant y \leqslant z$, then $x y \leqslant 1, (x-1)(y-1) \geqslant 0$.
If $z \geqslant 1$, then
$$x+y+z \leqslant x y z+2 \Leftrightarrow 2(x-1)(y-1)(z-1)+(x+y+z-2)^{2} \geqslant 0$$
If $z<1$, then
$$x+y+z \leqslant x y z+2 \Leftrightarrow(x-1)(y-1)+(z-1)(x y-1) \geqslant 0$$
Combining both cases, the in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,925 |
Example 1.28 Let $x, y, z$ be positive real numbers satisfying $x^{2}+y^{2}+z^{2}=3$. Prove that
$$\frac{x}{\sqrt{y^{2}+3}+x}+\frac{y}{\sqrt{z^{2}+3}+y}+\frac{z}{\sqrt{x^{2}+3}+z} \leqslant 1$$ | Given the condition $3 \geqslant x+y+z$, we have
$$\sum \frac{x}{x+\sqrt{y^{2}+3}} \leqslant \sum \frac{x}{x+\frac{y+3}{2}} \leqslant \sum \frac{x}{x+\frac{y+(x+y+z)}{2}}=2 \sum \frac{x}{3 x+2 y+z}$$
It suffices to prove $\square$
$$\begin{array}{l}
\sum \frac{x}{3 x+2 y+z} \leqslant \frac{1}{2} \Leftrightarrow \sum\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,926 |
Example 9.8 (2007 Indian Mathematical Olympiad) $a, b, c \geqslant 0$, if $b+c \leqslant 1+a, a+c \leqslant$ $1+b, a+b \leqslant 1+c$, prove that
$$\sum_{c c} a^{2} \leqslant 2 a b c+1$$ | Assume without loss of generality that $a \leqslant b \leqslant c$, from $b+c \leqslant 1+a \leqslant 1+b$, we have $c \leqslant 1$.
(1) If $a \leqslant b c$, we have
$$\begin{aligned}
\left(a^{2}+b^{2}+c^{2}\right)-(2 a b c+1)= & a^{2}+(b+c)^{2}-2(1+a) b c-1 \leq \\
& a^{2}+(1+a)^{2}-2(1+a) b c-1 \leq \\
& 2(1+a)(a-b ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,927 |
Example $9.9 a, b, c \geqslant 0, a^{2}+b^{2}+c^{2}=3$, prove that
$$12+9 a b c \geqslant 7 \sum a b$$ | Proof: Let $a=1+x, b=1+y, c=1+z$, then
$$\sum x^{2}+2 \sum x=0 \Leftrightarrow 9 x y z+\frac{5}{2} \sum x^{2}+2 \sum x y \geqslant 0$$
Assume without loss of generality that $a \geqslant b \geqslant c$, then $1 \geqslant x \geqslant y \geqslant z \geqslant-1$, and $x \geqslant 0, z \leqslant 0$.
When $x \geqslant y \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,928 |
Example $9.11$ a, b, c are non-negative real numbers, prove that
$$\frac{a^{3}}{2 a^{2}-a b+2 b^{2}}+\frac{b^{3}}{2 b^{2}-b c+2 c^{2}}+\frac{c^{3}}{2 c^{2}-c a+2 a^{2}} \geqslant \frac{a+b+c}{3}$$ | Prove that the principle of tangent method for completing the square $\Leftrightarrow$
$$\sum\left(\frac{a^{3}}{2 a^{2}-a b+2 b^{2}}-\frac{a}{3}-\frac{a-b}{3}\right) \geqslant 0 \Leftrightarrow \sum \frac{(a-b)^{2}(2 b-a)}{2 a^{2}-a b+2 b^{2}} \geqslant 0$$
Assume without loss of generality that $a=\max \{a, b, c\}$.
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,930 |
Lemma 10.4 For any real numbers $a, b, c$ we have
$$\begin{array}{l}
\frac{\sum a b \sum a+\left(6 \sum a b-2\left(\sum a\right)^{2}\right) x_{1}}{9} \leqslant a b c \leqslant \\
\frac{\sum a b \sum a+\left(6 \sum a b-2\left(\sum a\right)^{2}\right) x_{2}}{9}
\end{array}$$
where $x_{1}=\frac{\sum a+\sqrt{\left(\sum a\... | Proof: Let real numbers $a, b, c$ satisfy $c \geqslant b \geqslant a$, and consider the function
$$f(x)=(x-a)(x-b)(x-c)=x^{3}-\sum a x^{2}+\sum a b x-a b c$$
Then
$$f^{\prime}=3 x^{2}-2 \sum a x+\sum a b$$
Let the roots of $f^{\prime}(x)=0$ be $x_{1}, x_{2}$ with $x_{1} \geqslant x_{2}$, then we can easily find
$$x_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,933 |
Theorem 10.6 expresses a symmetric polynomial $F(a, b, c)$ in terms of real numbers $a, b, c$ as $f\left(\sum a, \sum a b, a b c\right)$, where $f(x, y, z)$ is a polynomial in $x, y, z$.
We denote $f^{\prime}(a b c)=\frac{\partial f}{\partial(a b c)}\left(\sum a, \sum a b, a b c\right)$, then:
(1) If $f^{\prime}(a b c... | Proof We first prove Theorem 10.6(1), by considering the triples $\left(x_{1}, x_{1}, y_{1}\right),\left(x_{2}, x_{2}, y_{2}\right)$, where
$$\begin{array}{l}
x_{1}=\frac{\sum a+\sqrt{\left(\sum a\right)^{2}-3 \sum a b}}{3}, y_{1}=\frac{\sum a-2 \sqrt{\left(\sum a\right)^{2}-3 \sum a b}}{3} \\
x_{2}=\frac{\sum a-\sqrt{... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,934 |
Example 1.29 (2009 Iran Mathematical Olympiad) Positive numbers $a, b, c$ satisfy $a+b+c=3$, prove
$$\frac{1}{2+a^{2}+b^{2}}+\frac{1}{2+b^{2}+c^{2}}+\frac{1}{2+c^{2}+a^{2}} \leqslant \frac{3}{4}$$ | By symmetry, without loss of generality, assume \(a \geqslant b \geqslant c\), hence \(a \geqslant 1\). Let
\[ f(a, b, c) = \frac{1}{2 + a^2 + b^2} + \frac{1}{2 + b^2 + c^2} + \frac{1}{2 + c^2 + a^2} \]
We can prove that \( f(a, b, c) \leqslant f\left(a, \frac{b+c}{2}, \frac{b+c}{2}\right) \), because
\[ f(a, b, c) - ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,937 |
Theorem 10.12 For non-negative real numbers $(x, y, z)$, define $f(x, y, z)$ as a function of $(x, y, z)$, where $y$ and $z$ are symmetric. Denote $f(x, y, z)$ as $f(x)$, and $f(x)$ is a differentiable function. We keep $\sum x, \sum x^{m}$ unchanged, for functions of the form $F(x, y, z)=f(x)+f(y)+f(z)$, denote $g\lef... | Prove by first discussing the case where $g(x)$ is a convex function.
We control $x+y+z, x^{m}+y^{m}+z^{m}$ unchanged (where $m$ is a real number not equal to 1), at this point there exist non-negative real numbers $a \geqslant b \geqslant c$ (when $m>0$, such $a, b, c$ have at most one zero), satisfying $x+y+z=a+b+c, ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,941 |
Theorem 10.14 For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$, define $f\left(x_{1} ; x_{2}, \cdots, x_{n}\right)$ as a function of $x_{1}, x_{2}, \cdots, x_{n}$, where $x_{2}, \cdots, x_{n}$ are symmetric. Denote $f\left(x_{1} ; x_{2}, \cdots, x_{n}\right)$ as $f\left(x_{1}\right)$, and $f(x)$ is different... | First, prove the case where $g(x)$ is a convex function.
When $m>0$, first prove that when the function $F\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ reaches its minimum value, it must have $x_{1} \leqslant x_{2}=$ $x_{3}=\cdots=x_{n}$, or $d$ numbers are 0, and at least $n-d-1$ positive numbers are equal. We adjust thre... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,944 |
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