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Example $1.30 a, b, c \geqslant 0$, prove that
$$\begin{array}{c}
\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)[(a+b)(b+c)(c+a)-8 a b c] \geqslant \\
9 a b c(a-b)^{2}(b-c)^{2}(c-a)^{2}
\end{array}$$ | Prove that by using $a, b, c$ to replace $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$, the inequality becomes
$$\begin{array}{l}
\left(a^{3}+b^{3}+c^{3}\right)[(a+b)(b+c)(c+a)-8 a b c] \geqslant \\
9(a-b)^{2}(b-c)^{2}(c-a)^{2}
\end{array}$$
Now, without loss of generality, assume $c=\min \{a, b, c\}$, so we have
$$\begin{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,948 |
Theorem 10.19 For a ternary quartic homogeneous cyclic symmetric inequality defined on $\mathbf{R}_{+}^{3}$, the necessary and sufficient condition for its validity is $f(x, 1,1) \geqslant 0, f(x, 1,0) \geqslant 0, x \in[0,+\infty)$. If the equation about $\sigma_{3}$, $A \sigma_{3}^{2}+B \sigma_{3}+C=0$, has real solu... | Prove that we control $\sigma_{1}, \sigma_{2}$ unchanged, and change the value of $\sigma_{3}$.
$$\begin{array}{l}
F^{\prime}\left(\sigma_{3}\right)=k_{4} \sigma_{1}+k_{0} \sigma_{1}\left(-\frac{3}{2}+\frac{\left(18 \sigma_{1} \sigma_{2}-4 \sigma_{1}^{3}-53 \sigma_{3}\right)}{4 \sqrt{\sigma_{1}^{2} \sigma_{2}^{2}-4 \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,950 |
Example 10.1 (Tiks) For $a, b, c \geqslant 0$, and not all zero, prove
$$\frac{a^{2}}{(2 a+b)(2 a+c)}+\frac{b^{2}}{(2 b+c)(2 b+a)}+\frac{c^{2}}{(2 c+a)(2 c+b)} \leqslant \frac{1}{3}$$ | Prove that the original inequality is equivalent to
$$\frac{\sum\left[a^{2}(2 b+c)(2 b+a)(2 c+a)(2 c+b)\right]}{\prod[(2 a+b)(2 a+c)]} \leqslant \frac{1}{3}$$
Let \( p=a+b+c, q=ab+bc+ca, r=abc \), we provide the following identities:
$$\begin{array}{c}
\prod[(2 a+b)(2 a+c)]=8 p^{2} q^{2}+4 p^{3} r+4 q^{3}-18 p q r+27 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,951 |
Example 10.2 (1996 Iran Mathematical Olympiad) Let $a, b, c$ be positive numbers, prove that
$$(a b+b c+c a)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right] \geqslant \frac{9}{4}$$ | Prove that the inequality, after being brought to a common denominator and rearranged, is equivalent to
$$4 \sum a b \sum(a+b)^{2}(b+c)^{2}-9(a+b)^{2}(b+c)^{2}(c+a)^{2} \geqslant 0$$
Let
$$\begin{aligned}
f\left(\sum a, \sum a b, a b c\right)= & 4 \sum a b \sum(a+b)^{2}(b+c)^{2}- \\
& 9(a+b)^{2}(b+c)^{2}(c+a)^{2}
\end... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,952 |
Example 10.3 (Han Jingjun) Determine the minimum value of $k$ such that when non-negative real numbers $a, b, c \geqslant k$, and $a + b + c = 9$, we have
$$\sqrt{a b + b c + c a} \leqslant \sqrt{a} + \sqrt{b} + \sqrt{c}$$ | Let $a=\frac{13-4 \sqrt{10}}{3}, b=c=\frac{7+2 \sqrt{10}}{3}$, at this time
$$\sqrt{a b+b c+c a}=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
Thus, $k \geqslant \frac{13-4 \sqrt{10}}{3}$. Next, we prove that the inequality holds when $k=\frac{13-4 \sqrt{10}}{3}$.
We keep $a+b+c$ and $a b+b c+c a$ unchanged, which means $a^{2}+b^{2}+c... | \frac{13-4 \sqrt{10}}{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,953 |
Example 10.4 (Han Jingjun) Let positive numbers \(a, b, c\) satisfy \(abc = 1\). Prove that
\[
\begin{array}{l}
\frac{1}{a^{3}(b+c)}+\frac{1}{b^{3}(c+a)}+\frac{1}{c^{3}(a+b)} \geqslant \\
\frac{3}{2}+\frac{3}{4}\left[\frac{a(b-c)^{2}}{b+c}+\frac{b(c-a)^{2}}{c+a}+\frac{c(a-b)^{2}}{a+b}\right]
\end{array}
\] | Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, then the original inequality becomes
$$\sum \frac{x^{2}}{y+z} \geqslant \frac{3}{2}+\frac{3}{4} \sum \frac{(y-z)^{2}}{y+z}$$
Let $p=x+y+z, q=xy+yz+zx, r=xyz=1$, then the above inequality is equivalent to
$$4 p^{4}-18 p^{2} q+12 q^{2}+34 p r-6 p q+6 r \geqslant 0$$
Le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,954 |
Example 10.5 Non-negative real numbers $a, b, c$ satisfy $a+b+c=1$, and no two of them are zero at the same time, prove that
$$\frac{1}{\sqrt{a^{2}+a b+b^{2}}}+\frac{1}{\sqrt{b^{2}+b c+c^{2}}}+\frac{1}{\sqrt{c^{2}+c a+a^{2}}} \geqslant 4+\frac{2}{\sqrt{3}}$$ | Assume without loss of generality that $a \leqslant b \leqslant c$. We control $A=a+b+c, B=a^{2}+b^{2}+c^{2}$ to be constant, then
$$\frac{1}{\sqrt{a^{2}+a b+b^{2}}}=\frac{1}{\sqrt{\frac{1}{2} A^{2}+\frac{1}{2} B-A c}}$$
At this point, $m=2$, then by Theorem 10.12, we can consider the function
$$g(x)=f^{\prime}(x)=\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,955 |
Example 10.6 (Yang Xuezhi Conjecture) Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, and $\sum_{i=1}^{n} x_{i}^{2} \leqslant n$, prove that
$$2+(n-2) \prod_{i=1}^{n} x_{i} \geqslant \prod_{i=1}^{n} x_{i} \sum_{i=1}^{n} \frac{1}{x_{i}}$$ | Prove (Han Jingjun) When $n=1$, the condition is the inequality to be proved.
When $n=2$, from the condition $x_{1}^{2}+x_{2}^{2} \leqslant 2$ we know $2 x_{1} x_{2} \leqslant 2$, which is the inequality to be proved.
Below, we prove the case for $n \geqslant 3$, let $\prod_{i=1}^{n} x_{i} \cdot \frac{1}{x_{i}}=\sqrt{\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,956 |
Example 10.7 (Han Jingjun) $x_{1}, x_{2}, \cdots, x_{n}$ are positive real numbers, $\sum_{i=1}^{n} x_{i}=n$, find the maximum value of $\sum_{i=1}^{n} \prod_{j \neq i} x_{j}^{t}$.
Solution When $x \leqslant 1$
$$\frac{\sum_{i=1}^{n} \prod_{j \neq i} x_{j}^{t}}{n} \leqslant\left(\frac{\sum_{i=1}^{n} \prod_{j \neq i} x_... | When $x>1$, note that $\sum_{i=1} \prod_{j \neq i} x_{j}^{t}=\left(\prod_{i=1} x_{i}^{t}\right)\left(\sum_{i=1} \frac{1}{x_{i}^{t}}\right)$. We control $\sum_{i=1} x_{i}, \prod_{i=1} x_{i}$ to be constant, then by Corollary 10.15(2), the maximum value of $\sum_{i=1}^{n} \frac{1}{x_{i}^{\prime}}$ must be achieved when $... | null | Inequalities | proof | Yes | Yes | inequalities | false | 731,957 |
Example $1.4$ If $a, b, c$ are positive numbers, prove that
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}+2(a+b+c) \geqslant \frac{(a+b+c)^{3}}{a b+b c+c a}$$ | Prove that the original inequality is equivalent to
$$(a b+b c+c a)\left[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}+2(a+b+c)\right] \geqslant(a+b+c)^{3}$$
Expanding, we need to prove
$$a^{2} b^{4}+b^{2} c^{4}+c^{2} a^{4} \geqslant a b^{3} c^{2}+b c^{3} a^{2}+c a^{3} b^{2}$$
Let \( x=a b^{2}, y=b c^{2}, z=c a^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,959 |
Example 1.31 (Cai Jianxing) Let $x, y, z \geqslant 0, a, b, c$ be positive real constants, and define the function
$$F(x, y, z)=\frac{x}{a x+b y+c z}+\frac{y}{a y+b z+c x}+\frac{z}{a z+b x+c y}$$
Prove:
(1) If $b^{2} \geqslant c a$ and $c^{2} \geqslant a b$, then $F(x, y, z) \geqslant \frac{3}{a+b+c}$;
(2) If $b^{2} \... | To prove that when $a=b=c$, the inequality obviously holds. Now, let's consider the case where at least two of $a, b, c$ are not equal. We make the following substitution:
$$\left\{\begin{array}{l}
u=a x+b y+c z \\
v=a y+b z+c x \\
w=a z+b x+c y
\end{array}\right.$$
Solving the above system of equations, we get:
$$\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,960 |
Theorem $10.320 \leqslant x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n+1}(n \in \mathrm{N})$, and the equality holds at most in one place. $f\left(x_{1} ; x_{2}, \cdots, x_{n+1}\right)$ is a function of $x_{1}, x_{2}, \cdots, x_{n+1}$, and is symmetric with respect to $x_{2}, x_{3}, \cdots, x_{n+1}$. We denote... | Proof We only prove (1)
$$
\begin{array}{c}
0<x_{1}<x_{2}<\cdots<x_{n}, \quad x_{2 i-1}=x_{2 i}, \quad x_{2 i}<x_{2 i+1}, \quad i=1,2, \cdots, n \\
x_{1}>0, \quad x_{2 i-1}=x_{2 i}, \quad x_{2 i}<x_{2 i+1}, \quad i=1,2, \cdots, n
\end{array}
$$
Also, $D_{1}>0$, so by Lemma 10.31, there exists $\xi, x_{1}<\xi<x_{n}$, ... | proof | Calculus | proof | Yes | Yes | inequalities | false | 731,961 |
Theorem $10.34 f\left(x_{1} ; x_{2}, \cdots, x_{n}\right)$ is a function of $x_{1}, x_{2}, \cdots, x_{m}$, and is symmetric with respect to $x_{2}$, $x_{3}, \cdots, x_{m}$. We denote $f\left(x_{1} ; x_{2}, \cdots, x_{m}\right)$ as $f\left(x_{1}\right)$. Similarly, we define $f\left(x_{i}\right)(i=2,3, \cdots, m)$. Fix ... | Prove: We only prove the case when $F$ reaches its minimum value, the others are completely similar.
Since $s_{(m, i)}(i=1,2, \cdots, n ; m \geqslant n+1)$ are fixed, then at this time $\left(x_{1}, x_{2}, \cdots, x_{m}\right)$ is a compact set on $\mathbf{R}_{+}^{m}$, and $\boldsymbol{F}$ is continuous on $\mathbf{R}_... | proof | Calculus | proof | Yes | Yes | inequalities | false | 731,964 |
Theorem 10.37 A symmetric inequality of degree $m$ in $n$ variables $F(x) \geqslant 0$ holds on $\mathbf{R}_{+}^{n}$ if and only if it holds for $\left\{x \mid x \in \mathbf{R}_{+}^{n}, v(x)^{*} \leqslant \max \left(\left[\frac{m}{2}\right], 1\right)\right\}$. | To prove the necessity is obvious, we now prove the sufficiency.
When $m \geqslant 2 n$ and $m=1$, the theorem is obviously true. We now prove it for $2 \leqslant m \leqslant 2 n-1$.
Let $\left[\frac{m}{2}\right]=t$. We fix $s_{(n, i)}(i=1,2, \cdots, t)$, then at this point $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,968 |
Example 11.1 If $a, b, c \in\left[\frac{1}{3}, 3\right]$, prove
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac{7}{5}$$ | Prove that the original inequality is equivalent to
$$a b c+3 \sum a^{2} b \geqslant 2 \sum a^{2} c$$
(1) If $a \geqslant b \geqslant c$, then
$$(a-b)(b-c)(c-a) \leqslant 0 \Leftrightarrow \sum a^{2} b \geqslant \sum a^{2} c$$
In this case, the original inequality clearly holds.
(2) If $a \geqslant c \geqslant b$, rew... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,969 |
Example 11. 2 Let $f(x, y)=a x^{2}+2 b x y+c y^{2}, a, b, c \in \mathbf{R}, D=a c-b^{2}$, if $D>0$, then there exists an integer $(u, v) \neq(0,0)$, such that
$$|f(u, v)| \leqslant \sqrt{\frac{4 D}{3}}$$ | Prove that since $f(x, y)=0$ is an ellipse, for any large integer $A,|f(u, v)|<A$ has only a finite number of integer solutions.
Therefore, $|f(u, v)|_{\text {min }}$ exists, denoted by $m$, and let $f(u, v)=m$, then it is clear that $(u, v)=$ 1. Thus, there exist $r, s \in \mathbb{Z}$, such that $u r-v s=1$.
Let $x=... | m \leqslant \sqrt{\frac{4 D}{3}} | Algebra | proof | Yes | Yes | inequalities | false | 731,970 |
Example 1.32 (Han Jingjun) $x, y, z \geqslant 0$, prove that
$$\sum \sqrt{(x+y)(x+z)} \geqslant \sum x+\sqrt{3 \sum x y}$$ | $$\begin{array}{l}
\sum \sqrt{(x+y)(x+z)}-\sum x \geqslant \sqrt{3 \sum x y} \Leftrightarrow \\
\sum x^{2}+3 \sum x y+\left(\sum x\right)^{2}-2 \sum x \sum \sqrt{(x+y)(x+z)} \geqslant 3 \sum x y \Leftrightarrow \\
\sum x^{2}+\sum x y \geqslant \sum x \sqrt{(x+y)(x+z)}
\end{array}$$
The above inequality is evident by t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,971 |
Example 11.3 (Vasile) \(x=a+\frac{1}{b}-1, y=b+\frac{1}{c}-1, z=c+\frac{1}{a}-1, a, b, c>0\), prove that
\[x y+y z+z x \geqslant 3\] | Proof: By the Pigeonhole Principle, among $x, y, z$, there must exist two numbers, say $x, y$, such that $(x-1)(y-1) \geqslant 0$. Therefore,
$$\begin{aligned}
x y+y z+z x= & (x-1)(y-1)+x+y+y z+z x-1 \geqslant \\
& x+y+y z+z x-1= \\
& (x+y)(z+1)-1= \\
& \left(b+\frac{1}{b}-2+a+\frac{1}{c}\right)\left(\frac{1}{a}+c\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,972 |
Example 11.4 Let $0<m_{1} \leqslant a_{i} \leqslant M_{1}, 0<m_{2} \leqslant b_{i} \leqslant M_{2}, i=1,2, \cdots, n$, then
$$\left[\sqrt{\frac{m_{2} M_{2}}{m_{1} M_{1}}} \sum_{i=1}^{n} a_{i}^{2}\right]+\left[\sqrt{\frac{m_{1} M_{1}}{m_{2} M_{2}}} \sum_{i=1}^{n} b_{i}^{2}\right] \leqslant\left[\sqrt{\frac{M_{1} M_{2}}{... | Prove that because $m_{i} \leqslant a_{i} \leqslant M_{1}, m_{2} \leqslant b_{i} \leqslant M_{2}$, we have
$$\frac{m_{1}}{M_{2}} \leqslant \frac{a_{i}}{b_{i}} \leqslant \frac{M_{1}}{m_{2}}, i=1,2, \cdots, n$$
Thus,
$$\begin{array}{l}
\sum_{i=1}^{n}\left[m_{2} M_{2} a_{i}^{2}-\left(m_{1} m_{2}+M_{1} M_{2}\right) a_{i} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,973 |
Example 11.5 (1991 IMO Shortlist) $x, y \in [0,1], n, m \in \mathbf{N}, x+y=1$, prove that
$$\left(1-x^{n}\right)^{m}+\left(1-y^{m}\right)^{n} \geqslant 1$$ | Proof Consider an $m \times n$ grid, each cell is colored either black or white, with the probability of coloring black being $x$, and the probability of coloring white being $y,\left(1-x^{n}\right)^{m}$ corresponds to no row being entirely black, $\left(1-y^{m}\right)^{n}$ corresponds to no column being entirely white... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,974 |
Example 11.7 The $n$-dimensional real number arrays $\left(a_{1}, a_{2}, \cdots, a_{n}\right),\left(b_{1}, b_{2}, \cdots, b_{n}\right),\left(c_{1}, c_{2}, \cdots, c_{n}\right)$ satisfy
$$\left\{\begin{array}{l}
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1 \\
b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1 \\
c_{1}^{2}+c_{2}^{2}+\cdot... | Proof: Let $\bar{a}=\left(a_{1}, a_{2}, \cdots, a_{n}\right), \bar{b}=\left(b_{1}, b_{2}, \cdots, b_{n}\right), \bar{c}=\left(c_{1}, c_{2}, \cdots, c_{n}\right)$, then $|\bar{a}|=1,|\bar{b}|=1,|\bar{c}|=1 ; b_{1} c_{1}+b_{2} c_{2}+\cdots+b_{n} c_{n}=0 \Rightarrow \bar{b} \cdot \bar{c}=0 \Rightarrow \bar{b} \perp \bar{c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,976 |
Example 11.8 (Crux) $a, b, c, d \geqslant 0$, prove that
$$(a+b)^{3}(b+c)^{3}(c+d)^{3}(d+a)^{3} \geqslant 16 a^{2} b^{2} c^{2} d^{2}(a+b+c+d)^{4}$$ | Assume $a+b+c+d=1$, note that $(a+b)(b+c)(c+d)(d+a)$ is cyclic symmetric rather than fully symmetric, we attempt to bring it closer to a symmetric polynomial.
$$\begin{array}{l}
(a+b)(b+c)(c+d)(d+a)= \\
a^{2} c^{2}+b^{2} d^{2}+2 a b c d+\sum a b c(a+b+c)= \\
(a c-b d)^{2}+\sum a b c(a+b+c+d) \geqslant \\
a b c+b c d+c ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,977 |
Example 11.9 $a, b, c \geqslant 0$, at most one of them is 0, prove that
$$\frac{a(b+c)}{b^{2}+c^{2}}+\frac{b(c+a)}{c^{2}+a^{2}}+\frac{c(a+b)}{a^{2}+b^{2}} \geqslant \frac{a(b+c)}{a^{2}+b c}+\frac{b(c+a)}{b^{2}+c a}+\frac{c(a+b)}{c^{2}+a b}$$ | To prove the inequality is equivalent to
$$\sum \frac{a(b+c)\left(a^{2}+b c-b^{2}-c^{2}\right)}{\left(b^{2}+c^{2}\right)\left(a^{2}+b c\right)} \geqslant 0$$
For given positive real numbers \(x, y, z, x \geqslant y\), we have
$$\begin{array}{l}
\frac{1}{\left(x^{2}+y z\right)\left(y^{2}+z^{2}\right)}-\frac{1}{\left(y^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,978 |
Example $11.10 a, b, c, d \geqslant 0$, no two are zero at the same time, and $a+b+c=1$, prove that
$$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+d}}+\frac{d}{\sqrt{d+a}} \leqslant \frac{3}{2}$$ | Assume without loss of generality that $a+c \geqslant b+d$, then $x=a+c \geqslant \frac{1}{2}$. By Jack Garfunkel's inequality, we have
$$\begin{array}{c}
\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}} \leqslant \frac{5}{4} \sqrt{a+b+c}=\frac{5}{4} \sqrt{1-d} \Rightarrow \\
\frac{a}{\sqrt{a+b}}+\frac{b}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,979 |
Example 11.11 (Han Jingjun) $a, b, c, d \geqslant 0$, not all three are zero at the same time, prove that
$$\frac{a}{\sqrt{a+b+c}}+\frac{b}{\sqrt{b+c+d}}+\frac{c}{\sqrt{c+d+a}}+\frac{d}{\sqrt{d+a+b}} \leqslant \frac{5}{4} \sqrt{a+b+c+d}$$ | Assume $d=\min \{a, b, c, d\}$, and let $x=a+d$, then
$$\frac{a}{\sqrt{a+b+c}}+\frac{d}{\sqrt{d+a+b}} \leqslant \frac{a}{\sqrt{a+b+d}}+\frac{d}{\sqrt{d+a+b}}=\frac{x}{\sqrt{x+b}}$$
It is also clear that
$$\frac{b}{\sqrt{b+c+d}} \leqslant \frac{b}{\sqrt{b+c}}$$
Thus, it suffices to prove
$$\frac{x}{\sqrt{x+b}}+\frac{b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,980 |
Example 1.33
$$\begin{array}{c}
\text { Let } x, y, z \in \mathbf{R}^{+} \text {, and }(x+y+z)^{3}=32 x y z \text {, find } \\
p=\frac{x^{4}+y^{4}+z^{4}}{(x+y+z)^{4}}
\end{array}$$
the maximum and minimum values. | Let $\sum x=1$, then $x y z=\frac{1}{32}$, denote $s=x y+y z+z x$, then
that is
$$\begin{array}{c}
p=2\left(\sum x y-1\right)^{2}-\frac{7}{8} \\
p=2(s-1)^{2}-\frac{7}{8}
\end{array}$$
On the other hand
$$(1-x)^{2}=(y+z)^{2} \geqslant 4 y z=\frac{1}{8 x} \Rightarrow \frac{3-\sqrt{5}}{4} \leqslant x \leqslant \frac{1}{... | \frac{383-165 \sqrt{5}}{256} \leqslant p \leqslant \frac{9}{128} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,982 |
Example 11.13 Let $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative real numbers, satisfying $a_{1}+a_{2}+\cdots+a_{n}=4$. Prove that
$$a_{1}^{3} a_{2}+a_{2}^{3} a_{3}+\cdots+a_{n}^{3} a_{1} \leqslant 27$$ | We prove this using mathematical induction.
When $n=3$, since the inequality is cyclic symmetric about $a_{1}, a_{2}, a_{3}$, without loss of generality, assume $a_{1}$ is the largest. If $a_{2}<a_{3}$, then
$$\begin{array}{c}
a_{1}^{3} a_{2}+a_{2}^{3} a_{3}+a_{3}^{3} a_{1}-\left(a_{1} a_{2}^{3}+a_{2} a_{3}^{3}+a_{3} a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,983 |
Example 11.15 (Suranyi) \(a_{i} \geqslant 0, i=1,2, \cdots, n\), prove that
$$(n-1)\left(a_{1}^{n}+\cdots+a_{n}^{n}\right)+n a_{1} \cdots a_{n} \geqslant\left(a_{1}+\cdots+a_{n}\right)\left(a_{1}^{n-1}+\cdots+a_{n}^{n-1}\right)$$
In particular, when \(n=3\), it is the 3rd degree Schur inequality. | To prove that due to the symmetry of the inequality, we may assume $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{k} \geqslant a_{k+1}$, and $a_{1}+a_{2}+\cdots+a_{k}=1$. We will use mathematical induction to prove this inequality. When $n=1$, the original inequality clearly holds. Assume that when $n=k$, the ine... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,985 |
Example 11.16 (2004 China National Team Training Problem) $a, b, c, x, y, z \in \mathbf{R}$, satisfying
$$(a+b+c)(x+y+z)=3,\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)=4$$
Prove
$$a x+b y+c z \geqslant 0$$ | To prove that if $a, b, c$ are permuted, the condition remains unchanged.
Thus, if $a x + b y + c z \geqslant 0$, then it must also be true that $a y + b z + c x \geqslant 0$ and $a z + b x + c y \geqslant 0$. Given the condition $\sum_{c y c} a x + \sum_{c y c} a y + \sum_{c y c} a z = 3$.
If one of $\sum_{c y c} a x,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,986 |
Example 11.19 Non-negative real numbers $a_{1}, a_{2}, \cdots, a_{n}, n \geqslant 5$, satisfy $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=1$, prove that
$$a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\cdots+a_{n}^{2} a_{1}<\sqrt{\frac{5+2 \sqrt{7}}{33+6 \sqrt{7}}}$$ | Let $a_{0}=a_{n}, a_{n+1}=a_{1}$, and set $S=$ LHS.
Construct the Lagrange function
$$F=a_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\cdots+a_{n}^{2} a_{1}+\lambda\left(1-a_{1}^{2}-a_{2}^{2}-\cdots-a_{n}^{2}\right)$$
By the necessary condition for extremum, we get
$$\left\{\begin{array}{l}
\frac{\partial F}{\partial a_{k}}=a_{k-1}... | \sqrt{\frac{5+2 \sqrt{7}}{33+6 \sqrt{7}}} | Inequalities | proof | Yes | Yes | inequalities | false | 731,990 |
Example 11.20 (Fan Inequality) $b_{i} \in \mathbf{R}, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} b_{i}=0$, then $\sum_{i=1}^{n} b_{i} b_{i+1} \leqslant \cos \frac{2 \pi}{n} \sum_{i=1}^{n} b_{i}^{2}$ | Prove that when $b_{i}=0(i=1,2, \cdots, n)$, the original inequality obviously holds. We only need to consider the case where $b_{i} \neq 0$.
By homogeneity, we may assume $\sum_{i=1}^{n} b_{i}^{2}=1$ (note that there is already a constraint on $b_{i}$, think about what kind of constraint condition allows us to assume... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,991 |
Note: In the section on AM-GM inequality, we introduced the Shapiro inequality, knowing that it only holds for certain $n$. Vasile et al. discussed restricting the range of $x_{i}$ to make the Shapiro inequality hold. With the help of Fan's inequality, the following conclusion was obtained.
$$a_{n}=\frac{1}{\sqrt{2 \co... | To prove the desired inequality is equivalent to
$$\sum_{i=1}^{n} \frac{x_{i}-\frac{x_{i+1}+x_{i+2}}{2 a_{n}^{2}}}{x_{i+1}+x_{_{i+2}}} \geqslant \frac{n\left(a_{n}^{2}-1\right)}{2 a_{n}^{2}}$$
By the Cauchy inequality, it suffices to prove
$$\frac{\left(\sum_{i=1}^{n} x_{i}-\frac{1}{2 a_{n}^{2}} \sum_{i=1}^{n}\left(x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,992 |
Example 1.34 Find the largest positive real number $k$, such that for any positive real numbers $a, b, c, d$, there is always $(a+b+c)\left[3^{4}(a+b+c+d)^{5}+2^{4}(a+b+c+2 d)^{5}\right] \geqslant k a b c d^{3}$ | Let $t=a+b+c, u=\frac{t}{d}$, and set $a=b=c=1, d=3$, we can get $k \leqslant 2^{4} \times 3^{7} \times 5$.
To prove
$$\begin{aligned}
(a+b+c)\left[3^{4}(a+b+c+d)^{5}+2^{4}(a+b+c+2 d)^{5}\right] \geqslant \\
2^{4} \times 3^{7} \times 5 a b c d^{3}
\end{aligned}$$
For this, we can prove a stronger conclusion:
$$\begin{... | 2^{4} \times 3^{7} \times 5 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,993 |
Example 11.21 (Hilbert Inequality) Let $a_{0}, a_{2}, \cdots, a_{N}>0$, then we have
$$\sum_{m, n=0}^{N} \frac{a_{m} a_{n}}{m+n+1}<\pi \sum_{0}^{N} a_{n}^{2}$$ | Proof: Assuming $a_{k}(k=0,1, \cdots, N)$ has at least two numbers as 0, then the original inequality obviously holds. Now let's consider
$$F(a)=F\left(a_{0}, a_{2}, \cdots, a_{N}\right)=\sum_{m, n=0}^{N} \frac{a_{m} a_{n}}{m+n+1}$$
where $a_{0}, a_{1}, \cdots, a_{N}$ satisfy
$$G(a)=\sum_{0}^{N} a_{n}^{2}=t$$
Here, $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,994 |
Example $12.2 a, b, c \geqslant 0$, at most only 1 is 0, prove that
$$\sum \frac{\sqrt{a b+4 b c+4 a c}}{a+b} \geqslant \frac{9}{2}$$ | The inequality mentioned above has been introduced in the previous chapter ${ }^{(1)}$. It is not easy to prove, being a strengthened version of the 1996 Iran inequality. The equality holds when $a=b=c, a=b, c=0$ and their cyclic permutations.
To maintain the characteristics of this inequality, we use the AM-GM inequa... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,996 |
Example 12.3 Let $a, b, c$ be non-negative real numbers, with at most one of them being 0, prove that
$$a^{3}+b^{3}+c^{3}+3 a b c \geqslant \frac{\left(a^{2} b+b^{2} c+c^{2} a\right)^{2}}{a b^{2}+b c^{2}+c a^{2}}+\frac{\left(a b^{2}+b c^{2}+c a\right)^{2}}{a^{2} b+b^{2} c+c^{2} a}$$
Similarly, let's first look at its ... | Proof If $a, b, c$ has one number as 0, it is obviously true.
If $a b c>0$, let $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$, we need to prove
$$\begin{array}{c}
\frac{x}{z}+\frac{y}{x}+\frac{z}{y}+3 \geqslant \frac{(x y+y z+z x)^{2}}{x y z(x+y+z)}+\frac{(x+y+z)^{2}}{x y+y z+z x} \Leftrightarrow \\
\frac{x}{z}+\frac{y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,997 |
Example $12.4 ~ a, b, c \geqslant 0, a+b+c=3$, prove that
$$\sqrt{3-a b}+\sqrt{3-b c}+\sqrt{3-c a} \geqslant 3 \sqrt{2}$$ | Prove that the above inequality is known, then what is the situation when the coefficients 1 of $ab, bc, ca$ inside each radical are changed to $\lambda$? In 2008 and 2009, the IMO full scorer Wei Dongyi obtained the following conclusion.
Let $x, y, z$ be non-negative real numbers, and satisfy $x+y+z=1$. Define
$$f(x,... | f_{\min }=\min \left(\sqrt{9 \lambda-1}, 2 \sqrt{\lambda}+\sqrt{\lambda-\frac{1}{4}}\right) | Inequalities | proof | Yes | Yes | inequalities | false | 731,998 |
For example, in $\triangle ABC$, $D$ is on $AB$, $E$ is on $BC$, and $F$ is on $AC$. Let the areas of $\triangle BDE$, $\triangle CEF$, $\triangle ADF$, and $\triangle DEF$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S$, respectively. Prove that
$$\frac{1}{S_{1}^{2}}+\frac{1}{S_{2}^{2}}+\frac{1}{S_{3}^{2}} \geqslant \frac{3}{S^... | Prove that the problem above appears in "Toward IMO 2003," as a geometry training problem for the national training team.
In all the books the author has seen, it is set as $\frac{B E}{E C}=x, \frac{F C}{A F}=y, \frac{A D}{D B}=z (x, y, z>0)$.
Then, the algebraic inequality to be proved is
$$\frac{1}{x^{2}(1+y)^{2}}+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,999 |
Example 12.6 (Sun Shibao) Let $x, y, z \geqslant 0$, with no two being 0 at the same time, prove
$$1 \leqslant \sum \frac{x^{2}}{\sqrt{\left(x^{2}+y^{2}+x y\right)\left(x^{2}+z^{2}+z x\right)}} \leqslant \frac{2 \sqrt{3}}{3}$$
This problem has been discussed before, here we provide another proof. | To prove the left side of the inequality. Let \( PA = x, PB = y, PC = z \), and let point \( P \) satisfy \( \angle APB = \angle BPC = \angle CPA = \frac{2\pi}{3} \). Consider the plane of \( \triangle ABC \) as the complex plane, and let \( P, A, B, C \) correspond to the complex numbers \( z, z_1, z_2, z_3 \).
Notic... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,000 |
Example 12.7 There is an existing conclusion.
$a, b, c>0$, then we have
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{a+1}{b+1}+\frac{b+1}{c+1}+\frac{c+1}{a+1}$$ | Prove that the proof can be provided using the following lemma introduced in the previous chapters.
$x, y, z, u, v, w > 0$ and $x y z = u v w, x \leqslant y \leqslant z, u \leqslant v \leqslant w, x \leqslant u, z \geqslant w$.
Then $x + y + z \geqslant u + v + w$.
View $\frac{a}{b}, \frac{b}{c}, \frac{c}{a}, \frac{a+1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,001 |
Example 12.8 We know that when $a, b, c \geqslant 0, a+b+c=2$, we have
$$\sqrt{a^{2}+b c}+\sqrt{b^{2}+c a}+\sqrt{c^{2}+a b} \leqslant 3$$
Subtracting $a+b+c$ from both sides is equivalent to
$$\frac{b c}{a+\sqrt{a^{2}+b c}}+\frac{c a}{b+\sqrt{b^{2}+c a}}+\frac{a b}{c+\sqrt{c^{2}+a b}} \leqslant \frac{a+b+c}{2}$$ | And we have
$$\begin{array}{c}
\left(a+\sqrt{a^{2}+b c}\right)^{2}=2 a^{2}+b c+2 a \sqrt{a^{2}+b c} \leqslant 2 a^{2}+b c+a+a^{3}+a b c \\
\left(a+\sqrt{a^{2}+b c}\right)^{2}=2 a^{2}+b c+2 a \sqrt{a^{2}+b c} \leqslant 2 a^{2}+b c+2 a^{2}+b c \\
\left(a+\sqrt{a^{2}+b c}\right)^{2}=2 a^{2}+b c+2 a \sqrt{a^{2}+b c} \leqsl... | \sum \frac{a}{2} \geqslant \sum \frac{b c}{\sqrt{(a+b)(a+c)}} \geqslant \frac{\sqrt{3 \sum a}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 732,002 |
Example 12.12
Given that $a, b, c$ are positive numbers, prove that
$$\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}} \geqslant \frac{a+b+c}{2}$$ | Prove (Han Jingjun) Given that $a, b, c$ are positive numbers, prove that
$$\frac{3 a^{4}+a^{2} b^{2}}{a^{3}+b^{3}}+\frac{3 b^{4}+b^{2} c^{2}}{b^{3}+c^{3}}+\frac{3 c^{4}+c^{2} a^{2}}{c^{3}+a^{3}} \geqslant 2(a+b+c)$$
The proof is quite challenging. If using the difference allocation method, it is equivalent to proving... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,003 |
Example $1.35 a, b, c, d \geqslant 0$, prove that
$$\frac{a-b}{a+2 b+c}+\frac{b-c}{b+2 c+d}+\frac{c-d}{c+2 d+a}+\frac{d-a}{d+2 a+b} \geqslant 0$$ | To prove the inequality is equivalent to
$$\sum\left(\frac{a-b}{a+2 b+c}+\frac{1}{2}\right) \geqslant 2 \Leftrightarrow \sum \frac{3 a+c}{a+2 b+c} \geqslant 4$$
By the Cauchy-Schwarz inequality, we have
$$\left(\sum \frac{3 a+c}{a+2 b+c}\right)\left(\sum(3 a+c)(a+2 b+c)\right) \geqslant 16(a+b+c+d)^{2}$$
In fact,
$$\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,004 |
Example 12.13 Let \(x, y, z\) be positive real numbers. Prove that
\[
\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>2 \sqrt[3]{x^{3}+y^{3}+z^{3}}
\] | This is a problem that appeared in the 2008 National Training Team test, and its proof is quite simple.
In fact, let \( a^{2}=\frac{y z}{x}, b^{2}=\frac{z x}{y}, c^{2}=\frac{x y}{z} \Rightarrow a, b, c>0 \).
The original inequality is equivalent to
\[
\begin{array}{l}
a^{2}+b^{2}+c^{2}>\sqrt[3]{a^{3} b^{3}+b^{3} c^{3}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,005 |
Example 12.14 We explore the intermediate quantities of $\sqrt{2}\left(a^{2}+b^{2}+c^{2}\right), \sqrt{2}(a b+b c+c a)$, and obtain the following inequality.
(Han Jingjun) For non-negative real numbers $a, b, c$, prove
$$\sqrt{2}\left(a^{2}+b^{2}+c^{2}\right) \geqslant a \sqrt{a^{2}+b c}+b \sqrt{b^{2}+c a}+c \sqrt{c^{2... | Prove by squaring both sides, it suffices to show
$$\sum a^{4}+2 \sum a b \sqrt{\left(a^{2}+b c\right)\left(b^{2}+c a\right)} \geqslant 2 \sum a^{2} b^{2}+3 \sum a^{2} b c$$
By the Cauchy-Schwarz inequality, we have
$$\sum a b \sqrt{\left(a^{2}+b c\right)\left(b^{2}+c a\right)} \geqslant \sum a b(a b+c \sqrt{a b})$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,006 |
Example $13.1 x_{i} \geqslant 0(i=1,2, \cdots, n), s_{(n, \alpha)}\left(x_{1}, x_{2}, \cdots, x_{n}\right)=$ $\sum_{i=1}^{n} x_{i}^{\alpha}, \alpha \in \mathbf{Z}, n \geqslant 3$, let $f_{k}=\sum_{i=1}^{n} x_{i}^{k} \sum_{i=1}^{n} \frac{1}{x_{i}^{k}}$, then we have
$$\sqrt{f_{2}} \leqslant \sqrt{f_{1}}\left(\sqrt{f_{1}... | Prove that fixing the values of $s_{(n, 2)}, s_{(n, 1)}, s_{(n,-1)}$, by the decision theorem (2), the maximum value of $s_{(n,-2)}$ can be achieved when $0 < x_{1} \leq x_{2} = \cdots = x_{n-1} \leq x_{n}$ or $x_{1} = x_{2} = \cdots = x_{a} \leq x_{a+1} = \cdots = x_{n}$ $(0 \leq a \leq n-1)$. By the homogeneity of th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,007 |
Theorem 13.2 If $x \geqslant 0, y \geqslant 0, z \geqslant 0, k$ is a non-negative real number, then
(1) $\sum(y z)^{k}(x-y)(x-z) \geqslant 0(k \geqslant 0)$;
(2) $\sum x^{k}(y+z)(x-y)(x-z) \geqslant 0(k \geqslant 1)$;
(3) $\sum(y z)^{k}(y+z)(x-y)(x-z) \geqslant 0(k \geqslant 0)$. | (1) can be transformed into
$$(x y z)^{k} \sum x^{-k}(x-y)(x-z) \geqslant 0$$
By Schur's inequality, the above inequality holds.
(2), (3) are left to the reader as exercises. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,009 |
Theorem 13.4 A ternary homogeneous quartic symmetric inequality $f(x, y, z)$ can be uniquely represented as
$$f(x, y, z)=a g_{4,1}+b g_{4,2}+c g_{4,3}+d g_{4,4}$$
where
$$\begin{array}{c}
g_{4,1}=\sum x^{2}(x-y)(x-z) \\
g_{4,2}=\sum x(y+z)(x-y)(x-z) \\
g_{4,3}=\sum y z(x-y)(x-z) \\
g_{3,3}=x y z(x+y+z)
\end{array}$$
... | Here, we provide a simple method to determine the coefficients
$$a=f(1,0,0), c=f(1,1,0), d=\frac{f(1,1,1)}{3}, b=a+\frac{c-f(-1,0,1)}{4}$$ | not found | Inequalities | proof | Yes | Yes | inequalities | false | 732,011 |
Theorem 13.5 A ternary homogeneous symmetric polynomial of degree five $f(x, y, z)$ can be uniquely represented as
$$f(x, y, z)=a g_{5,1}+b g_{5,2}+c g_{5,3}+d g_{5,4}+e g_{5,5}$$
where
$$\begin{array}{c}
g_{5,1}=\sum x^{3}(x-y)(y-z) \\
g_{5,2}=\sum x^{2}(y+z)(x-y)(y-z) \\
g_{5,3}=\sum y z(y+z)(x-y)(x-z) \\
g_{5,4}=x ... | $$\begin{array}{c}
a=f(1,0,0), c=\frac{f(1,1,0)}{2}, e=\frac{f(1,1,1)}{3} \\
b=\frac{f(1, \mathrm{i}, 0)}{2(1+\mathrm{i})}+\frac{c}{2}, d=\frac{f(-1, \mathrm{i}, 1) \mathrm{i}+8 b+e-2 a}{2}
\end{array}$$
The following is the translation:
$$\begin{array}{c}
a=f(1,0,0), c=\frac{f(1,1,0)}{2}, e=\frac{f(1,1,1)}{3} \\
b=\... | not found | Inequalities | proof | Yes | Yes | inequalities | false | 732,012 |
Example 13.2 Let $a, b, c$ be the three sides of a triangle, satisfying $a+b+c=3$, find the maximum value of the following expression:
$$a^{2}+b^{2}+c^{2}+\frac{4}{3} a b c$$ | Let's make the substitution: let $a=x+y, b=y+z, c=z+x$, then $x+y+z=\frac{3}{2}$. After homogenization, the original problem is equivalent to finding the maximum value of $k$ such that the following inequality always holds:
$$\begin{aligned}
f(x, y, z)= & {\left[(x+y)^{2}+(y+z)^{2}+(z+x)^{2}\right] \frac{2(x+y+z)}{3}+}... | \frac{13}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,013 |
Example 1.36 (Vasile) $a, b, c, x, y, z$ are positive real numbers and satisfy
$$(a+b+c)(x+y+z)=\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)=4$$ | Proof
$$a b c x y z<\frac{1}{36}$$
Proof By the AM - GM inequality, we can obtain
$$\begin{array}{l}
4(a b+b c+c a)(x y+y z+z x)= \\
{\left[(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)\right]\left[(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)\right]=} \\
20-(a+b+c)^{2}\left(x^{2}+y^{2}+z^{2}\right)-\left(a^{2}+b^{2}+c^{2}\... | 36 a b c x y z \leqslant 1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,015 |
For example, let $a, b, c$ be the side lengths of $\triangle ABC$. Prove that
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9+\frac{(a-c)^{2}}{b^{2}}$$ | Prove (Yang Xuezhi) When $a \geqslant b \geqslant c$, we have $\frac{a-c}{b} \geqslant \frac{a-b}{c} \geqslant \frac{b-c}{a}$. Therefore, we only need to prove the case when $a \geqslant b \geqslant c$. The original inequality is equivalent to
$$\begin{array}{l}
\sum\left(\frac{a}{b}+\frac{b}{a}-2\right) \geqslant \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,016 |
Example 13.5 (Huang Chendi) Given $a \geqslant b \geqslant c \geqslant d, a+b+c+d=1$, and $k$ is a given positive real number, find the maximum value of $\lambda$ such that
$$\frac{a-b}{k+c+d}+\frac{b-c}{k+d+a}+\frac{c-d}{k+a+b}+\frac{d-a}{k+b+c} \geqslant \lambda(a-b)(b-c)(c-d)$$ | Solving this problem is not easy. First, we observe the original problem and try to guess the maximum value of $\boldsymbol{\lambda}$, which is quite difficult. Notice that both sides of the inequality have $a-b, b-c, c-d, d-a$, so we consider a direct difference substitution. We start by bounding the left side of the ... | \frac{64}{(2 k+1)^{3}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,017 |
Theorem 13.6 (1) If the form $F\left(x_{1}, x_{2}, \cdots, x_{m}\right)$ is strictly positive for $x_{i} \geqslant 0, i=1,2, \cdots, m, \sum_{i=1}^{m} x_{i}>0$, then $F$ can be expressed as $F=\frac{G}{H}$, where $G$ and $H$ are both forms with positive coefficients. In particular, we can assume
$$H=\left(x_{1}+x_{2}+\... | Proof, for the sake of simplicity in writing, assume $m=3$, for a general $m$, there is no essential difference.
The function $F(x, y, z)$ is positive and continuous in the closed domain $x, y, z \geqslant 0, x+y+z=1$, hence it has a positive minimum $\mu$ in this domain. Let
$$F(x, y, z)=\sum_{n} A_{\alpha \beta \gam... | proof | Algebra | proof | Yes | Yes | inequalities | false | 732,018 |
Theorem 13.8 (Yang Lu) Let $u_{1}, u_{2}, \cdots, u_{n}, t(n \geqslant 2)$ be non-negative real numbers, and let the polynomial $f(y)$ of $y$ be defined as
$$f(y)=\prod_{j_{1} j_{2}, \cdots j_{n-1} \in|1,2|}\left(y-\left[\sqrt{u_{1}}+(-1)^{j_{1}} \sqrt{u_{2}}+\cdots+(-1)^{j_{n-1}} \sqrt{u_{n}}\right]^{2}\right)$$
Then... | Prove the necessity. From the given, for $j_{1}, j_{2}, \cdots, j_{n} \in\{1,2\}$, there are $2^{n}$ inequalities:
$$\begin{array}{l}
\sqrt{t}-\left[(-1)^{j_{1}} \sqrt{u_{1}}+(-1)^{j_{2}} \sqrt{u_{2}}+\cdots+(-1)^{j_{n}} \sqrt{u_{n}}\right] \geqslant \\
\sqrt{t}-\left(\sqrt{u_{1}}+\sqrt{u_{2}}+\cdots+\sqrt{u_{n}}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,020 |
Example 14.1
(Han Jingjun) $a, b, c \geqslant 0, a^{2}+b^{2}+c^{2}=3$, prove that
$$\sum_{c \mathrm{c}} a \sqrt{3 b^{3}+3 b+3} \leqslant a b c+2+\sum a$$ | Prove using
$$\begin{array}{c}
\left(b^{2}+2\right)^{2}=b^{4}+4 b^{2}+4 \geqslant 4 b^{3}+4 b-2 b^{2}+3 \geqslant 3 b^{3}+3 b+3 \\
a^{2} c+b^{2} a+c^{2} b \leqslant a b c+2
\end{array}$$
that is to prove the original problem. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,023 |
Example $14.2 a, b, c>0$, prove that
$$\sum_{\text {cyc }} \frac{(a+b)^{3}}{3 a^{2}+3 a b+b^{2}} \geqslant \frac{8}{7}(a+b+c)$$ | Proof Using Tangent Line Method to Prove Local Inequality
$$\frac{(a+b)^{3}}{3 a^{2}+3 a b+b^{2}} \geqslant \frac{12 a+44 b}{49}$$
By adding similar three inequalities, we obtain the desired inequality. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,024 |
Example $14.3 a, b, c>0, k \geqslant 2$, prove that
$$\sum \frac{1}{b^{2}+k b c+c^{2}} \geqslant \frac{9}{(k+2)(a b+b c+c a)}$$ | Prove that for the function $f(k)=\frac{k+2}{b^{2}+k b c+c^{2}}$, we have
$$f^{\prime}(k)=\frac{(b-c)^{2}}{\left(b^{2}+k b c+c^{2}\right)^{2}} \geqslant 0$$
Thus, $f(k)$ is increasing, and it suffices to prove the case $k=2$, which is the 1996 Iran Inequality. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,025 |
Example 1.37 For any positive real numbers $x, y, z$, prove
$$\frac{x^{2}+y z}{(y+z)^{2}}+\frac{y^{2}+z x}{(z+x)^{2}}+\frac{z^{2}+x y}{(x+y)^{2}} \geqslant \frac{3}{2}$$ | Proof:
If we multiply both sides of the inequality by 2, the inequality becomes
$$\frac{2\left(x^{2}+y z\right)}{(y+z)^{2}}+\frac{2\left(y^{2}+z x\right)}{(z+x)^{2}}+\frac{2\left(z^{2}+x y\right)}{(x+y)^{2}} \geqslant 3$$
Now let $a=y+z, b=z+x, c=x+y$, after some simple calculations we have $2 b c-c a-a b+a^{2}=2\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,026 |
Example $14.4 a, b, c \in \mathbf{R}, a+b+c=1$, prove that
$$\frac{a}{1+a^{2}}+\frac{b}{1+b^{2}}+\frac{c}{1+c^{2}} \leqslant \frac{9}{10}$$ | Prove that if $a, b, c \geqslant-\frac{3}{4}$, using the tangent line method, we have
$$\frac{x}{x^{2}+1} \leqslant \frac{3}{10}+\frac{6}{25}(3 x-1), \forall x \geqslant-\frac{3}{4}$$
The proposition is proved in this case!
If $a \leqslant-\frac{3}{4}$, then by the AM-GM inequality, we have
$$\frac{c}{c^{2}+1} \leqsla... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,027 |
Example $14.5 a, b, c>0, a+b+c=1$, prove that
$$\sqrt{4 a^{2}+b c}+\sqrt{4 b^{2}+c a}+\sqrt{4 c^{2}+a b} \leqslant \frac{5}{2}$$ | Proof: Let $a \geqslant b \geqslant c$, then we have
$$\sqrt{4 a^{2}+b c} \leqslant 2 a+\frac{c}{4}, \sqrt{4 b^{2}+c a}+\sqrt{4 c^{2}+a b} \leqslant \frac{a}{2}+\frac{5 b}{2}+\frac{9 c}{4}$$
Adding them up, we get! | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,028 |
Example 14.6 (2008 Ukrainian Mathematical Olympiad) $a, b, c, d>0$, prove that
$$\frac{(a+b)(b+c)(c+d)(d+a)}{(1+a)(1+b)(1+c)(1+d)} \geqslant \frac{16 a b c d}{(1+\sqrt[4]{a b c d})^{4}}$$ | Prove that for $x, y>0$, we have
$$\frac{x+y}{(1+x)(1+y)} \geqslant \frac{2 \sqrt{x y}}{(1+\sqrt{x y})^{2}}$$
Thus,
$$\begin{array}{l}
\frac{(a+b)}{(1+a)(1+b)} \cdot \frac{(c+d)}{(1+c)(1+d)} \cdot(b+c)(d+a) \geqslant \\
\frac{4 \sqrt{a b c d}(\sqrt{a b}+\sqrt{c d})^{2}}{(1+\sqrt{a b})^{2}(1+\sqrt{c d})^{2}} \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,029 |
Example 14.8 (Vasile) \(a, b, c \in \mathbf{R}\), satisfying \(a^{2}+b^{2}+c^{2}=1\), prove that
\[
\frac{1}{3+a^{2}-2 b c}+\frac{1}{3+b^{2}-2 c a}+\frac{1}{3+c^{2}-2 a b} \leqslant \frac{9}{8}
\] | Proof
$$\Leftrightarrow \sum \frac{1}{4-(b+c)^{2}} \leqslant \frac{9}{8} \Leftrightarrow \sum \frac{(b+c)^{2}}{4-(b+c)^{2}} \leqslant \frac{3}{2}$$
By Cauchy inequality, we have
$$\begin{array}{l}
\sum \frac{(b+c)^{2}}{4-(b+c)^{2}} \leqslant \sum \frac{(b+c)^{2}}{4-2\left(b^{2}+c^{2}\right)}= \\
\frac{1}{2} \sum \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,031 |
Example 14.9 $a, b, c \geqslant 0, p \in \mathbf{R}$, prove that
$$\sum a^{3} b+p^{2} \sum a b^{3} \geqslant(p-1)^{2} a b c \sum a+2 p \sum a^{2} b^{2}$$ | Prove the inequality is equivalent to
$$\sum a b(a-p b)^{2} \geqslant(p-1)^{2} a b c \sum a$$
By Cauchy's inequality, we have
$$\begin{array}{l}
{\left[\sum a b(a-p b)^{2}\right]\left(\sum c\right) \geqslant\left[\sum \sqrt{a b c}(a-p b)\right]^{2}=} \\
(p-1)^{2} a b c\left(\sum a\right)^{2} \Rightarrow \sum a b(a-p b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,032 |
Example 14. 10 (Vasile) $a, b, c \in \mathbf{R}, p>0$, prove that
$$\sum \frac{a^{2}-b c}{2 p a^{2}+p^{2} b^{2}+c^{2}} \geqslant 0$$ | Prove that the expanded inequality is equivalent to
$$\sum(p b-c)^{2}\left[\left(p a b+c a-b^{2}-p c^{2}\right)^{2}+3 p\left(a^{2}-b c\right)^{2}\right] \geqslant 0$$
Obviously, it holds. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,033 |
Example $14.11 x, y, z \geqslant 0, x+y+z=1$, prove that
$$\sqrt{x+(y-z)^{2}}+\sqrt{y+(z-x)^{2}}+\sqrt{z+(x-y)^{2}} \geqslant \sqrt{3}$$ | Prove that after squaring and rearranging both sides of the inequality, we have $\Leftrightarrow$
$$\begin{array}{l}
2 \sum_{c y c} \sqrt{\left(x+(y-z)^{2}\right)\left(y+(x-z)^{2}\right)}+\sum_{c y c}(x-y)^{2} \geqslant 2 \Leftrightarrow \\
\sum_{c y c} \sqrt{\left(x+(y-z)^{2}\right)\left(y+(x-z)^{2}\right)} \geqslant ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,034 |
Example $14.12 a, b, c>0$ and $a b c=1$, prove that
$$\frac{12 a+7}{2 a^{2}+1}+\frac{12 b+7}{2 b^{2}+1}+\frac{12 c+7}{2 c^{2}+1} \leqslant 19$$ | $$\begin{aligned}
\Leftrightarrow & \sum_{c y}\left(9-\frac{12 a+7}{2 a^{2}+1}\right) \geqslant 8 \Leftrightarrow \sum_{c y c} \frac{2(3 a-1)^{2}}{2 a^{2}+1} \geqslant 8 \Leftrightarrow \\
& \sum_{c c} \frac{(3 a-1)^{2}}{2 a^{2}+1} \geqslant 4
\end{aligned}$$
By the generalized Cauchy inequality, we have
$$\text { LHS... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,035 |
Example $14.13 a, b, c, x, y, z \in \mathbf{R}$ and satisfy $(x+y) c-(a+b) z=\sqrt{6}$, find the minimum value of the following expression.
$$a^{2}+b^{2}+c^{2}+x^{2}+y^{2}+z^{2}+a x+b y+c z$$ | Proof: Let $a+b=2t, x+y=2s$, by Cauchy inequality, we have
$$\begin{array}{l}
a^{2}+b^{2}+x^{2}+y^{2}+a x+b y=\left(a+\frac{x}{2}\right)^{2}+\left(b+\frac{y}{2}\right)^{2}+\frac{3}{4}\left(x^{2}+y^{2}\right) \geqslant \\
\frac{1}{2}\left(a+b+\frac{x+y}{2}\right)^{2}+\frac{3}{8}(x+y)^{2}=2\left(t^{2}+t s+s^{2}\right)
\e... | 3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 732,036 |
Example 1.38 For non-negative real numbers $a, b, c$ where not two of them are zero at the same time, prove
$$\begin{array}{l}
\frac{a}{b+c}\left(a^{2}-b^{2}-c^{2}+b c\right)+\frac{b}{c+a}\left(b^{2}-c^{2}-a^{2}+c a\right)+ \\
\frac{c}{a+b}\left(c^{2}-a^{2}-b^{2}+a b\right) \geqslant 0
\end{array}$$ | Prove that the original inequality is equivalent to
$$\begin{array}{l}
\sum \frac{a}{b+c}\left[a^{2}+3 b c-(b+c)^{2}\right] \geqslant 0 \Leftrightarrow \\
\frac{a\left(a^{2}+3 b c\right)}{b+c}+\frac{b\left(b^{2}+3 c a\right)}{c+a}+\frac{c\left(c^{2}+3 a b\right)}{a+b} \geqslant 2(a b+b c+c a)
\end{array}$$
Using the C... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,037 |
Example $14.14 a, b, c, d \geqslant 0$, satisfying $a+b+c+d=4$, prove that $a \sqrt{b c}+b \sqrt{c d}+c \sqrt{d a}+d \sqrt{a b} \leqslant 2+\sqrt{a b c d}$ | Proof: Let $a=x^{2}, b=y^{2}, c=z^{2}, d=t^{2}$, our problem becomes $x^{2}+y^{2}+z^{2}+t^{2}=4$, then
$$x^{2} y z+y^{2} z t+z^{2} t x+t^{2} x y \leqslant 2+2 x y z t$$
Let $(X, Y, Z, T)$ be a permutation of $(x, y, z, t)$ such that $X \geqslant Y \geqslant Z \geqslant T$.
We have
$$\begin{array}{l}
X^{2} Y Z+Y^{2} X ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,038 |
Example 14.15 Let $a, b, c$ be non-negative real numbers such that $a+b+c=2$. Prove that
(1) $\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right) \leqslant 2$;
(2) $\left(3 a^{2}-2 a b+3 b^{2}\right)\left(3 b^{2}-2 b c+3 c^{2}\right)\left(3 c^{2}-2 c a+3 a^{2}\right) \leqslant 36$. | Proof: Without loss of generality, let $c=\min \{a, b, c\}$, and set $x=a+\frac{c}{2}, y=b+\frac{c}{2}$. Then we have
$$\begin{array}{c}
a^{2}+b^{2} \leqslant x^{2}+y^{2}, b^{2}+c^{2} \leqslant y^{2}, c^{2}+a^{2} \leqslant x^{2} \\
3 a^{2}-2 a b+3 b^{2} \leqslant 3 x^{2}-2 x y+3 y^{2} \\
3 b^{2}-2 b c+3 c^{2} \leqslant... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,039 |
For example, $14.16 a, b, c, d>0$, satisfying $a b c d=1$, prove that
$$\sum \frac{1}{1+a+a^{2}+a^{3}} \geqslant 1$$ | Prove the local inequality
$$\frac{1}{1+a+a^{2}+a^{3}}+\frac{1}{1+b+b^{2}+b^{3}} \geqslant \frac{1}{1+\sqrt{(a b)^{3}}}$$
Just add them up. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,040 |
Example $14.17 a, b, c, k>0$ and $a+b+c=2$, prove that
$$\sum\left(\frac{a+b}{a^{2}+a b+b^{2}}\right)^{k} \geqslant 2+\left(\frac{2}{3}\right)^{k}$$ | Proof Without loss of generality, let $c=\min \{a, b, c\}$, we have
$$\begin{array}{c}
\frac{a+c}{a^{2}+a c+c^{2}} \geqslant \frac{1}{a+\frac{c}{2}} \\
\frac{b+c}{b^{2}+b c+c^{2}} \geqslant \frac{1}{b+\frac{c}{2}} \\
\frac{a+b}{a^{2}+a b+b^{2}} \geqslant \frac{a+b+c}{\left(a+\frac{c}{2}\right)^{2}+\left(a+\frac{c}{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,041 |
Example $14.18 a, b, c>0, a+b+c=3$, prove that
$$\frac{3}{2} \leqslant \frac{a}{a b+1}+\frac{b}{b c+1}+\frac{c}{c a+1} \leqslant \frac{3}{a b c+1}$$ | Prove that
$$\sum\left(a-\frac{a}{a b+1}\right) \leqslant \sum \frac{a^{2} b}{2 \sqrt{a b}} \leqslant \frac{1}{2} \cdot \frac{1}{3}(a+b+c)^{2}=\frac{3}{2}$$
Then the left side of the inequality is proved.
The right side of the inequality is equivalent to
$$\sum\left(a-\frac{a^{2} b}{a b+1}\right) \leqslant \frac{3}{a ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,042 |
Example 14. Given $19 a, b, c>0$ and $a+b+c=1$, prove that
$$\frac{a^{2}+b c}{a^{2}+1}+\frac{b^{2}+a c}{b^{2}+1}+\frac{c^{2}+b a}{c^{2}+1} \leqslant \frac{13}{20}$$ | Prove that since $0 \leqslant a \leqslant 1$, using the method of local inequalities, we have
$$\begin{array}{l}
\frac{a}{a^{2}+1} \leqslant \frac{12}{25} a+\frac{4}{25} \\
\frac{1}{a^{2}+1} \leqslant 1-\frac{1}{2} a^{2}
\end{array}$$
From these two inequalities, we have
$$\begin{array}{l}
\sum \frac{a^{2}}{a^{2}+1} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,043 |
Example 14.20 Let $\triangle ABC$ be an acute triangle, prove that
$$\sum \frac{\cos ^{2} A}{\cos A+1} \geqslant \frac{1}{2}$$ | To prove that the problem is equivalent to $1 \geqslant a, b, c \geqslant 0$ and satisfies $a^{2}+b^{2}+c^{2}+2 a b c=1$ when $\frac{a^{2}}{a+1}+\frac{b^{2}}{b+1}+\frac{c^{2}}{c+1} \geqslant \frac{1}{2}$, the inequality is equivalent to
$$\begin{array}{l}
\frac{2 a^{2}}{a+1}-a^{2}+\frac{2 b^{2}}{b+1}-b^{2}+\frac{2 c^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,044 |
Example 14.21 (2008 IMO Shortlist) $a, b, c, d \in \mathbf{R}^{+}$, prove that
$$\frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+\frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geqslant 0$$ | To prove the desired inequality is equivalent to
$$\begin{array}{l}
\frac{(a-c)^{2}}{c+d+a}+\frac{(b-d)^{2}}{d+a+b}+ \\
(a-c)(b-d)\left(\frac{2 b+d}{(a+b+d)(b+c+d)}-\frac{2 a+c}{(a+b+c)(c+d+a)}\right) \geqslant 0
\end{array}$$
Thus, we only need to prove
$$\begin{array}{l}
\left(\frac{2 b+d}{(a+b+d)(b+c+d)}-\frac{2 a+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,045 |
Example 14.22 Try to find the minimum value of $M$, such that for any complex numbers $a, b, c$, we have
$$\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leqslant M\left(|a|^{2}+|b|^{2}+|c|^{2}\right)^{2}$$ | Notice
$$\begin{array}{l}
a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)= \\
(a-b)(b-c)(c-a)(a+b+c)
\end{array}$$
Thus, the original inequality is transformed into
$$|(a-b)(b-c)(c-a)(a+b+c)| \leqslant M\left(|a|^{2}+|b|^{2}+|c|^{2}\right)^{2}$$
Let $a-b=\omega, b-c=\omega^{2}, c-a... | \frac{9}{16} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,046 |
Example 14. $23 a, b, c \in[-1,1]$, satisfying $a+b+c+a b c=0$, prove that
$$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+4 a b c \geqslant 3$$ | Proof: Let $a=\frac{1-x}{1+x}, b=\frac{1-y}{1+y}, c=\frac{1-z}{1+z}$, where $x, y, z \geqslant 0$.
Then we have $x y z=1$, and the inequality becomes
$$\begin{array}{l}
\sum \frac{1}{1-\frac{1-x}{1+x}}+\frac{4(1-x)(1-y)(1-z)}{(1+x)(1+y)(1+z)} \geqslant 3 \Leftrightarrow \\
\sum \frac{1+x}{2 x}+\frac{4(1-x)(1-y)(1-z)}{(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,047 |
Example 1.39 (Cai Jianxing) For non-negative real numbers $a, b, c$ where not two of them are zero simultaneously, determine the constant $k$ such that the following inequality always holds:
$$\sqrt{S+k} \leqslant \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}} \leqslant \sqrt{S+k+1}$$
where $\frac{a}{b... | Proof: Let $a=b=c$, we have $2 \leqslant k \leqslant 3$, and let $a=0, b=c$, we have $1 \leqslant k \leqslant 2$, hence $k=2$. We first give two identities:
$$\begin{array}{c}
\sum(a+b)(a+c)=(a+b+c)^{2}+a b+b c+c a \\
\sum a(a+b)(a+c)=\left(a^{2}+b^{2}+c^{2}\right)(a+b+c)+3 a b c
\end{array}$$
For the right side of th... | 2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,048 |
Example $14.24 a, b, c>0, a+b+c=1$. Prove that
$$\frac{\sqrt{a^{2}+a b c}}{b+c a}+\frac{\sqrt{b^{2}+a b c}}{c+a b}+\frac{\sqrt{c^{2}+a b c}}{a+b c} \leqslant \frac{1}{2 \sqrt{a b c}}$$ | Prove that by Cauchy inequality,
$$\begin{array}{l}
{\left[\sum \frac{\sqrt{a^{2}+a b c}}{(b+c)(b+a)}\right]^{2} \leqslant\left[\sum \frac{a}{(a+b)(b+c)}\right]\left(\sum \frac{a+c}{b+c}\right)=} \\
\frac{\sum a^{2}+\sum a b}{(a+b)(b+c)(c+a)}\left(\sum \frac{a+c}{b+c}\right)
\end{array}$$
In fact,
$$\sum \frac{a+c}{b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,049 |
Example $14.25 a, b, c>0, k \geqslant 2$, prove that
$$\sum_{\text {cyc }} \sqrt{a^{2}+k a b+b^{2}} \leqslant \sqrt{4\left(a^{2}+b^{2}+c^{2}\right)+(3 k+2)(a b+a c+b c)}$$ | Prove that by Cauchy inequality we have
$$\begin{aligned}
\left(\sum \sqrt{a^{2}+k a b+b^{2}}\right)^{2} \leqslant & \left(\sum(a+b)\right)\left(\sum \frac{a^{2}+k a b+b^{2}}{a+b}\right)= \\
& 2\left(\sum a\right)\left(\sum \frac{a^{2}+k a b+b^{2}}{a+b}\right)
\end{aligned}$$
Therefore, it suffices to prove
$$\begin{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,050 |
Example 14.26 Let $a, b, c$ be non-negative real numbers, not all zero, prove that
$$\sqrt{\frac{a}{b+3 c}}+\sqrt{\frac{b}{c+3 a}}+\sqrt{\frac{c}{a+3 b}} \geqslant \frac{3}{2}$$ | Proof: Let
$$\frac{a}{b+3 c}=\frac{x^{2}}{4}, \frac{b}{c+3 a}=\frac{y^{2}}{4}, \frac{c}{a+3 b}=\frac{z^{2}}{4}$$
where $x, y, z \geqslant 0$.
At this point, the original inequality is equivalent to proving that when $16=7 x^{2} y^{2} z^{2}+3\left(x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2}\right)$, we have
$$x+y+z \geqslant 3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,051 |
Example 14.27 Prove that for all real numbers $a, b, c$,
$$\frac{1}{4 a^{2}-a b+4 b^{2}}+\frac{1}{4 b^{2}-b c+4 c^{2}}+\frac{1}{4 c^{2}-c a+4 a^{2}} \geqslant \frac{9}{7\left(a^{2}+b^{2}+c^{2}\right)}$$ | Prove that the complete expansion of the above expression is equivalent to
$$\begin{array}{l}
\sum_{s y m}\left(56 a^{6}+28 a^{5} b+128 a^{4} b^{2}+44 a^{3} b^{3}+\right. \\
\left.\frac{95}{2} a^{4} b c+31 a^{3} b^{2} c-\frac{45}{2} a^{2} b^{2} c^{2}\right) \geqslant 0
\end{array}$$
Using the 3rd and 6th Schur inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,052 |
Example $14.28 a, b, c \geqslant 0$, no two are zero at the same time, prove that
$$\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b} \geqslant \frac{3}{2} \sqrt[5]{\frac{a^{5}+b^{5}+c^{5}}{3}}$$ | Notice that
$$\sum \frac{a^{2}}{b+c}=\sum \frac{(b+c-a)^{2}}{b+c}$$
By Cauchy's inequality, we have
$$\sum \frac{(b+c-a)^{2}}{b+c} \geqslant \frac{\left(\sum(b+c-a)^{2}\right)^{2}}{\sum(b+c)(b+c-a)^{2}}$$
Then, the elementary inequality method can be used to complete the proof. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,053 |
Example $14.29 a, b, c \geqslant 0$, prove that
$$\sum_{c y} \frac{a}{\sqrt{a+b}} \leqslant \frac{5}{4} \sqrt{a+b+c}$$ | Prove: It is easy to prove the following lemma using the method of symmetric differentiation
$$9 \sum a^{3}+75 \sum a b^{2} \geqslant 53 \sum a^{2} b+93 a b c$$
Squaring both sides of the inequality, we get
$$\begin{array}{l}
\sum \frac{a^{2}}{a+b}+2 \sum \frac{a b}{\sqrt{(a+b)(b+c)}} \leqslant \frac{25}{16} \sum a \L... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,054 |
Let $w_{a}, w_{b}, w_{c}$ be the lengths of the three angle bisectors of $\triangle A B C$. $a, b, c$ are the three sides of $\triangle A B C$. Prove that
$$\frac{1}{w_{a} w_{b}}+\frac{1}{w_{b} w_{c}}+\frac{1}{w_{a} w_{c}} \geqslant \frac{4}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}$$ | We have the following local inequalities:
$$\begin{array}{l}
\frac{1}{w_{a} w_{b}} \geqslant \frac{a b(b+c)(c+a)(a+b)}{2(a+b+c)(a b c)^{2}} \\
\frac{1}{w_{b} w_{c}} \geqslant \frac{b c(c+a)(a+b)(b+c)}{2(a+b+c)(a b c)^{2}} \\
\frac{1}{w_{c} w_{a}} \geqslant \frac{c a(a+b)(b+c)(c+a)}{2(a+b+c)(a b c)^{2}}
\end{array}$$
T... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,055 |
Example 14.31 Positive numbers \(a, b, c\) satisfy \(abc = 1\), prove
\[3(a+b+c) \geqslant \sqrt{8a^2+1} + \sqrt{8b^2+1} + \sqrt{8c^2+1}\] | Prove without loss of generality, let $c=\min (a, b, c)$, then $c \leqslant 1$, using the Cauchy inequality
$$\sqrt{8 a^{2}+1}+\sqrt{8 b^{2}+1} \leqslant \sqrt{(a+b)\left(\frac{8 a^{2}+1}{a}+\frac{8 b^{2}+1}{b}\right)}=(a+b) \sqrt{c+8}$$
Then it suffices to prove
$$(a+b)(3-\sqrt{c+8}) \geqslant \sqrt{8 c^{2}+1}-3 c$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,056 |
Example 14.33 For $\forall n \in \mathbf{N}, a, b, c \in \mathbf{R}$, we have
$$(a+b)^{2 n}+(b+c)^{2 n}+(c+a)^{2 n} \geqslant \frac{2^{2 n}}{3^{2 n-1}+1}\left(a^{2 n}+b^{2 n}+c^{2 n}+(a+b+c)^{2 n}\right)$$ | Prove that without loss of generality, $a(a+b+c) \geqslant 0$. Let $f(a, b, c)=$ LHS - RHS, then we have
$$f(a, b, c) \geqslant f\left(a, \frac{b+c}{2}, \frac{b+c}{2}\right)$$
Then prove
$$f\left(a, \frac{b+c}{2}, \frac{b+c}{2}\right) \geqslant 0$$
The equality holds if and only if $(a, b, c) \sim(1,1,1),(a, b, c) \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,058 |
For example, $2.1 a, b, c>0$, prove that
$$\frac{63}{2}+\frac{(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)}{a b c} \geqslant \frac{27}{2} \frac{a+b+c}{\sqrt[3]{a b c}}$$ | Assume $a b c=1$, let
$$f(a, b, c)=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)+\frac{63}{2}-\frac{27}{2}(a+b+c)$$
By the AM - GM inequality, we have
$$\begin{array}{l}
f(a, b, c)-f(a, \sqrt{b c}, \sqrt{b c})= \\
b^{3}+c^{3}-2(\sqrt{b c})^{3}+b c(b+c-2 \sqrt{b c})+a\left(b^{2}+c^{2}-2 b c\right)+ \\
a^{2}(b+c-2 \sqrt{b c})-\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,059 |
Example 14.34 Let $l_{a}, l_{b}, l_{c}$ be the angle bisectors of the corresponding sides of $\triangle A B C$, and $S$ be the area. Prove that
$$l_{a} l_{b}+l_{b} l_{c}+l_{c} l_{a} \geqslant 3 \sqrt{3} S$$ | Proof: Let the sides of a triangle be $a, b, c$, and let $p=\frac{1}{2}(a+b+c), x=p-a, y=p-b$, $z=p-c$, then
$$\begin{array}{c}
l_{a}=\frac{2 \sqrt{b c p(p-a)}}{b+c}=\frac{2 \sqrt{x(x+y+z)(x+y)(x+z)}}{2 x+y+z} \\
S=\sqrt{p(p-a)(p-b)(p-c)}=\sqrt{x y z(x+y+z)}
\end{array}$$
The original inequality is equivalent to
$$\be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,060 |
Example 14.35 Non-negative real numbers $a, b, c$ are not all zero at the same time, prove
$$\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}} \geqslant \frac{a+b+c}{2}$$ | Prove that using the Cauchy inequality, we can obtain
$$\sum \frac{a^{4}}{a^{3}+b^{3}}\left[\sum a^{2}\left(a^{3}+b^{3}\right)\right] \geqslant\left(\sum a^{3}\right)^{2}$$
Then we need to prove
$$2\left(\sum a^{3}\right)^{2} \geqslant\left(\sum a\right)\left[\sum a^{2}\left(a^{3}+b^{3}\right)\right]$$
Using Vasile's... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,061 |
Example 1.5 For positive numbers $a, b, c$, prove
$$a b c(a+2)(b+2)(c+2) \leqslant\left[1+\frac{2}{3}(a b+b c+c a)\right]^{3}$$ | Prove that for the right side of the inequality, by the AM-GM inequality, we have
$$\begin{aligned}
{\left[1+\frac{2}{3}(a b+b c+c a)\right]^{3}=} & {\left[\frac{(1+a b+b c)+(1+b c+c a)+(1+c a+a b)}{3}\right]^{3} \geqslant } \\
& (1+a b+b c)(1+b c+c a)(1+c a+a b)
\end{aligned}$$
And by the Cauchy inequality, we have
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,062 |
For example, $2.2 a, b, c>0$, determine the range of $k$ such that the following inequality always holds.
$$\frac{b c+c a+a b}{a^{2}+b^{2}+c^{2}}+k \frac{(b+c)(c+a)(a+b)}{8 a b c} \geqslant 1+k$$ | Prove that we have obtained $k_{\text {min }}=\sqrt{2}$. When $k=\sqrt{2}$, we need to prove
$$\frac{b c+c a+a b}{a^{2}+b^{2}+c^{2}}+\sqrt{2} \frac{(b+c)(c+a)(a+b)}{8 a b c} \geqslant 1+\sqrt{2}$$
Let
$$f(a, b, c)=\frac{b c+c a+a b}{a^{2}+b^{2}+c^{2}}+\sqrt{2} \frac{(b+c)(c+a)(a+b)}{8 a b c}-(1+\sqrt{2})$$
Then we ne... | proof | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 732,063 |
Example 2.3 $a, b, c \geqslant 0$, prove
$$\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}} \geqslant \frac{10}{(a+b+c)^{2}}$$ | Suppose $a+b+c=1, a=\max \{a, b, c\}$, let
$$f(a, b, c)=\sum_{c y c} \frac{1}{a^{2}+b^{2}}-10$$
We need to prove $f(a, b, c) \geqslant f(a, b+c, 0)$. This is equivalent to
$$\begin{array}{l}
\left(\frac{1}{b^{2}+c^{2}}-\frac{1}{(b+c)^{2}}\right)+\left(\frac{1}{a^{2}+b^{2}}+\frac{1}{a^{2}+c^{2}}-\frac{1}{a^{2}}-\frac{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,064 |
For example, $2.4 a, b, c$ are non-negative real numbers and satisfy $a+b+c=3$, prove
$$\left(a+b^{2}\right)\left(b+c^{2}\right)\left(c+a^{2}\right) \leqslant \frac{1}{4}\left(7+\frac{4}{27}\right)^{2}$$ | Prove that
$$\begin{array}{l}
\left(a+b^{2}\right)\left(b+c^{2}\right)\left(c+a^{2}\right)-\left(a^{2}+b\right)\left(b^{2}+c\right)\left(c^{2}+a\right)= \\
(a-b)(b-c)(c-a)(a b+b c+c a-a-b-c)
\end{array}$$
Given that \(a+b+c \geqslant \sqrt{3(a b+b c+c a)}\), it is clear that when \(a \geqslant b \geqslant c\), we have... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,065 |
Example 2.5 (Hungkhtn) $a, b, c, d \geqslant 0, a+b+c+d=3$. Prove that
$$\begin{array}{l}
a b(a+2 b+3 c)+b c(b+2 c+3 d)+ \\
c d(c+2 d+3 a)+d a(d+2 a+3 b) \leqslant 6 \sqrt{3}
\end{array}$$ | Let $f(a, b, c, d)=a b(a+2 b+3 c)+b c(b+2 c+3 d)+c d(c+2 d+3 a)+d a(d+2 a+3 b)$. We have
$$\begin{array}{c}
f(a, b, c, d)-f(a+c, b, 0, d)=c(b-d)(a+c-b-d) \\
f(a, b, c, d)-f(0, b, a+c, d)=-a(b-d)(a+c-b-d)
\end{array}$$
From the above two equations, we get
$$f(a, b, c, d) \leqslant \max \{f(a+c, b, 0, d), f(0, b, a+c, d... | 6 \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 732,066 |
Example 2.6 Prove: The $n$-variable 3rd degree homogeneous symmetric inequality $F\left(x_{1}, x_{2}, \cdots, x_{n}\right) \geqslant 0$ on $\mathbf{R}_{+}^{n}\left({ }^{(1)}\right.$ holds if and only if $F(1,0,0, \cdots, 0) \geqslant 0, F(1,1,0,0, \cdots, 0) \geqslant 0, \cdots, F(1,1,1, \cdots, 1, 0) \geqslant 0, F(1,... | Proof: Let $t=\frac{x_{1}+x_{2}}{2}, x=x_{1}, y=x_{2}$, and
we have
$$\begin{array}{c}
F=a \sum_{i=1}^{n} x_{i}^{3}+b \sum_{i<j}^{n} x_{i} x_{j}\left(x_{i}+x_{j}\right)+c \sum_{i<j<k} x_{i} x_{j} x_{k} \\
A=\sum_{i=3}^{n} x_{j} ; B=\sum_{i=3}^{n} x_{j}^{2} ; C=\sum_{2<i<j} x_{i} x_{j}
\end{array}$$
$$\begin{array}{l}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,067 |
Example 2.7 Positive real numbers $a, b, c, d$ satisfy $a b c d=1, a, b, c, d \neq \frac{1}{3}$, prove
$$\frac{1}{(3 a-1)^{2}}+\frac{1}{(3 b-1)^{2}}+\frac{1}{(3 c-1)^{2}}+\frac{1}{(3 d-1)^{2}} \geqslant 1$$ | Proof: Let
$$f(a, b, c, d)=\frac{1}{(3 a-1)^{2}}+\frac{1}{(3 b-1)^{2}}+\frac{1}{(3 c-1)^{2}}+\frac{1}{(3 d-1)^{2}}$$
If $\min \{a, b, c, d\}>\frac{1}{3}$, then by the AM - GM inequality we have
$$\begin{aligned}
f(a, b, c, d)-f(\sqrt{a d}, b, c, \sqrt{a d}) \geqslant & \frac{2}{(3 a-1)(3 d-1)}-\frac{2}{(3 \sqrt{a d}-1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,068 |
Example $2.8 x_{1}, x_{2}, \cdots, x_{n}>0$, satisfy $x_{1}+x_{2}+\cdots+x_{n}=n$, then $x_{1} x_{2} \cdots x_{n}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}-n+3\right) \leqslant 3$ | Proof: Let $t=x_{1}+x_{2}-1$, where $x_{1}, x_{2}$ are the maximum and minimum numbers of $x_{1}, \cdots, x_{n}$, respectively. We will prove this using mathematical induction.
When $n=2$, it is obviously true. If it is true for $n-1$.
Let $f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=x_{1} x_{2} \cdots x_{n}\left(\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 732,069 |
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