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Example 5.1.7 Let $-1<x_{1}<x_{2}<\cdots<x_{n}<1$ and $y_{1}<y_{2}<\cdots<y_{n}$ be real numbers, and satisfy $x_{1}+x_{2}+\cdots+x_{n}=x_{1}^{13}+x_{2}^{13}+\cdots+x_{n}^{13}$, prove that $x_{1}^{13} y_{1}+x_{2}^{13} y_{2}+\cdots+x_{n}^{13} y_{n}<x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{n} y_{n} \quad$ (Russia MO 2000) | Proof: According to Abel's formula, we notice that
$$\begin{array}{l}
\sum_{i=1}^{n} y_{i}\left(x_{i}^{13}-x_{i}\right)=\left(y_{1}-y_{2}\right)\left(x_{1}^{13}-x_{1}\right)+\left(y_{2}-y_{3}\right)\left(x_{1}^{13}+x_{2}^{13}-x_{1}-x_{2}\right)+\cdots \\
+\left(y_{n-1}-y_{n}\right)\left(\sum_{i=1}^{n-1} x_{i}^{13}-\sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,472 |
Example 1.1.10 Let $a, b, c>0, abc=1$, prove: $\sqrt{\frac{a+b}{a+1}}+\sqrt{\frac{b+c}{b+1}}+\sqrt{\frac{c+a}{c+1}} \geq 3$ | Proof: By the AM-GM inequality, we have
$$L H S \geq 3 \sqrt[3]{\sqrt{\frac{a+b}{a+1}} \cdot \sqrt{\frac{b+c}{b+1}} \cdot \sqrt{\frac{c+a}{c+1}}}=3 \sqrt[6]{\frac{(a+b)(b+c)(c+a)}{(a+1)(b+1)(c+1)}}$$
Thus, we only need to prove
$$(a+b)(b+c)(c+a) \geq(a+1)(b+1)(c+1)$$
Since $a b c=1$, $(*)$ is equivalent to
$$a b(a+b)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,473 |
Theorem 9 (Rearrangement Inequality) Let $\left(a_{1}, a_{2}, \cdots, a_{n}\right),\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ be two increasing sequences of real numbers. Suppose $\left(i_{1}, i_{2}, \cdots, i_{n}\right)$ is any permutation of $(1,2, \cdots, n)$, then
$$a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} \geq a_... | Notice that $a_{1} \leq a_{2} \leq \cdots \leq a_{n}$ and $b_{1} \leq b_{2} \leq \cdots \leq b_{n}$, so according to Abel's formula,
$$\begin{array}{l}
\sum_{k=1}^{n} a_{k} b_{k}-\sum_{k=1}^{n} a_{k} b_{i_{k}}=\sum_{k=1}^{n} a_{k}\left(b_{k}-b_{i_{k}}\right)=\left(a_{1}-a_{2}\right)\left(b_{1}-b_{i_{1}}\right)+\left(a_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,474 |
Example 5.2.1 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, prove: $a_{1}+a_{2}+\cdots+a_{n} \geq n \sqrt[n]{a_{1} a_{2} \cdots a_{n}}$ | Proof: Without loss of generality, assume $a_{1} a_{2} \cdots a_{n}=1$ (normalization). Let $a_{1}=\frac{x_{1}}{x_{2}}, a_{2}=\frac{x_{2}}{x_{3}}, \cdots, a_{n-1}=\frac{x_{n-1}}{x_{n}}$, where $x_{1}, x_{2}, \cdots, x_{n}>0$, then $a_{n}=\frac{x_{n}}{x_{1}}$. The inequality becomes
$$\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,475 |
Example 5.2.2 Let $a, b, c$ be the lengths of the sides of a triangle, prove that
$$a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geq 0$$ | Proof: Since $a, b, c$ are the three sides of a triangle, $a \geq b$ implies $a^{2}+b c \geq b^{2}+c a$. Based on this property, if $a \geq b \geq c$, then $a^{2}+b c \geq b^{2}+c a \geq c^{2}+a b$; additionally, $\frac{1}{a} \leq \frac{1}{b} \leq \frac{1}{c}$. According to the rearrangement inequality, we have
$$\sum_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,476 |
Example 5.2.3 Let $a, b, c>0$, prove: $\frac{a^{2}+b c}{b+c}+\frac{b^{2}+c a}{c+a}+\frac{c^{2}+a b}{a+b} \geq a+b+c$ | Prove: For the sequences $\left(a^{2}, b^{2}, c^{2}\right)$ and $\left(\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}\right)$ (if $a \geq b \geq c$, then both sequences are increasing) applying the rearrangement inequality, we get
$$\sum_{c y c} \frac{a^{2}}{b+c} \geq \sum_{c y c} \frac{b^{2}}{b+c}$$
This implies
$$\sum_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,477 |
Example 5.2.4 Let $a, b, c>0$, prove:
$$\frac{a+b}{a+c}+\frac{a+c}{b+c}+\frac{b+c}{a+b} \leq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$$ | Prove: The inequality is equivalent to
$$\sum_{c y c}\left(\frac{a}{b}-\frac{a}{b+c}\right) \geq \sum_{c y c} \frac{a}{a+c} \Leftrightarrow \sum_{c y c} \frac{a c}{b(b+c)} \geq \sum_{c y c} \frac{a}{a+c}$$
Consider the expressions \( P=\sum_{c y c} \frac{a c}{b(b+c)}, Q=\sum_{c y c} \frac{b c}{a(b+c)} \). By the rearr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,478 |
Example 5.2.5 Let $a, b, c>0$, prove:
$$\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b} \leq \frac{(a+b+c)^{2}}{a b+b c+c a}$$ | Prove: The inequality is equivalent to
$$\begin{array}{l}
\sum_{c y c} \frac{(a+b)[a(b+c)+b c]}{b+c} \leq(a+b+c)^{2} \Leftrightarrow \sum_{c y c} a(a+b)+\sum_{c y c} \frac{b c(a+b)}{b+c} \leq(a+b+c)^{2} \\
\Leftrightarrow \sum_{c y c}\left(\frac{b c}{b+c}\right)(a+b) \leq a b+b c+c a
\end{array}$$
The last inequality ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,479 |
Example 5.2.6 Let $a, b, c, d \geq 0$ and satisfy $a+b+c+d=4$, prove that $a^{2} b c+b^{2} c d+c^{2} d a+d^{2} a b \leq 4$ (Song Yoon Kim) | Proof: Suppose $(x, y, z, t)$ is a permutation of $(a, b, c, d)$, satisfying $x \geq y \geq z \geq t$, then $x y z \geq x y t \geq x z t \geq y z t$. By the rearrangement inequality, we infer
$$x \cdot x y z+y \cdot x y t+z \cdot x z t+t \cdot y z t \geq a^{2} b c+b^{2} c d+c^{2} d a+d^{2} a b$$
According to the AM-GM... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,480 |
Example 5.2.7 Let $a, b, c, d>0$, prove
$$\left(\frac{a}{a+b+c}\right)^{2}+\left(\frac{b}{b+c+d}\right)^{2}+\left(\frac{c}{c+d+a}\right)^{2}+\left(\frac{d}{d+a+b}\right)^{2} \geq \frac{4}{9}$$ | Proof: Without loss of generality, we can assume $a+b+c+d=1$. Let $(x, y, z, t)$ be a permutation of $(a, b, c, d)$ such that $x \geq y \geq z \geq t$, then
$$\frac{1}{x+y+z} \geq \frac{1}{x+y+t} \geq \frac{1}{x+z+t} \geq \frac{1}{y+z+t}.$$
By the rearrangement inequality, we deduce
$$\begin{array}{l}
\sum_{c y c}\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,481 |
Example 6.0.8 Let $x, y, z>0$, and satisfy $x y+y z+z x=1$. Prove:
$$10 x^{2}+10 y^{2}+z^{2} \geq 4$$ | To prove: Before we provide a general and natural solution, let's first look at a beautiful, short, and somewhat magical solution. By the AM-GM inequality, we have
$$2 x^{2}+2 y^{2} \geq 4 x y, \quad 8 x^{2}+\frac{1}{2} z^{2} \geq 4 x z, \quad 8 y^{2}+\frac{1}{2} z^{2} \geq 4 y z$$
Adding these inequalities, we get
$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,482 |
Example 6.1.1 Let $k$ be a positive real number. Find the minimum value of the following expression
$$k\left(x^{2}+y^{2}\right)+z^{2}$$
where $x, y, z>0$, and satisfy $x y+y z+z x=1$ | Solution: We decompose $k=l+(k-l)(0 \leq l \leq k)$, and apply the AM-GM inequality, obtaining the following inequalities:
$$\begin{array}{l}
l x^{2}+l y^{2} \geq 2 l x y \\
(k-l) x^{2}+\frac{1}{2} z^{2} \geq \sqrt{2(k-l)} x z \\
(k-l) y^{2}+\frac{1}{2} z^{2} \geq \sqrt{2(k-l)} y z
\end{array}$$
Combining these result... | \frac{-1+\sqrt{1+8 k}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,483 |
Example 1.1.11 Let $a, b, c$ be the lengths of the sides of a triangle, prove:
$$(a+b-c)^{a}(b+c-a)^{b}(c+a-b)^{c} \leq a^{a} b^{b} c^{c}$$ | Proof: By the weighted AM-GM inequality, we have
$$\begin{array}{l}
\left(\frac{a+b-c}{a}\right)^{\frac{a}{a+b+c}} \cdot\left(\frac{b+c-a}{b}\right)^{\frac{b}{a+b+c}} \cdot\left(\frac{c+a-b}{c}\right)^{\frac{c}{a+b+c}} \\
\leq \frac{a}{a+b+c} \cdot \frac{a+b-c}{a}+\frac{b}{a+b+c} \cdot \frac{b+c-a}{b}+\frac{c}{a+b+c} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,484 |
Example 6.1.2 Let real numbers $x, y, z, t$ satisfy the condition $x y + y z + z t + t x = 1$. Find the minimum value of the following expression:
$$5 x^{2} + 4 y^{2} + 5 z^{2} + t^{2}$$ | Solution: We choose a positive number $l<5$, and apply the AM-GM inequality, we have the following inequalities:
$$\begin{array}{l}
l x^{2}+2 y^{2} \geq 2 \sqrt{2 l} x y \\
2 y^{2}+l z^{2} \geq 2 \sqrt{2} y z \\
(5-l) z^{2}+\frac{1}{2} t^{2} \geq \sqrt{2(5-l)} z t \\
\frac{1}{2} t^{2}+(5-l) x^{2} \geq \sqrt{2(5-l)} t x... | 2 \sqrt{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,485 |
Example 6.1.3 Let $x, y, z>0$, and satisfy $x+y+z=3$, find the minimum value of the following expression
$x^{2}+y^{2}+z^{3} \quad$ (Pham Kim Hung) | Solution: Let $a$ and $b$ be two positive real numbers. By the AM-GM inequality, we have
$$\begin{array}{l}
x^{2}+a^{2} \geq 2 a x \\
y^{2}+a^{2} \geq 2 a y \\
z^{3}+b^{3}+b^{3} \geq 3 b^{2} z
\end{array}$$
Combining these inequalities, we get $x^{2}+y^{2}+z^{3}+2\left(a^{2}+b^{3}\right) \geq 2 a(x+y)+3 b^{2} z$, with... | 6 a-\left(2 a^{2}+b^{3}\right) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,486 |
Example 6.1.4 Let $a, b, c$ be three positive constants, and $x, y, z$ be three positive variables, satisfying $a x+b y+c z=x y z$. Prove: If there exists a unique positive number $d$ such that $\frac{2}{d}=\frac{1}{a+d}+\frac{1}{b+d}+\frac{1}{c+d}$, then the minimum value of the expression $x+y+z$ is
$$\sqrt{d(d+a)(d+... | To avoid the complex condition $a x+b y+c z=x y z$, we will find the minimum value of the following homogeneous expression:
$$\frac{(a x+b y+c z)(x+y+z)^{2}}{x y z}$$
Of course, if the minimum value of the above expression is $\mathbf{C}$, then the minimum value of the expression $x+y+z$ is also equal to C.
Assume $m... | \sqrt{d(d+a)(d+b)(d+c)} | Algebra | proof | Yes | Yes | inequalities | false | 731,487 |
Example 6.1.5 Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers, prove that
$$x_{1}+\sqrt{x_{1} x_{2}}+\cdots+\sqrt[n]{x_{1} x_{2} \cdots x_{n}} \leq e\left(x_{1}+x_{2}+\cdots+x_{n}\right)$$ | Proof: Suppose $a_{1}, a_{2}, \cdots, a_{n}$ are positive real numbers. According to the AM-GM inequality, we have
$$\begin{array}{l}
\sqrt[k]{\left(a_{1} x_{1}\right)\left(a_{2} x_{2}\right) \cdots\left(a_{k} x_{k}\right)} \leq \frac{a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{k} x_{k}}{k} \\
\Rightarrow \sqrt[k]{x_{1} x_{2} \c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,488 |
Example 6.2.1 Let $x, y, z$ be three positive real numbers, and satisfy $x+y+z=3$. Find the minimum value of the following expression:
$$x^{4}+2 y^{4}+3 z^{4}$$ | Solution: Let $a, b, c$ be three positive real numbers, and satisfy $a+b+c=3$. According to Holder's inequality, we get
$$\left(x^{4}+2 y^{4}+3 z^{4}\right)\left(a^{4}+2 b^{4}+3 c^{4}\right)^{3} \geq\left(a^{3} x+2 b^{3} y+3 c^{3} z\right)^{4}$$
We choose $a, b, c$ such that $a^{3}=2 b^{3}=3 c^{3}=k^{3}$. Then we have... | null | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,489 |
Example 6.2.2 Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, prove:
$$\frac{1}{a_{1}}+\frac{2}{a_{1}+a_{2}}+\cdots+\frac{n}{a_{1}+a_{2}+\cdots+a_{n}}<2\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)$$ | Proof: Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers (to be determined). According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(a_{1}+a_{2}+\cdots+a_{k}\right)\left(\frac{x_{1}^{2}}{a_{1}}+\frac{x_{2}^{2}}{a_{2}}+\cdots+\frac{x_{k}^{2}}{a_{k}}\right) \geq\left(x_{1}+x_{2}+\cdots+x_{k}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,490 |
Example 6.2.3 Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive real numbers. Prove that
$$x_{1}^{2}+\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\cdots+\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^{2} \leq 4\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)$$ | Proof: Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers (to be determined). According to the Cauchy-Schwarz inequality, we have
$$\left(\frac{x_{1}^{2}}{a_{1}}+\frac{x_{2}^{2}}{a_{2}}+\cdots+\frac{x_{k}^{2}}{a_{k}}\right)\left(a_{1}+a_{2}+\cdots+a_{k}\right) \geq\left(x_{1}+x_{2}+\cdots+x_{k}\right)^{2}$$
Th... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,491 |
Example 6.2.4 Find the smallest $t$ such that the following inequality holds for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$.
$$x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \leq t\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \quad \text { (MYM 2004) }$$ | Solution: Let $c_{1}, c_{2}, \cdots, c_{n}$ be positive real numbers (to be chosen later). According to the Cauchy-Schwarz inequality, we have
\[
\left(\sum_{i=1}^{k} x_{i}\right)^{2} \leq S_{k}\left(\sum_{i=1}^{k} \frac{x_{i}^{2}}{c_{i}}\right)
\]
(*)
Here, $S_{1}, S_{2}, \cdots, S_{n}$ are determined by
\[
S_{k}=\su... | \frac{1}{4 \sin ^{2} \frac{\pi}{2(2 n+1)}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,492 |
Example 7.1.1 If $x$ is a positive number, find the minimum value of the expression $x^{x}$.
The above text is translated into English, preserving the original text's line breaks and format. | Consider the function $f(x)=x^{x}=e^{x \ln x}$. Its derivative is $f^{\prime}(x)=e^{x \ln x}(\ln x+1)$. Clearly, $f^{\prime}(x)=0 \Leftrightarrow \ln x=-1 \Leftrightarrow x=\frac{1}{e}$. If $x \in\left(0, \frac{1}{e}\right]$, then $f(x)$ is decreasing; if $x \in\left[\frac{1}{e},+\infty\right)$, then $f(x)$ is increasi... | \frac{1}{e^{\frac{1}{e}}} | Calculus | math-word-problem | Yes | Yes | inequalities | false | 731,493 |
Example 7.1.2 Let $a, b, c$ be positive real numbers, prove that
$$\frac{a^{3}}{b^{3}+c^{3}}+\frac{b^{3}}{c^{3}+a^{3}}+\frac{c^{3}}{a^{3}+b^{3}} \geq \frac{a^{2}}{b^{2}+c^{2}}+\frac{b^{2}}{c^{2}+a^{2}}+\frac{c^{2}}{a^{2}+b^{2}}$$ | Proof: Let's prove a general problem. Suppose real numbers $s \geq t \geq 0$, then
$$\frac{a^{s}}{b^{s}+c^{s}}+\frac{b^{s}}{c^{s}+a^{s}}+\frac{c^{s}}{a^{s}+b^{s}} \geq \frac{a^{t}}{b^{t}+c^{t}}+\frac{b^{t}}{c^{t}+a^{t}}+\frac{c^{t}}{a^{t}+b^{t}}$$
Thus, it suffices to prove that the following function is increasing:
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,494 |
Example 1.1.12 Let $a, b, c \geq 0, a+b+c=2$, prove: $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \leq 2$ | Prove: From the identity
$$(a b+b c+c a)\left(a^{2}+b^{2}+c^{2}\right)=a b\left(a^{2}+b^{2}\right)+b c\left(b^{2}+c^{2}\right)+c a\left(c^{2}+a^{2}\right)+a b c(a+b+c)$$
we have
$$\begin{array}{l}
(a b+b c+c a)\left(a^{2}+b^{2}+c^{2}\right) \geq a b\left(a^{2}+b^{2}\right)+b c\left(b^{2}+c^{2}\right)+c a\left(c^{2}+a^... | a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \leq 2 | Inequalities | proof | Yes | Yes | inequalities | false | 731,495 |
Example 7.1.3 Let $a, b, c, d$ be positive real numbers, prove that
$$\sqrt{\frac{a b+a c+a d+b c+b d+c d}{6}} \geq \sqrt[3]{\frac{a b c+b c d+c d a+d a b}{4}}$$ | Proof: Consider the function $f(x)=(x-a)(x-b)(x-c)(x-d)$
Here
$$A=\sum_{sym} a, B=\sum_{sym} ab, C=\sum_{sym} abc, D=abcd$$
Since the equation $f(x)=0$ has 4 positive real roots, we have (by Rolle's Theorem) that the equation $f'(x)=0$ has 3 positive real roots, denoted as $m, n, p > 0$, then
$$f'(x)=4(x-m)(x-n)(x-p)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,496 |
Example 7.1.4 If real numbers $a, b, c$ satisfy the conditions $a \leq b \leq c, a+b+c=6, ab+bc+ca=9$, prove: $0 \leq a \leq 1 \leq b \leq 3 \leq c \leq 4$ (British MO) | Let $p=a b c$, and consider the function
$$f(x)=(x-a)(x-b)(x-c)=x^{3}-6 x^{2}+9 x-p$$
We have $f^{\prime}(x)=3 x^{2}-12 x+9=3(x-1)(x-3)$. Therefore, $f^{\prime}(x)=0 \Rightarrow x=1,3$
Since $f(x)$ has three roots $a \leq b \leq c$, we have
$$1 \leq b \leq 3, \quad f(1) f(3) \leq 0$$
Notice that $f(1)=f(4)=4-p$ and ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 731,497 |
Example 7.2.1 Let $a, b, c$ be positive real numbers, prove that
$$a^{3}+b^{3}+c^{3}+3 a b c \geq a b(a+b)+b c(b+c)+c a(c+a)$$ | Proof: Without loss of generality, assume $a \geq b \geq c$. Consider the function of $a$:
$$f(a)=a^{3}+b^{3}+c^{3}+3 a b c-a b(a+b)-b c(b+c)-c a(c+a)$$
We have
$$f^{\prime}(a)=3 a^{2}+3 b c-2 a b-b^{2}-2 a c-c^{2}$$
Notice that $f^{\prime \prime}(a)=6 a-2 b-2 c \geq 0$ and $f^{\prime \prime}(b) \geq 0$, so $f^{\prim... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,498 |
Example 7.2.2 Let \(a, b, c, d\) be positive real numbers, and satisfy
\[2(ab + bc + cd + ac + bd) + abc + bcd + cda + dab = 16\]
Prove the following inequality
\[\left.a + b + c + d \geq \frac{2}{3}(ab + bc + cd + da + ac + bd) \text{ (Viennam M0 } 1996\right)\] | Proof: For similar reasons as in Example 7.1.3, we infer that there exist three positive numbers \( x, y, z \) satisfying the following conditions:
\[
\sum_{\text{sym}} a = \frac{4}{3} \sum_{\text{sym}} x, \quad \sum_{\text{sym}} ab = 2 \sum_{\text{sym}} xy, \quad \sum_{\text{sym}} abc = 4xyz
\]
The remaining part is ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,499 |
Example 7.2.3 Let $a, b, c \geq 0$, prove that
$a^{3}+b^{3}+c^{3}+9 a b c+4(a+b+c) \leq 8(a b+b c+c a)$
(Le Trung Kien) | Let
$$f(b)=b^{3}+b(4+9 a c-8 a-8 c)+a^{3}+c^{3}+4(a+c)-8 a c$$
By the AM-GM inequality, $\left(a^{3}+4 a\right)+\left(c^{3}+4 c\right) \geq 4 a^{2}+4 c^{2} \geq 8 a c$, so the problem is reduced to proving $4+9 a c \geq 8(a+c)$, with equality holding if $a=c=2, b=0$ or $a=b=c=0$. Otherwise, let $x=a+c, y=a c$, then $8... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,500 |
Example 7.2.4 Assume that $n \geq 2$ is a positive integer, and $n$ positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy the condition
$$\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}\right) \leq(n+\sqrt{10}-3)^{2}$$
Prove: For every triplet $\left(x_{i}, x_{j}... | To prove: It is sufficient to prove the following result (which can directly solve the problem)
Assume $x_{1} \geq x_{2} \geq \cdots \geq x_{n}>0$, and satisfy $x_{1}>x_{2}+x_{3}$, then
$$\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots \frac{1}{x_{n}}\right)>(n+\sqrt{10}-3)^{2}$$
In ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,501 |
Example 7.2.5 Let $a, b, c>0$, and satisfy $12 \geq 21 a b+2 b c+8 c a$, prove: $\frac{1}{a}+\frac{2}{b}+\frac{3}{c} \leq \frac{15}{2}$ | To prove: Although this problem has already been proven using the balancing coefficient method, we will provide a proof using derivatives here. Let $x=\frac{1}{a}, y=\frac{2}{b}, z=\frac{3}{c}$. We will prove an equivalent problem:
If $x, y, z > 0$ and $12xyz \geq 2x + 8y + 21z$, then $P(x, y, z) = x + 2y + 3z \leq \f... | \frac{15}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,502 |
Example 7.2.6 Let $a, b, c$ be three positive real numbers, and satisfy $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16$, find the minimum and maximum values of the following expression.
$$P=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \text { (Pham Kim Hung) }$$ | First, we find the minimum value. For this, assume $a \geq b \geq c$, so we have
$$\sum_{c y c} \frac{a}{b}-\sum_{c y c} \frac{b}{a}=\frac{(a-b)(a-c)(c-b)}{a b c} \leq 0$$
Let $x=\frac{a}{b} \geq 1, y=\frac{b}{c} \geq 1$, then the given condition becomes $x+y+\frac{1}{x y}+\frac{1}{x}+\frac{1}{y}+x y=13$
Let $x+y=s, ... | \frac{13-\sqrt{5}}{2} \text{ and } \frac{13+\sqrt{5}}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,503 |
Theorem 10 (Schur's Inequality) Let $a, b, c \geq 0$, then $a^{3}+b^{3}+c^{3}+3 a b c \geq a b(a+b)+b c(b+c)+c a(c+a)$ | Proof: Due to symmetry, we can assume $a \geq b \geq c$. Let $x=a-b, y=b-c$, then the inequality becomes the following form: $\square$
$$\begin{array}{l}
\sum_{c y c} a(a-b)(a-c) \geq 0 \Leftrightarrow c(x+y) y-(c+y) x y+(c+x+y) x(x+y) \geq 0 \\
\Leftrightarrow c\left(x^{2}+x y+y^{2}\right)+x^{2}(x+2 y) \geq 0
\end{arr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,504 |
Let $a, b, c$ be three real numbers, prove that
$$a^{6}+b^{6}+c^{6}+a^{2} b^{2} c^{2} \geq \frac{2}{3}\left[a^{5}(b+c)+b^{5}(c+a)+c^{5}(a+b)\right]$$ | Prove: According to the AM-GM inequality and Schur's inequality, we have
$$3 \sum_{c y c} a^{6}+3 a^{2} b^{2} c^{2} \geq 2 \sum_{c y c} a^{6}+\sum_{c y c} a^{4}\left(b^{2}+c^{2}\right)=\sum_{c y c}\left(a^{6}+a^{4} b^{2}\right)+\sum_{c y c}\left(a^{6}+a^{4} c^{2}\right) \geq 2 \sum_{c y c} a^{5}(b+c)$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,505 |
Example 1.1.13 Let $a, b, c, d>0$, prove: $\frac{1}{a^{2}+a b}+\frac{1}{b^{2}+b c}+\frac{1}{c^{2}+c d}+\frac{1}{d^{2}+d a} \geq \frac{4}{a c+b d}$ | Notice
$$\frac{a c+b d}{a^{2}+a b}=\frac{a^{2}+a b+a c+b d}{a^{2}+a b}-1=\frac{a(a+c)+b(d+a)}{a(a+b)}-1=\frac{a+c}{a+b}+\frac{b(d+a)}{a(a+b)}-1$$
According to the AM-GM inequality, we get
$$(a c+b d) \sum_{c y c} \frac{1}{a^{2}+a b}=\sum_{c y c} \frac{a+c}{a+b}+\sum_{c y c} \frac{b(d+a)}{a(a+b)}-4 \geq \sum_{c y c} \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,506 |
Theorem 11 (General Schur's Inequality) Let $a, b, c \geq 0, \mathrm{k}$ be a positive constant, then
$$a^{k}(a-b)(a-c)+b^{k}(b-a)(b-c)+c^{k}(c-a)(c-b) \geq 0$$ | Proof: Of course, we can assume $a \geq b \geq c$, in which case, we have
$$\begin{array}{c}
c^{k}(c-a)(c-b) \geq 0 \\
a^{k}(a-b)(a-c)+b^{k}(b-a)(b-c)=(a-b)\left[\left(a^{k+1}-b^{k+1}\right)+c\left(a^{k}-b^{k}\right)\right] \geq 0
\end{array}$$
Adding the two inequalities, we get the result. The equality holds when $a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,507 |
Example 8.1.1 Let $a, b, c \geq 0$, and $a+b+c=2$, prove: $a^{4}+b^{4}+c^{4}+a b c \geq a^{3}+b^{3}+c^{3}$ | Prove: According to the 4th Schur's Inequality, we have
$$a^{4}+b^{4}+c^{4}+a b c(a+b+c) \geq\left(a^{3}+b^{3}+c^{3}\right)(a+b+c)$$
Substituting $a+b+c=2$ into the final inequality, we get the desired result. The equality holds if $a=b=c=\frac{2}{3}$ or $a=b=1, c=0$ and their permutations. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,508 |
Example 8.1.2 Let $a, b, c>0$, prove:
$\frac{a^{2}}{\sqrt{(b+c)\left(b^{3}+c^{3}\right)}}+\frac{b^{2}}{\sqrt{(c+a)\left(c^{3}+a^{3}\right)}}+\frac{c^{2}}{\sqrt{(a+b)\left(a^{3}+b^{3}\right)}} \geq \frac{3}{2}$ (Pham Kim Hung) | Proof: By Holder's inequality, we have
$$\left(\sum_{c y c} \frac{a^{2}}{\sqrt{(b+c)\left(b^{3}+c^{3}\right)}}\right)^{2}\left(\sum_{c y c} a^{2}(b+c)\left(b^{3}+c^{3}\right)\right) \geq\left(\sum_{c y c} a^{2}\right)^{3}$$
Therefore, it suffices to prove
$$\begin{array}{l}
4\left(\sum_{c y c} a^{2}\right)^{3} \geq 9 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,509 |
Example 8.1.3 Let $a, b, c \geq 0$, prove:
$\frac{a^{2}}{2 b^{2}-b c+2 c^{2}}+\frac{b^{2}}{2 c^{2}-c a+2 a^{2}}+\frac{c^{2}}{2 a^{2}-a b+2 b^{2}} \geq 1$ (Vasile Cirtoaje) | Proof: According to the Cauchy-Schwarz inequality, we have
$$\sum_{c y c} \frac{a^{2}}{2 b^{2}-b c+2 c^{2}} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a^{2}\left(2 b^{2}-b c+2 c^{2}\right)+b^{2}\left(2 c^{2}-c a+2 a^{2}\right)+c^{2}\left(2 a^{2}-a b+2 b^{2}\right)}$$
Thus, it suffices to prove
$$\left(\sum_{c y c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,510 |
Example 8.1.4 Let $a, b, c \geq 0$, prove: $\frac{a^{3}}{b^{2}-b c+c^{2}}+\frac{b^{3}}{c^{2}-c a+a^{2}}+\frac{c^{3}}{a^{2}-a b+b^{2}} \geq a+b+c$ | Prove: By applying the Cauchy-Schwarz inequality, we have
\[
\sum_{c y c} \frac{a^{3}}{b^{2}-b c+c^{2}}=\sum_{c y c} \frac{a^{4}}{a\left(b^{2}-b c+c^{2}\right)} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\sum_{c y c} a\left(b^{2}-b c+c^{2}\right)},
\]
thus it suffices to prove
\[
\left(\sum_{c y c} a^{2}\right)^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,511 |
Example 8.1.5 Let $a, b, c \geq 0$, prove:
$$a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}$$ | Proof: According to the AM-GM inequality, we have
$$\sum_{c y c} a^{2} \sqrt{b^{2}-b c+c^{2}}=\sum_{c y c} a \sqrt{a^{2}\left(b^{2}-b c+c^{2}\right)} \leq \frac{1}{2} \sum_{c y c} a\left(a^{2}+b^{2}+c^{2}-b c\right)$$
Thus, by the third-degree Schur inequality, we have
$$2 \sum_{c y c} a^{3}-\sum_{c y c} a\left(a^{2}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,512 |
Example 8.1.6 Let $a, b, c \geq 0$, prove:
$\frac{a^{3}}{\sqrt{b^{2}-b c+c^{2}}}+\frac{b^{3}}{\sqrt{c^{2}-c a+a^{2}}}+\frac{c^{3}}{\sqrt{a^{2}-a b+b^{2}}} \geq a^{2}+b^{2}+c^{2}$ (Vo Quoc Ba Can) | Prove: By applying the Cauchy-Schwarz inequality, we have
$$\sum_{c y c} \frac{a^{3}}{\sqrt{b^{2}-b c+c^{2}}} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{\sum_{c y c} a \sqrt{b^{2}-b c+c^{2}}}$$
Therefore, it suffices to prove
$$\sum_{c y c} a \sqrt{b^{2}-b c+c^{2}} \leq a^{2}+b^{2}+c^{2}$$
Again, by the Cauchy-Sc... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,513 |
Example 8.2.1 Let $a, b, c>0$, and $a+b+c=2$, prove: $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+a b c \leq 1$ (Pham Kim Hung) | Proof: Since $a+b+c=2$, the inequality is equivalent to
$$(a b+b c+c a)^{2} \leq 1+3 a b c$$
Let $x=a b+b c+c a, y=a b c$. If $x \leq 1$, the inequality obviously holds; otherwise, $x \geq 1$, by the AM-GM inequality, we have
$$\prod_{c y c}(a+b-c) \leq a b c \Rightarrow 8 \prod_{c y c}(1-a) \leq a b c \Rightarrow 8+9... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,514 |
Example 8.2.2 Let $a, b, c \geq 0$, and $a^{2}+b^{2}+c^{2}=1$, prove: $a+b+c \leq \sqrt{2}+\frac{9 a b c}{4}$ | Proof: Let $x=a+b+c, y=ab+bc+ca, z=abc$. By the 4th degree Schur's inequality, we have
$$\begin{array}{l}
\sum_{cyc} a^{4} + abc \sum_{cyc} a \geq \sum_{cyc} a^{3}(b+c) \Leftrightarrow \left(\sum_{cyc} a^{2}\right)^{2} - 2\left(\sum_{cyc} ab\right)^{2} + 6abc\left(\sum_{cyc} a\right) \geq \left(\sum_{cyc} a^{2}\right)\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,515 |
Example 8.2.3 Let $a, b, c>0$, and satisfy $abc=1$, prove:
$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leq \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} \quad$ (Bulgarian MO 1998) | Proof: Let $S=\sum_{cyc} a, P=\sum_{cyc} ab, Q=abc$, through not too complicated calculations, we have
$$LHS=\sum_{cyc} \frac{1}{S+1-a}=\frac{S^{2}+4S+3+P}{S^{2}+2S+PS+P}, \quad RHS=\sum_{cyc} \frac{1}{2+a}=\frac{12+4S+P}{9+4S+2P}$$
Therefore, we only need to prove
$$\frac{S^{2}+4S+3+P}{S^{2}+2S+PS+P} \leq \frac{12+4S... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,516 |
Example 1.1.14 Let $a, b, c, d, e \geq 0$, and $a+b+c+d+e=5$, prove:
$$a b c+b c d+c d e+d e a+e a b \leq 5$$ | Prove: Without loss of generality, let $e=\min (a, b, c, d, e)$.
According to the AM-GM inequality, we have
$$\begin{array}{l}
a b c+b c d+c d e+d e a+e a b=e(a+c)(b+d)+b c(a+d-e) \\
\leq e\left(\frac{a+b+c+d}{2}\right)^{2}+\left(\frac{b+c+a+d-e}{3}\right)^{3}=\frac{e(5-e)^{2}}{4}+\frac{(5-2 e)^{3}}{27}
\end{array}$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,517 |
Example 8.2.4 Let $a, b, c>0$, prove:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a b c}{2\left(a^{3}+b^{3}+c^{3}\right)} \geq \frac{5}{3}$ (Pham Kim Hung) | Proof: Without loss of generality, we assume $a+b+c=3$, and let $x=ab+bc+ca, y=abc$, then we have
$$\frac{abc}{a^{3}+b^{3}+c^{3}}=\frac{y}{27+3y-9x}, \quad \sum_{cyc} \frac{a}{b+c}=\frac{27+3y-6x}{3x-y}$$
We only need to prove
$$\frac{27+3y-6x}{3x-y}+\frac{y}{2(27+3y-9x)} \geq \frac{5}{3}$$
By the AM-GM inequality, $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,518 |
Example 8.2.5 Let $a, b, c$ be real numbers, and $a+b+c=3$, prove:
$$\left(1+a+a^{2}\right)\left(1+b+b^{2}\right)\left(1+c+c^{2}\right) \geq 9(ab+bc+ca) \text{ (Pham Kim Hung) }$$ | Proof: Let $x=a+b+c, y=ab+bc+ca, z=abc$. According to the problem, we have $x=3$. Therefore, the inequality is equivalent to
$$\begin{array}{l}
z^{2}+z+1+\sum_{sym}\left(a+a^{2}\right)+\sum_{sym} ab+\sum_{sym} a^{2} b^{2}+abc\left(\sum_{sym} a+\sum_{sym} ab\right)+\sum_{sym} a^{2}(b+c) \geq 9 y \\
\Leftrightarrow z^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,519 |
Example 8.2.6 Let $a, b, c>0$, and $abc=1$, prove:
$\frac{1}{1+3a}+\frac{1}{1+3b}+\frac{1}{1+3c}+\frac{1}{1+a+b+c} \geq 1 \quad$ (Pham Kim Hung) | Proof: Let $x=a+b+c, y=ab+bc+ca$, then the inequality can be rewritten as follows:
$$\frac{3+6x+9y}{28+3x+9y}+\frac{1}{1+x} \geq 1 \Leftrightarrow \frac{1}{1+x} \geq \frac{25-3x}{28+3x+9y} \Leftrightarrow 3x^{2}-19x+9y+3 \geq 0$$
Let $z=\sqrt{\frac{x}{3}}$, since $y^{2}=(ab+bc+ca)^{2} \geq 3abc(a+b+c)=9z^{2}$, thus $y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,520 |
Example 8.2.7 Let $a, b, c \geq 0$, and $a+b+c=1$, prove:
\[
\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}+16 a b c} \geq 8\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \quad (\text{Pham Kim Hung, MYM})
\] | Proof: Let $x=4(ab+bc+ca), y=8abc$, then we have
$$a^{2}+b^{2}+c^{2}=1-\frac{x}{2} ; \quad a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=\frac{x^{2}}{16}-\frac{y}{4}$$
Thus, the inequality is equivalent to
$$\begin{array}{l}
2 x \geq(4-2 x+8 y)\left(x^{2}-4 y\right) \Leftrightarrow x(x-1)^{2} \geq 4 y[(x-1)(x+2)-4 y] \\
\Leftri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,521 |
Example 8.2.8 Let $a, b, c \geq 0$, and $a+b+c=2$, prove:
$a^{2}+b^{2}+c^{2} \geq 2\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}+4 a^{2} b^{2} c^{2}\right)$ (Pham Kim Hung) | Proof: Let $p=ab+bc+ca, q=abc$, then the inequality can be written in the following form
$$\begin{array}{l}
(a+b+c)^{2}-2(ab+bc+ca) \geq 2(ab+bc+ca)^{3}-6abc(a+b+c)(ab+bc+ca)+14a^{2}b^{2}c^{2} \\
\Leftrightarrow 2-p \geq p^{3}-6pq+7q^{2}
\end{array}$$
Let
$r=\max \left\{0, \frac{8p-8}{9}\right\}$, then according to Sc... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,522 |
Example 8.3.1 Let $a, b, c$ be three real numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove the following inequality
$$a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b) \leq 6$$ | Proof: Of course, the inequality is non-homogeneous. However, the condition \(a^{2}+b^{2}+c^{2}=3\) can help us transform the inequality into a homogeneous one:
\[a^{3}(b+c)+b^{3}(c+a)+c^{3}(a+b) \leq \frac{2}{3}\left(a^{2}+b^{2}+c^{2}\right)^{2}\]
Rearranging, the inequality is equivalent to:
\[
\begin{array}{l}
2 \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,523 |
Example 8.3.2 Let $a, b, c \geq 0$, prove: $\sqrt{\frac{a b+b c+c a}{3}} \leq \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}$ | Proof: Without loss of generality, assume $a b+b c+c a=3$. By the AM-GM inequality, we have $a+b+c \geq 3$ and $a b c \leq 1$. Therefore,
$$\begin{array}{l}
(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)-a b c=3(a+b+c)-a b c \geq 8 \\
\Rightarrow \sqrt{\frac{a b+b c+c a}{3}} \leq 1 \leq \sqrt[3]{\frac{(a+b)(b+c)(c+a)}{8}}
\end{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,524 |
Example 8.3.3 Let $a, b, c \geq 0$, prove:
$$\frac{(2 a+b+c)^{2}}{2 a^{2}+(b+c)^{2}}+\frac{(2 b+c+a)^{2}}{2 b^{2}+(c+a)^{2}}+\frac{(2 c+a+b)^{2}}{2 c^{2}+(a+b)^{2}} \leq 8 \text { (USA M0 2003) }$$ | Prove: Using the expression $a+b+c=3$ for normalization, we get the form of the left side of the inequality as follows:
$$\frac{(3+a)^{2}}{2 a^{2}+(3-a)^{2}}+\frac{(3+b)^{2}}{2 b^{2}+(3-b)^{2}}+\frac{(3+c)^{2}}{2 c^{2}+(3-c)^{2}}$$
Notice that
$$\frac{3(3+a)^{2}}{2 a^{2}+(3-a)^{2}}=\frac{a^{2}+6 a+9}{a^{2}-2 a+3}=1+\f... | 8 | Inequalities | proof | Yes | Yes | inequalities | false | 731,525 |
Example 8.3.4 Let $a, b, c \geq 0$, prove: $\frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+b^{2}}+\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}} \geq \frac{3}{5}$ | Proof: Without loss of generality, we assume $a+b+c=3$, then the inequality becomes
$$\sum_{c y c} \frac{(3-2 a)^{2}}{a^{2}+(3-a)^{2}} \geq \frac{3}{5} \Leftrightarrow \sum_{c y c} \frac{1}{2 a^{2}-6 a+9} \leq \frac{3}{5}$$
Just note that
$$\begin{array}{l}
\sum_{c y c}\left(\frac{5}{2 a^{2}-6 a+9}-1\right)=\sum_{c y ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,526 |
Example 8.3.5 Let $a, b, c>0$, prove:
$\frac{(2 a+b+c)^{2}}{4 a^{3}+(b+c)^{3}}+\frac{(2 b+c+a)^{2}}{4 b^{3}+(c+a)^{3}}+\frac{(2 c+a+b)^{2}}{4 c^{3}+(a+b)^{3}} \leq \frac{12}{a+b+c}$ (Pham Kim Hung) | Prove: Assume $a+b+c=3$. Then the inequality becomes
$$\sum_{c y c} \frac{(3+a)^{2}}{4 a^{3}+(3-a)^{3}} \leq 4$$
Notice that
$$\begin{array}{l}
\frac{(3+a)^{2}}{4 a^{3}+(3-a)^{3}}-\frac{4}{3}=\frac{(a-1)\left(-4 a^{2}-15 a+27\right)}{4 a^{3}+(3-a)^{3}}= \\
(a-1)\left(\frac{2}{3}+\frac{(a-1)\left(-2 a^{2}-12 a-9\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,527 |
Example 1.1.15 Let $a, b, c, d>0$, prove:
$$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^{2} \geq \frac{1}{a^{2}}+\frac{4}{a^{2}+b^{2}}+\frac{9}{a^{2}+b^{2}+c^{2}}+\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}$$ | Proof: Let's prove
$$\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}+\sum_{sym} \frac{2}{a b} \geq \frac{4}{a^{2}+b^{2}}+\frac{9}{a^{2}+b^{2}+c^{2}}+\frac{16}{a^{2}+b^{2}+c^{2}+d^{2}}$$
By the AM-GM inequality, we have
$$\begin{array}{l}
\frac{2}{a b} \geq \frac{4}{a^{2}+b^{2}} \\
\frac{2}{a c}+\frac{2}{b c} \geq \fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,528 |
Example 8.3.6 Let $a, b, c, d \geq 0$, prove that
$$\frac{a}{b^{2}+c^{2}+d^{2}}+\frac{b}{c^{2}+d^{2}+a^{2}}+\frac{c}{d^{2}+a^{2}+b^{2}}+\frac{d}{a^{2}+b^{2}+c^{2}} \geq \frac{3 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}}$$ | Prove: Without loss of generality, assume $a^{2}+b^{2}+c^{2}+d^{2}=1$, then the inequality becomes
$$\frac{a}{1-a^{2}}+\frac{b}{1-b^{2}}+\frac{c}{1-c^{2}}+\frac{d}{1-d^{2}} \geq \frac{3 \sqrt{3}}{2}$$
By the AM-GM inequality, we have
$$2 a^{2}\left(1-a^{2}\right)\left(1-a^{2}\right) \leq\left(\frac{2}{3}\right)^{3} \R... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,529 |
Example 8.3.7 Let $a, b, c \geq 0$, prove:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3 \sqrt[3]{a b c}}{2(a+b+c)} \geq 2 \text { (Pham Kim Hung) }$$ | Prove: By applying the Cauchy-Schwarz inequality, we have
$$\sum_{c y c} \frac{a}{b+c} \geq \frac{(a+b+c)^{2}}{2(a b+b c+c a)}$$
Using the normalization \(a+b+c=1\), it suffices to prove
$$\frac{1}{2 x}+\frac{3}{2} \sqrt[3]{a b c} \geq 2$$
where \(x=a b+b c+c a \leq \frac{1}{3}\). If \(x \leq \frac{1}{4}\), the inequ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,530 |
Example 8.3.8 Let $a, b, c \geq 0$, prove:
$$\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}} \geq \frac{9}{4(a b+b c+c a)}(\text{Iran TST 1996})$$ | Proof: We use $ab + bc + ca = 1$ to normalize, then the inequality becomes
\[4 \sum_{cyc} (a+b)^2 (c+a)^2 \geq 9 (a+b)^2 (b+c)^2 (c+a)^2\]
or
\[4\left(1+a^2\right)^2 + 4\left(1+b^2\right)^2 + 4\left(1+c^2\right)^2 \geq 9(a+b+c-abc)^2\]
Let $s = a + b + c$, then the inequality becomes the following form:
\[4\left(s^4 -... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,531 |
Example 8.3.9 Let $a, b, c, d>0$, prove:
$$\begin{array}{l}
\frac{a b c}{(d+a)(d+b)(d+c)}+\frac{b c d}{(a+b)(a+c)(a+d)}+ \\
\frac{c d a}{(b+a)(b+c)(b+d)}+\frac{d a b}{(c+a)(c+b)(c+d)} \geq \frac{1}{2} \text { (Nguyen Van Thach) }
\end{array}$$ | Proof: Let $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}, t=\frac{1}{d}$, then the inequality becomes
$$\frac{x^{3}}{(x+y)(x+z)(x+t)}+\frac{y^{3}}{(y+x)(y+z)(y+t)}+\frac{z^{3}}{(z+x)(z+y)(z+t)}+\frac{t^{3}}{(t+x)(t+y)(t+z)} \geq \frac{1}{2}$$
Without loss of generality, assume $x+y+z+t=4$. By the AM-GM inequality, we h... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,532 |
Example 8.4.1 Let $a, b, c>0$, and satisfy $a b c=1$, prove:
$\frac{1}{3 a^{2}+(a-1)^{2}}+\frac{1}{3 b^{2}+(b-1)^{2}}+\frac{1}{3 c^{2}+(c-1)^{2}} \geq 1 \quad$ (Le Huu Dien Khue) | Proof: We imagine finding a real constant $k$ that satisfies
$$\frac{1}{3 a^{2}+(a-1)^{2}} \geq \frac{1}{3}+k \ln a$$
If such a valid constant $k$ exists, we get (noting that $\ln a+\ln b+\ln c=0$)
$$\sum_{c y c} \frac{1}{3 a^{2}+(a-1)^{2}} \geq 1+k\left(\sum_{c y c} \ln a\right)=1$$
Let $f(x)=\frac{1}{3 x^{2}+(x-1)^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,533 |
Example 8.4.2 Let $a, b, c>0$, and $abc=1$, prove:
$\frac{1}{a^{2}-a+1}+\frac{1}{b^{2}-b+1}+\frac{1}{c^{2}-c+1} \leq 3 \quad($ Vu Dinh Quy $)$ | Proof: First, we will prove
$$f(x)=\frac{1}{x^{2}-x+1}+\ln x-1 \geq 0 \quad (x \in(0,1.8])$$
In fact, we have
$$f^{\prime}(x)=\frac{-2 x+1}{\left(x^{2}-x+1\right)^{2}}+\frac{1}{x}=\frac{(x-1)\left(x^{3}-x^{2}-1\right)}{\left(x^{2}-x+1\right)^{2}}$$
The equation $x^{3}=x^{2}+1$ has exactly one real root in $(0,2]$, so... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,534 |
Example 8.4.3 Let $a, b, c, d, e, f > 0$, and satisfy $a b c d e f = 1$. Prove:
\[
\frac{2 a+1}{a^{2}+a+1}+\frac{2 b+1}{b^{2}+b+1}+\frac{2 c+1}{c^{2}+c+1}+\frac{2 d+1}{d^{2}+d+1}+\frac{2 e+1}{e^{2}+e+1}+\frac{2 f+1}{f^{2}+f+1} \leq 4
\]
(Pham Kim Hung) | Consider the following function
$$f(x)=\frac{1+2 x}{1+x+x^{2}}+\frac{\ln x}{3}-1$$
Of course, we have
$$f^{\prime}(x)=\frac{-2 x^{2}-2 x+1}{\left(1+x+x^{2}\right)^{2}}+\frac{1}{3 x}=\frac{(x-1)\left(x^{3}-3 x^{2}-6 x-1\right)}{3 x\left(1+x+x^{2}\right)^{2}}$$
Notice that the equation $x^{3}=3 x^{2}+6 x+1$ has only on... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,535 |
Example 8.4.4 Let $a, b, c, d>0$, and $a b c d=1$, prove:
$\frac{1+a}{1+a^{2}}+\frac{1+b}{1+b^{2}}+\frac{1+c}{1+c^{2}}+\frac{1+d}{1+d^{2}} \leq 4 \quad$ (Vasile Cirtoaje) | Consider the following function
$$f(x)=\frac{1+x}{1+x^{2}}+\frac{\ln x}{2}-1 \quad (x>0)$$
We have
$$f^{\prime}(x)=\frac{(x-1)\left(x^{3}-x^{2}-3 x-1\right)}{2 x\left(1+x^{2}\right)^{2}}$$
Since the equation $x^{3}=x^{2}+3 x+1$ has exactly one real root $x_{0}$, and $4 \geq x_{0}>1$, it is easy to see that
$$\max _{0... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,536 |
Example 8.4.5 Let $a, b, c, d, e, f>0$, and satisfy $a b c d e f=1$, prove:
$$\frac{a-1}{a^{2}+a+1}+\frac{b-1}{b^{2}+b+1}+\frac{c-1}{c^{2}+c+1}+\frac{d-1}{d^{2}+d+1}+\frac{e-1}{e^{2}+e+1}+\frac{f-1}{f^{2}+f+1} \leq 0$$ | Consider the following function
$$f(x)=\frac{x-1}{x^{2}+x+1}-\frac{\ln x}{3}$$
Notice that
$$f^{\prime}(x)=\frac{(x-1)\left(-x^{3}-6 x^{2}-3 x+1\right)}{3 x\left(x^{2}+x+1\right)^{2}}$$
The equation $x^{3}+6 x^{2}+3 x=1$ has exactly one real root $x_{0}$, and it is clear that $x_{0}>\frac{1}{5}$, therefore
$$\max _{1... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,537 |
Example 8.4.6 Let $a, b, c, d, e, f, g>0$, and $a+b+c+d+e+f+g=7$, prove:
$$\left(a^{2}-a+1\right)\left(b^{2}-b+1\right)\left(c^{2}-c+1\right)\left(d^{2}-d+1\right)\left(e^{2}-e+1\right)\left(f^{2}-f+1\right)\left(g^{2}-g+1\right) \geq 1$$
(Pham Kim Hung) | Prove: The inequality is equivalent to
$$\sum_{c y c} \ln \left(a^{2}-a+1\right) \geq 0$$
Consider the function \( f(x)=\ln \left(x^{2}-x+1\right)-x+1 \). We have
$$f^{\prime}(x)=\frac{(x-1)(2-x)}{x^{2}+x+1}$$
It is easy to see that
$$\min _{0 \leq x \leq 2.75} f(x)=\min \{f(1), f(2.75)\}=0$$
If \( a, b, c, d, e, f,... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,538 |
Example 1.1.16 Find the smallest value of $M$ such that the following inequality holds for all real numbers $a, b, c$.
$$\left|a b\left(a^{2}-b^{2}\right)+b c\left(b^{2}-c^{2}\right)+c a\left(c^{2}-a^{2}\right)\right| \leq M\left(a^{2}+b^{2}+c^{2}\right)^{2}$$
(IMO 2006) | Let $x=a-b, y=b-c, z=c-a, s=a+b+c$, then the inequality can be written as follows:
$$9|s x y z| \leq M\left(s^{2}+x^{2}+y^{2}+z^{2}\right)^{2}$$
where $s, x, y, z$ are any real numbers, and satisfy $x+y+z=0$.
In fact, $s$ is an independent variable. First, we examine the relationship between $x y z$ and $x^{2}+y^{2}+z... | \frac{9 \sqrt{2}}{32} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,539 |
Example 8.4.7 Let $a, b, c, d > 0$, and $a + b + c + d = 4$, prove:
$$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}} \geq a^{2}+b^{2}+c^{2}+d^{2} \quad(\text{Pham Kim Hung})$$ | Proof: Consider the following function
$$f(x)=\frac{1}{x^{2}}-x^{2}+4 x-4(x>0)$$
We have
$$f^{\prime}(x)=\frac{-2}{x^{3}}-2 x+4=\frac{-2(x-1)\left(x^{3}-x^{2}-x-1\right)}{x^{3}}$$
The equation $f^{\prime}(x)=0$ has two positive real roots, one root is 1 and the other root is less than 1.
$$\max _{010$$
(1) The first ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,540 |
1. Let $a, b, c \geq 0$, and satisfy $a+b+c=3$, prove:
$$\left(a^{2}-a b+b^{2}\right)\left(b^{2}-b c+c^{2}\right)\left(c^{2}-c a+a^{2}\right) \leq 12$$
(Pham Kim Hung) | Proof: Without loss of generality, assume $a \geq b \geq c$. Of course, we have
$$\begin{array}{l}
b^{2}-b c+c^{2}=b^{2}-c(b-c) \leq b^{2} \\
c^{2}-c a+a^{2}=a^{2}-c(a-c) \leq a^{2}
\end{array}$$
Thus, it suffices to prove
$$M=a^{2} b^{2}\left(a^{2}-a b+b^{2}\right) \leq 12$$
Let $x=\frac{a-b}{2} \geq 0, s=\frac{a+b}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,541 |
2、Let $a, b, c>0$, prove:
$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \geq \frac{9}{1+a b c}$ (Wa 1 ther Janous)
2、Suppose $a, b, c>0$, prove:
$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right) \geq \frac{9... | Proof: By the rearrangement inequality, we have
$$\frac{1}{a(1+a)}+\frac{1}{b(1+b)}+\frac{1}{c(1+c)} \geq \frac{1}{b(1+c)}+\frac{1}{c(1+a)}+\frac{1}{a(1+b)}$$
Therefore, if the following inequalities hold, then the original inequality also holds.
$$\sum_{c y c} \frac{1}{b(1+c)} \geq \frac{3}{1+a b c} ; \quad \sum_{c y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,542 |
3. Let $a, b, c>0$, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove:
$\frac{a}{a^{2}+2 a+3}+\frac{b}{b^{2}+2 b+3}+\frac{c}{c^{2}+2 c+3} \leq \frac{1}{2}$ (Pham Kim Hung) | Proof: We obviously have $a^{2}+1 \geq 2 a, b^{2}+1 \geq 2 b, c^{2}+1 \geq 2 c$, so
$$\sum_{c y c} \frac{a}{a^{2}+2 b+3} \geq \sum_{c y c} \frac{a}{2(a+b+1)}$$
Thus, it suffices to prove
$$\sum_{c y c} \frac{a}{a+b+1} \leq 1 \Leftrightarrow \sum_{c y c} \frac{b+1}{a+b+1} \geq 2$$
Noting that $a^{2}+b^{2}+c^{2}=3$, we... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,543 |
4. Let $x_{1}, x_{2}, \cdots, x_{n}>0$, and satisfy $x_{1} x_{2} \cdots x_{n}=1$, prove:
$x_{1}+x_{2}+\cdots+x_{n} \geq \frac{2}{1+x_{1}}+\frac{2}{1+x_{2}}+\cdots+\frac{2}{1+x_{n}}$ (Pham Kim Hung) | Proof: Since $2-\frac{2}{1+x_{i}}=\frac{2 x_{i}}{1+x_{i}}$, the inequality becomes
$$\sum_{c y c} x_{i}+\sum_{c y c} \frac{2 x_{i}}{1+x_{i}} \geq 2 n$$
According to the AM-GM inequality, we have
$$\begin{array}{l}
-2 n+\sum_{c y c} x_{i}+\sum_{c y c} \frac{2 x_{i}}{1+x_{i}}=-2 n+\sum_{c y c}\left(\frac{x_{i}+1}{2}+\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,544 |
5. Prove that for all positive real numbers $a, b, c \in [1,2]$, we have
$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 6\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)$ (Tran Nam Dung, Vietnam TST 2006) | Proof: Instead of the condition $a, b, c \in [1,2]$, we will prove a stronger inequality with the condition that $a, b, c$ are the sides of a triangle. Using the following identities:
$$\left(\sum_{c y c} a\right)\left(\sum_{c y c} \frac{1}{a}\right)-9=\sum_{c y c} \frac{(a-b)^{2}}{a b} ; 6\left(\sum_{c y c} \frac{a}{b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,545 |
6. Let $a, b, c>0$, and satisfy $a+b+c=3$, prove: $a^{2}+b^{2}+c^{2} \geq \frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$ (Pham Kim Hung) | Proof: (Cauchy Inversion) By the AM-GM inequality, we have
$$\sum_{c y c} \frac{2+a}{2+b}=\sum_{c y c} \frac{2+a}{2}-\sum_{c y c} \frac{b(2+a)}{2(2+b)} \geq \frac{9}{2}-\frac{3 \sqrt[3]{a b c}}{2}$$
Thus, it suffices to prove $\square$
$$\sum_{c y c} a^{2}+\frac{3 \sqrt[3]{a b c}}{2} \geq \frac{(a+b+c)^{2}}{2} \Leftri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,546 |
7. Let $x, y, z \geq 0$, and $x+y+z=3$, prove:
$\sqrt{\frac{x}{1+2 y z}}+\sqrt{\frac{y}{1+2 z x}}+\sqrt{\frac{z}{1+2 x y}} \geq \sqrt{3}$ (Phan Thanh Viet) | Proof: According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\sum_{c y c} \sqrt{\frac{x}{1+2 y z}}=\sum_{c y c} \frac{x^{2}}{\sqrt{x} \sqrt{x^{2}+2 x^{2} y z}} \\
\geq \frac{(x+y+z)^{2}}{\sqrt{x} \sqrt{x^{2}+2 x^{2} y z}+\sqrt{y} \sqrt{y^{2}+2 y^{2} z x}+\sqrt{z} \sqrt{z^{2}+2 z^{2} x y}} \\
\geq \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,547 |
8. Let $a_{1}, a_{2}, \cdots, a_{n}>0$, and satisfy $a_{1} a_{2} \cdots a_{n}=1$, prove:
$$\sqrt{1+a_{1}^{2}}+\sqrt{1+a_{2}^{2}}+\ldots+\sqrt{1+a_{n}^{2}} \leq \sqrt{2}\left(a_{1}+a_{2}+\cdots+a_{n}\right)$$
(Gabriel Dospinescu) | Proof: From the obvious inequality $(\sqrt{x}-1)^{4} \geq 0$, we have
$$\frac{1+x^{2}}{2} \leq(x-\sqrt{x}+1)^{2} \Rightarrow \sqrt{\frac{1+x^{2}}{2}}+\sqrt{x} \leq 1+x$$
Based on this result, we of course have
$$\sum_{i=1}^{n} \sqrt{1+a_{i}^{2}} \leq \sqrt{2} \sum_{i=1}^{n} a_{i}+\sqrt{2}\left(n-\sum_{i=1}^{n} \sqrt{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,548 |
9. Let $a, b, c, k>0$, prove:
$\frac{a+b k}{a+k c}+\frac{b+k c}{b+k a}+\frac{c+k a}{c+k b} \leq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ (Nguyen Vient Anh) | Proof: Let $X=\frac{b}{1+k}, Y=\frac{c}{1+k}, Z=\frac{a}{1+k}$. According to Hölder's inequality, we have
$$\prod_{c y c}\left(1+\frac{k a}{b}\right) \geq(1+k)^{3}$$
or equivalently $X Y Z \geq 1$. Now the inequality becomes the following form
$$\begin{array}{l}
\sum_{c y c}\left(\frac{a}{b}-\frac{c+k a}{c+k b}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,549 |
Example 1.2.1 Let $a, b, c>0, a+b+c=3$, prove: $\frac{a}{1+b^{2}}+\frac{b}{1+c^{2}}+\frac{c}{1+a^{2}} \geq \frac{3}{2}$ | Prove: In fact, directly applying the AM-GM inequality to the denominator does not work, because the inequality changes direction.
$$\frac{a}{1+b^{2}}+\frac{b}{1+c^{2}}+\frac{c}{1+a^{2}} \leq \frac{a}{2 b}+\frac{b}{2 c}+\frac{c}{2 a} \geq \frac{3}{2} ?!$$
However, we can use the AM-GM inequality in a different form:
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,550 |
10. Let $a, b, c > 0$, and satisfy $abc = 1$, prove
$\frac{1}{(a+1)(a+2)}+\frac{1}{(b+1)(b+2)}+\frac{1}{(c+1)(c+2)} \geq \frac{1}{2} \quad($ Pham Kim Hung) | Proof: Given $a b c=1$, there exist three positive real numbers $x, y, z$ such that $a=\frac{y z}{x^{2}}, b=\frac{z x}{y^{2}}, c=\frac{x y}{z^{2}}$. The inequality becomes
$$\sum_{c y c} \frac{x^{4}}{\left(x^{2}+y z\right)\left(2 x^{2}+y z\right)} \geq \frac{1}{2}$$
By the Cauchy-Schwarz inequality, we have
$$L H S \g... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,551 |
11. Let $a, b, c, d \geq 0$, and satisfy $a+b+c+d=4$, prove
$$a^{2}+b^{2}+c^{2}+d^{2}-4 \geq 4(a-1)(b-1)(c-1)(d-1)$$
(Pham Kim Hung) | Prove: By applying the AM-GM inequality, we have
$$a^{2}+b^{2}+c^{2}+d^{2}-4=(a-1)^{2}+(b-1)^{2}+(c-1)^{2}+(d-1)^{2} \geq 4 \sqrt{|(a-1)(b-1)(c-1)(d-1)|}$$
Therefore, we only need to consider the inequality under the condition $a \geq b \geq 1 \geq c \geq d$ (to ensure $(a-1)(b-1)(c-1)(d-1) \geq 0$). Since $a+b \leq 4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,552 |
12. Let $x, y, z$ be distinct real numbers, prove: $\frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1 \quad$ (Le Huu Dien Khue) | $$\begin{array}{l}
\sum_{c y c}\left(1-\frac{x}{z}\right)^{2}\left(1-\frac{z}{y}\right)^{2}-\left(1-\frac{y}{x}\right)^{2}\left(1-\frac{z}{y}\right)^{2}\left(1-\frac{x}{z}\right)^{2} \\
=\sum_{c y c}\left(1-\frac{x}{z}-\frac{z}{y}+\frac{x}{y}\right)^{2}-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{y}{x}-\frac{z}{y}-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,553 |
13. Let $a, b, c, d \geq 0$, and satisfy $a+b+c+d=3$, prove that $a b(b+c)+b c(c+d)+c d(d+a)+d a(a+b) \leq 4 \quad$ (Pham Kim Hung) | Proof: Without loss of generality, we assume $b+d \leq a+c$. We have
$$\begin{array}{l}
a b(b+c)+b c(c+d)+c d(d+a)+d a(a+b) \\
=(a+c)(b c+d a)+(b+d)(a b+c d) \\
=(a+c)[(a+c)(b+d)-(a b+c d)]+(b+d)(a b+c d) \\
=(a+c)^{2}(b+d)+(a b+c d)(b+d-a-c) \leq(a+c)^{2}(b+d)
\end{array}$$
Finally, by the AM-GM inequality, we have
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,554 |
14. Let $a_{1}, a_{2}, \cdots, a_{n}$ be any real numbers, prove: $\sum_{i, j=1}^{n}\left|a_{i}+a_{j}\right| \geq n \sum_{i=1}^{n}\left|a_{i}\right|$ | Prove: Decompose the sequence $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ into two non-negative subsequences
$$\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}=\left\{b_{1}, b_{2}, \cdots, b_{n}\right\} \cup\left\{-c_{1},-c_{2}, \cdots,-c_{n}\right\}$$
satisfying $n=r+s, b_{i} \geq 0 \, i \in\{1,2, \cdots, r\}, c_{j}>0 \, j ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,555 |
15. Let $a, b, c \geq 0$, prove:
$3(a+b+c) \geq 2\left(\sqrt{a^{2}+b c}+\sqrt{b^{2}+c a}+\sqrt{c^{2}+a b}\right)$ (Pham Kim Hung) | Proof: Without loss of generality, we assume $a \geq b \geq c$. Therefore,
$$\sqrt{b^{2}+c a}+\sqrt{c^{2}+a b} \leq \sqrt{2\left(b^{2}+c^{2}\right)+2 a(b+c)}$$
Thus, we only need to prove
$$2 \sqrt{2\left(b^{2}+c^{2}\right)+2 a(b+c)}+2 \sqrt{a^{2}+b c} \leq 3(a+b+c)$$
Let $s=\frac{b+c}{2}$, squaring both sides, the i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,556 |
16. Let $a, b, c \geq 0$, prove:
$\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}} \geq \frac{10}{(a+b+c)^{2}}$ (Vasile Cirtoaje, Nguyen Vient Anh) | Proof: Without loss of generality, assume $c=\min (a, b, c)$, then
$$\begin{array}{l}
b^{2}+c^{2} \leq\left(b+\frac{c}{2}\right)^{2}=x^{2} \\
c^{2}+a^{2} \leq\left(a+\frac{c}{2}\right)^{2}=y^{2} \\
a^{2}+b^{2} \leq\left(a+\frac{c}{2}\right)^{2}+\left(b+\frac{c}{2}\right)^{2}=x^{2}+y^{2}
\end{array}$$
Thus, we have
$$\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,557 |
17. Let $a, b, c, d > 0$, and satisfy $a b c d = 1$, prove:
$\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+d^{2}\right) \geq (a+b+c+d)^{2}$ (Pham Kim Hung) | Proof: Since $a b c d=1$, then among $a, b, c, d$, there are at least two numbers that are not less than 1 or not greater than 1. Without loss of generality, assume these two numbers are $b, d$, then $(b-1)(d-1) \geq 0$. Applying the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(1+a^{2}\right)\left(1+b^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,558 |
18. Let $a, b, c > 0$, and satisfy $a + b + c = 1$. Prove:
$$\frac{a b}{\sqrt{a b + b c}} + \frac{b c}{\sqrt{b c + c a}} + \frac{c a}{\sqrt{c a + a b}} \leq \frac{1}{\sqrt{2}} \text{ (Chinese M0 2006) }$$ | Prove that the inequality is equivalent to
$$\sum_{c y c} a \cdot \sqrt{\frac{b}{a+c}} \leq \frac{1}{\sqrt{2}} \Leftrightarrow \sum_{c y c} \frac{a+b}{2} \cdot \sqrt{\frac{a^{2} b}{(a+c)(a+b)^{2}}} \geq \frac{1}{\sqrt{2}}$$
Using the weighted Jensen's inequality on the concave function \( f(x) = \sqrt{x} \), we have
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,559 |
19. Let $x, y, z \geq 0$, and satisfy $x+y+z=1$. Find the maximum value of the following expression $\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}$ (Pham Kim Hung) | First, we consider the case where $\min (x, y, z)=0$. Without loss of generality, assume $z=0$, then $x+y=1$, thus
$$\sum_{c y c} \frac{x-y}{\sqrt{x+y}}=\frac{x-y}{\sqrt{x+y}}+\sqrt{y}-\sqrt{x}=x-y+\sqrt{y}-\sqrt{x}=u(v-1)$$
where $u=\sqrt{x}-\sqrt{y}, v=\sqrt{x}+\sqrt{y}, \quad u^{2}+v^{2}=2$
Let $u^{2}(v-1)^{2}=\lef... | \sqrt{\frac{71-17 \sqrt{17}}{32}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,560 |
Example 1.2.2 Let $a, b, c, d>0, a+b+c+d=4$, prove: $\frac{a}{1+b^{2} c}+\frac{b}{1+c^{2} d}+\frac{c}{1+d^{2} a}+\frac{d}{1+a^{2} b} \geq 2$ | Prove: According to the AM-GM inequality, we have
$$\frac{a}{1+b^{2} c}=a-\frac{a b^{2} c}{1+b^{2} c} \geq a-\frac{a b^{2} c}{2 b \sqrt{c}}=a-\frac{a b \sqrt{c}}{2}=a-\frac{b \sqrt{a \cdot a c}}{2} \geq a-\frac{b(a+a c)}{4}$$
Based on this estimate, we have
$$\sum_{c y c} \frac{a}{1+b^{2} c} \geq \sum_{c y c} a-\frac{... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 731,561 |
20. Let $x, y, z \in[-1,1]$, and satisfy $x+y+z=0$, prove:
$\sqrt{1+x+y^{2}}+\sqrt{1+y+z^{2}}+\sqrt{1+z+x^{2}} \geq 3$ (Phan Thanh Nam) | Proof: First, we will prove that if $ab \geq 0$, then $\sqrt{1+a} + \sqrt{1+b} \geq 1 + \sqrt{1+a+b}$.
In fact, squaring both sides, the inequality becomes
$$2 + a + b + 2 \sqrt{(1+a)(1+b)} \geq 2 + a + b + 2 \sqrt{1+a+b} \Leftrightarrow (1+a)(1+b) \geq 1 + a + b \Leftrightarrow ab \geq 0$$
Notice that $x + y^2, y + z... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,562 |
21. Suppose $a, b, c \geq 0$, and satisfy $a b + b c + c a = 1$, prove: $\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \geq \frac{5}{2}$ (Berkeley Mathematics Circle) | Proof: Let $x=a+b+c, z=abc$, then the inequality becomes
$$\begin{array}{l}
2 \sum_{cyc}(a+b)(a+c) \geq 5 \prod_{cyc}(a+b) \Leftrightarrow 6+2 \sum_{cyc} a^{2} \geq 5(a+b+c-abc) \\
\Leftrightarrow 2 x^{2}-5 x+2+5 z \geq 0 \Leftrightarrow(x-2)(2 x-1)+5 z \geq 0
\end{array}$$
If $x \geq 2$, the inequality is obviously t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,563 |
22. Prove the following inequality
$$(\sqrt{2})^{n}\left(a_{1}+a_{2}\right)\left(a_{2}+a_{3}\right) \cdots\left(a_{n}+a_{1}\right) \leq\left(a_{1}+a_{2}+a_{3}\right)\left(a_{2}+a_{3}+a_{4}\right) \cdots\left(a_{n}+a_{1}+a_{2}\right)$$
where $a_{1}, a_{2}, \cdots, a_{n}$ are any positive real numbers (Russia MO) | Prove: According to the following inequalities
$$\begin{array}{l}
\left(a_{1}+a_{2}+a_{3}\right)^{2} \geq\left(2 a_{1}+a_{2}\right)\left(a_{2}+2 a_{3}\right) \\
\left(2 a_{1}+a_{2}\right)\left(2 a_{2}+a_{1}\right)=2 a_{1}^{2}+2 a_{2}^{2}+5 a_{1} a_{2} \geq 2\left(a_{1}+a_{2}\right)^{2}
\end{array}$$
we can immediately... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,564 |
Let $a, b, c \geq 0$, and $a b+b c+c a=1$, prove
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b+c} \geq 3$$ | Proof: If $a+b+c \leq 2$, then from the above result, the inequality holds. Now we assume $a+b+c \geq 2$ and $a \geq b \geq c$, then
$$\begin{array}{l}
\frac{1}{a+b+c}+\sum_{c y c} \frac{1}{a+b}=\frac{1}{a+b}+\frac{a b+b c+c a}{b+c}+\frac{a b+b c+c a}{c+a}+\frac{1}{a+b+c} \\
=\frac{1}{a+b}+a+b+\frac{c(1+a b)}{1+c^{2}}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,565 |
23. Let $a, b, c \geq 0$, and $a+b+c=3$, prove:
$\frac{a b+b c+c a}{a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}} \geq \frac{a^{3}+b^{3}+c^{3}}{36}$ (Pham Kim Hung, MYM) | To prove: Without loss of generality, assume $a \geq b \geq c$. Let
$$f(a, b, c)=36(a b+b c+c a)-\left(a^{3}+b^{3}+c^{3}\right)\left(a^{3} b^{3}+b^{3} c^{3}+c^{3} a^{3}\right)$$
We will prove that $f(a, b, c) \geq f(a, b+c, 0)$. In fact,
$$\begin{array}{l}
a^{3}+b^{3}+c^{3} \leq a^{3}+(b+c)^{3} ; a^{3} b^{3}+b^{3} c^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,566 |
24. Let real numbers $a, b, c, d$ satisfy $(1+) a^{2}\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+d^{2}\right)=16$, prove that $-3 \leq a b+b c+c a+d a+a c+b d-a b c d \leq 5$ (Titu Abdreescu and Gabriel Dospinescu) | Proof: We denote $S=a b+b c+c a+d a+a c+b d-a b c d$, then
$$S-1=(1-a b)(c d-1)+(a+b)(c+d)$$
Applying the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
(S-1)^{2} \leq\left[(1-a b)^{2}+(a+b)^{2}\right]\left[(1-c d)^{2}+(c+d)^{2}\right] \\
=(1+) a^{2}\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+d^{2}\right)=1... | -3 \leq S \leq 5 | Inequalities | proof | Yes | Yes | inequalities | false | 731,567 |
25. Let $a, b, c, d$ be positive real numbers, and satisfy $a b c d=1$, prove $\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq 1 \quad($ China TST 2004) | Prove: First note that, for any non-negative real numbers $x, y$, we have
$$\frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}} \geq \frac{1}{1+x y}$$
Expanding, the above inequality is equivalent to
$$\begin{array}{l}
\left(2+2 x+2 y+x^{2}+y^{2}\right)(1+x y) \geq\left(1+2 x+x^{2}\right)\left(1+2 y+y^{2}\right) \\
\Leftrightarro... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,568 |
26. Let $a, b, c$ be positive real numbers. Prove:
$\frac{a+2 b}{c+2 b}+\frac{b+2 c}{a+2 c}+\frac{c+2 a}{b+2 a} \geq 3$ (Pham Kim Hung) | Prove: By expanding, the inequality is equivalent to
$$\sum_{c y c}(a+2 b)(a+2 c)(b+2 a) \geq 3 \prod_{c y c}(c+2 b) \Leftrightarrow 2\left(a^{3}+b^{3}+c^{3}\right) \geq 3\left(a^{2} b+b^{2} c+c^{2} a\right)$$
Combining the third-degree Schur inequality and the AM-GM inequality, we get
$$2 \sum_{c y c} a^{3}+3 a b c-3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,569 |
27. Let $a, b, c$ be positive real numbers, prove that
$\frac{a^{4}}{a^{2}+a b+b^{2}}+\frac{b^{4}}{b^{2}+b c+c^{2}}+\frac{c^{4}}{c^{2}+c a+a^{2}} \geq \frac{a^{3}+b^{3}+c^{3}}{a+b+c}$ (Phan Thanh Viet) | Notice that
$$\frac{a^{3}+b^{3}+c^{3}}{a+b+c}=\frac{3 a b c}{a+b+c}+a^{2}+b^{2}+c^{2}-a b-b c-c a$$
Therefore, the inequality can be written in the following form:
$$\sum_{c y c}\left(\frac{a^{4}}{a^{2}+a b+b^{2}}-a^{2}+a b\right) \geq \frac{3 a b c}{a+b+c} \Leftrightarrow \sum_{c y c} \frac{a b^{3}}{a^{2}+a b+b^{2}} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,570 |
Example 1.1.1 Let $a, b, c>0$ and $a+b+c=3$, prove: $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq a b+b c+c a$ | Proof: Note the identity $2(a b+b c+c a)=(a+b+c)^{2}-\left(a^{2}+b^{2}+c^{2}\right)$, then the inequality is equivalent to
$$\sum_{c y c} a^{2}+2 \sum_{c y c} \sqrt{a} \geq 9$$
By the AM-GM inequality, we have
$$\sum_{c y c} a^{2}+2 \sum_{c y c} \sqrt{a}=\sum_{c y c}\left(a^{2}+\sqrt{a}+\sqrt{a}\right) \geq 3 \sum_{c ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,572 |
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