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Example 1.2.3 Let $a, b, c>0$, prove: $\frac{a^{3}}{a^{2}+b^{2}}+\frac{b^{3}}{b^{2}+c^{2}}+\frac{c^{3}}{c^{2}+d^{2}}+\frac{d^{3}}{d^{2}+a^{2}} \geq \frac{a+b+c+d}{2}$ | Prove: We have the following estimation
$$\frac{a^{3}}{a^{2}+b^{2}}=a-\frac{a b^{2}}{a^{2}+b^{2}} \geq a-\frac{a b^{2}}{2 a b}=a-\frac{b}{2}$$
Note: A similar result for 4 variables is
$$\frac{a^{4}}{a^{3}+2 b^{3}}+\frac{b^{4}}{b^{3}+2 c^{3}}+\frac{c^{4}}{c^{3}+2 d^{3}}+\frac{d^{4}}{d^{3}+2 a^{4}} \geq \frac{a+b+c+d}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,573 |
30. Let $a, b, c, d$ be non-negative real numbers, prove that
$\left(1+\frac{2 a}{b+c}\right)\left(1+\frac{2 b}{c+d}\right)\left(1+\frac{2 c}{d+a}\right)\left(1+\frac{2 d}{a+b}\right) \geq 9$ (Vasile Cirtoaje) | Prove: Rewrite the inequality into another form
$$\left(1+\frac{a+c}{a+b}\right)\left(1+\frac{a+c}{c+d}\right)\left(1+\frac{b+d}{b+c}\right)\left(1+\frac{b+d}{a+d}\right) \geq 9$$
For all positive real numbers \(x, y\), it is easy to see that
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \geq\left(1+\frac{2}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,575 |
31. Let $a, b, c$ be non-negative real numbers, prove that
$\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}}+\frac{1}{a^{2}+b^{2}+c^{2}} \geq \frac{6}{a b+b c+c a}$ (Pham Kim Hung) | Proof: Without loss of generality, assume $a \geq b \geq c$. Let $t=\sqrt{b^{2}+c^{2}}$ and
$$f(a, b, c)=\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}}+\frac{1}{a^{2}+b^{2}+c^{2}}-\frac{6}{a b+b c+c a}$$
We have
$$\begin{array}{l}
f(a, b, c)-f(a, t, 0)=\frac{c^{2}}{\left(a^{2}+b^{2}\right)\left(a^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,576 |
33. Let $a, b, c$ be non-negative real numbers, and satisfy $a+b+c=3$. Prove that $a \sqrt{1+b^{3}}+b \sqrt{1+c^{3}}+c \sqrt{1+a^{3}} \leq 5$ (Pham Kim Hung) | Proof: By the AM-GM inequality, we have
$$\sum_{c y c} a \sqrt{1+b^{3}}=\sum_{c y c} a \sqrt{(1+b)\left(1-b+b^{2}\right)} \leq \frac{1}{2} \sum_{c y c} a\left(1+b^{2}\right)$$
Thus, it suffices to prove
$$a b^{2}+b c^{2}+c a^{2} \leq 4$$
Without loss of generality, assume $b$ is the middle value among $a, b, c$. Then... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,578 |
34. Let $a, b, c$ be positive real numbers, and satisfy $a+b+c=1$. Prove:
$$\frac{a}{\sqrt{\frac{1}{b}-1}}+\frac{b}{\sqrt{\frac{1}{c}-1}}+\frac{c}{\sqrt{\frac{1}{a}-1}} \leq \frac{3 \sqrt{3}}{4} \cdot \sqrt{(1-a)(1-b)(1-c)}$$
(Do Hoang Giang) | Prove that the inequality is equivalent to
$$P=\sum_{c y c} \sqrt{\frac{a}{(a+c)(a+b)} \cdot \frac{a b}{(b+c)(c+a)}} \leq \frac{3 \sqrt{3}}{4}$$
Without loss of generality, assume $a=\min (a, b, c)$, and consider the following cases:
( i ) If $a \leq b \leq c$, then we have
$$\begin{array}{l}
\frac{a b}{(b+c)(c+a)} \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,579 |
36. Let $a, b, c$ be non-negative real numbers, prove that
$\frac{b+c}{\sqrt{a^{2}+b c}}+\frac{c+a}{\sqrt{b^{2}+c a}}+\frac{a+b}{\sqrt{c^{2}+a b}} \geq 4$ (Pham Kim Hung) | Prove: By applying the Hölder's inequality, we have
$$\left(\sum_{c y c} \frac{b+c}{\sqrt{a^{2}+b c}}\right)^{2}\left(\sum_{c y c}(b+c)\left(a^{2}+b c\right)\right) \geq 8\left(\sum_{c y c} a\right)^{3}$$
Therefore, it suffices to prove
$$(a+b+c)^{3} \geq 4 \sum_{c y c} a^{2}(b+c) \Leftrightarrow 6 a b c+\sum_{c y c} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,581 |
37. Let $a, b, c$ be nonnegative real numbers, and satisfy $a^{2}+b^{2}+c^{2}+a b c=4$. Prove: $2+a b c \geq a b+b c+c a \geq a b c$ (USA M0 2001) | To prove the right inequality, we only need to notice that among $a, b, c$, at least one, say $a$, is not greater than 1. Therefore, we have $a b + b c + c a \geq b c \geq a b c$. The equality holds when $(a, b, c) = (0, 0, 2)$ and its permutations. To prove the left inequality, notice that among $a, b, c$, there must ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,582 |
38. Let $a, b, c$ be non-negative real numbers, prove:
$\sqrt{\frac{a^{2}+2 b c}{b^{2}+c^{2}}}+\sqrt{\frac{b^{2}+2 c a}{c^{2}+a^{2}}}+\sqrt{\frac{c^{2}+2 a b}{a^{2}+b^{2}}} \geq 3$ (Vo Quoc Ba Can, Vu Dinh Quy) | To prove: Without loss of generality, assume $a \geq b \geq c$. First, we will prove
$$\sqrt{\frac{a^{2}+c^{2}}{b^{2}+c^{2}}}+\sqrt{\frac{b^{2}+c^{2}}{c^{2}+a^{2}}} \geq \sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}$$
In fact, this inequality is equivalent to
$$\frac{a^{2}+c^{2}}{b^{2}+c^{2}}+\frac{b^{2}+c^{2}}{c^{2}+a^{2}} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,583 |
Example 1.2.4 Let $a, b, c>0, a+b+c=3$, prove: $\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1$ | Prove: According to the AM-GM inequality, we have the following estimate:
$$\frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{\frac{2}{3}}}{3}$$
This implies
$$\sum_{c y c} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{c y c} a-\frac{2}{3} \sum_{c y c}(a b)^{\frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,584 |
39. Let $a, b, c$ be three distinct positive real numbers, prove:
$$\frac{1}{\left|a^{2}-b^{2}\right|}+\frac{1}{\left|b^{2}-c^{2}\right|}+\frac{1}{\left|c^{2}-a^{2}\right|}+\frac{8}{a^{2}+b^{2}+c^{2}} \geq \frac{28}{(a+b+c)^{2}}$$
(Pham Kim Hung) | Prove: Without loss of generality, we assume $a>b>c$, note that
$$\begin{array}{l}
\frac{1}{a^{2}-b^{2}}+\frac{1}{b^{2}-c^{2}}+\frac{1}{c^{2}-a^{2}}+\frac{8}{a^{2}+b^{2}+c^{2}}-\left(\frac{1}{a^{2}-b^{2}}+\frac{1}{b^{2}}+\frac{1}{a^{2}}+\frac{8}{a^{2}+b^{2}}\right) \\
=c^{2}\left[\frac{1}{a^{2}\left(a^{2}-c^{2}\right)}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,585 |
40. Find the best constant $k$ such that the following inequality holds for all positive real numbers $a, b, c$.
$$\frac{(a+b)(b+c)(c+a)}{a b c}+\frac{k(a b+b c+c a)}{a^{2}+b^{2}+c^{2}} \geq 8+k \quad(\text{Pham Kim Hung})$$ | Solution: We have
$$\begin{array}{l}
\frac{(a+b)(b+c)(c+a)}{a b c}-8=\frac{c(a-b)^{2}+a(b-c)^{2}+b(c-a)^{2}}{a b c} \\
1-\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2\left(a^{2}+b^{2}+c^{2}\right)}
\end{array}$$
So we only need to find a positive constant $k$ satisfying the condition
$$\... | 4 \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,586 |
41. Let $a, b, c$ be positive real numbers, and satisfy the condition $a+b+c+a b c=4$. Prove: $\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \geq \frac{a+b+c}{\sqrt{2}}$ (Cezar Lupu) | Proof: First, we will prove that $a+b+c \geq a b+b c+c a$. In fact, without loss of generality, we can assume $c \geq b \geq a$. We only need to prove
$$a+b-a b \geq \frac{4-a-b}{a b+1}(a+b-1) \Leftrightarrow (a+b-2)^{2} \geq a b(a-1)(b-1)$$
Applying the AM-GM inequality, we immediately get
$$(a+b-2)^{2} \geq 4|(a-1)(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,587 |
42、(i) Prove that for all non-negative real numbers $a, b, c$, we have
$$\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$$
(ii) Using the same conditions, prove
$$\sqrt{\frac{a^{2}+2 b c}{2 a^{2}+b c}}+\sqrt{\frac{b^{2}+2 c a}{2 b^{2}+... | Proof: (i) Since the inequality is homogeneous, we can assume $a b c=1$. The inequality becomes
$$\sqrt{\frac{2 x+1}{x+2}}+\sqrt{\frac{2 y+1}{y+2}}+\sqrt{\frac{2 z+1}{z+2}} \geq 2 \sqrt{2}$$
where $x=a^{3}, y=b^{3}, z=c^{3}, x y z=1$. Without loss of generality, assume $x \geq y \geq z$. Let $t=\sqrt{y z}$, then $t \l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,588 |
43. Let $x, y, z$ be non-negative real numbers, and satisfy $x+y+z=1$. Prove:
$\sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3}$ (Phan Thanh Nam, VME0 2004) | Proof: Let $z=\min \{x, y, z\}$. First, we will prove that if $u=y-z, v=x-z, k=\frac{1}{12}$, then
$$\sqrt{x+k u^{2}}+\sqrt{y+k v^{2}} \leq \sqrt{2(x+y)+k(u+v)^{2}}$$
In fact, this inequality is equivalent to
$$\begin{array}{l}
2 \sqrt{\left(x+k u^{2}\right)\left(y+k v^{2}\right)} \leq x+y+2 k u v \Leftrightarrow 4\le... | \sqrt{3} | Inequalities | proof | Yes | Yes | inequalities | false | 731,589 |
45. Let $a, b, c$ be positive real numbers, prove that
$\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{c}\right)^{2}+\left(2+\frac{c}{a}\right)^{2} \geq \frac{9(a+b+c)^{2}}{a b+b c+c a}$ (Pham Kim Hung) | Prove that the inequality is equivalent to
$$\sum_{c y c} \frac{a^{2}}{b^{2}}+4 \sum_{c y c} \frac{a}{b} \geq \frac{9\left(a^{2}+b^{2}+c^{2}\right)^{2}}{a b+b c+c a}+6$$
Consider the following identities:
$$\begin{array}{l}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{(a-b)^{2}}{a b}+\frac{(c-a)(c-b)}{a c} \\
a^{2}+b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,591 |
46. Let $a, b, c$ be non-negative real numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove:
$\frac{1}{3-a b}+\frac{1}{3-b c}+\frac{1}{3-c a}+\frac{1}{3-a^{2}}+\frac{1}{3-b^{2}}+\frac{1}{3-c^{2}} \geq 3$ (Pham Kim Hung) | Prove that the inequality is equivalent to
$$\sum_{c y c}\left(\frac{3}{3-a b}-1\right)+\sum_{c y c}\left(\frac{3}{3-c^{2}}-1\right) \geq 3 \Leftrightarrow \sum_{c y c} \frac{a b}{3-a b}+\sum_{c y c} \frac{c^{2}}{3-c^{2}} \geq 3$$
Applying the Cauchy-Schwarz inequality, we have
$$L H S \geq \frac{\left(\sum_{c y c} a ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,592 |
47. Let $a, b, c$ be three positive real numbers, prove that
$$\frac{1}{a \sqrt{a+b}}+\frac{1}{b \sqrt{b+c}}+\frac{1}{c \sqrt{c+a}} \geq \frac{3}{\sqrt{2 a b c}} \text { (Phan Thanh Nam, VME0 2005) }$$ | Proof: Let \( x = \sqrt{\frac{2 b c}{a(a+b)}}, y = \sqrt{\frac{2 c a}{b(b+c)}}, z = \sqrt{\frac{2 a b}{c(c+a)}} \). We need to prove that \( x + y + z \geq 3 \). However, we can prove the following stronger inequality:
\[ 3 \leq x y + y z + z x = \frac{2 c}{\sqrt{(a+b)(b+c)}} + \frac{2 a}{\sqrt{(b+c)(c+a)}} + \frac{2 b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,593 |
48. If real numbers $a, b, c, x, y, z$ satisfy the conditions $(a+b+c)(x+y+z)=3$; $\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)=4$, prove: $a x+b y+c z \geq 0$ (Math1inks Contest) | Proof: Let $\alpha=\sqrt[4]{\frac{a^{2}+b^{2}+c^{2}}{x^{2}+y^{2}+z^{2}}}$ and $a_{1}=\frac{a}{\alpha}, b_{1}=\frac{b}{\alpha}, c_{1}=\frac{c}{\alpha}, x_{1}=x \alpha, y_{1}=y \alpha, z_{1}=z \alpha$, then we have
$$\begin{array}{l}
a_{1}^{2}+b_{1}^{2}+c_{1}^{2}=\frac{a^{2}+b^{2}+c^{2}}{\alpha^{2}}=\sqrt{\left(a^{2}+b^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,594 |
Example 1.2.5 Let $a, b, c>0, a+b+c=3$, prove: $\frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1$ | Prove: Using the same technique as in Example 1.2.4, we only need to prove
$$b \sqrt[3]{a^{2}}+c \sqrt[3]{b^{2}}+a \sqrt[3]{c^{2}} \leq 3$$
According to the AM-GM inequality, we have,
$$3 \sum_{c y c} a \geq \sum_{c y c} a+2 \sum_{c y c} a b=\sum_{c y c}(a+a c+a c) \geq 3 \sum_{c y c} a \sqrt[3]{c^{2}}$$
Equality hol... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,595 |
49. Let $a, b, c, d$ be non-negative real numbers, and satisfy $a+b+c+d=4$. Prove: $\sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4$ (Pham Kim Hung) | Prove: According to the AM-GM inequality, we have
$$L H S \geq 4 \sqrt{\frac{(a+1)(b+1)(c+1)(d+1)}{(a b+1)(b c+1)(c d+1)(d a+1)}}$$
Thus, it suffices to prove
$$(a+1)(b+1)(c+1)(d+1) \geq(a b+1)(b c+1)(c d+1)(d a+1)$$
Expanding, the inequality becomes
$$\begin{array}{l}
a b c d+\sum_{s y m} a b c+\sum_{s y m} a b+\sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,596 |
50. Let $a, b, c$ be non-negative real numbers, prove: $\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}} \geq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{2}}$ | Proof: Let $x=\sqrt{a}, y=\sqrt{b}, z=\sqrt{c}$, then the inequality becomes
$$\begin{array}{l}
\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}+\frac{y^{2}}{\sqrt{y^{2}+z^{2}}}+\frac{z^{2}}{\sqrt{z^{2}+x^{2}}} \geq \frac{x+y+z}{\sqrt{2}} \\
\Leftrightarrow \sum_{c y c} \frac{2 x^{4}}{x^{2}+y^{2}}+\sum_{c y c} \frac{4 x^{2} y^{2}}{\sq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,597 |
51. Let $a, b, c$ be real numbers, prove: $\sqrt[3]{2 a^{2}-b c}+\sqrt[3]{2 b^{2}-c a}+\sqrt[3]{2 c^{2}-a b} \geq 0$ | Proof: First, note that the inequality only needs to consider $a, b, c$ as non-negative real numbers. Consider the identity
$$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$$
Therefore, $(x+y+z)\left(x^{3}+y^{3}+z^{3}-3 x y z\right) \geq 0$, so we can rewrite the inequality as
$$\begin{arr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,598 |
53. Let $a, b, c$ be three real numbers, and satisfy the condition $a^{2}+b^{2}+c^{2}=9$. Prove:
$3 \min (a, b, c) \leq 1+abc$ (Virgil Nicula) | Proof: Without loss of generality, we assume $c \geq b \geq a$. Consider the following cases:
( i ) $a \leq 0$: Let $d = -a$, $e = |b|$. We will prove
$$-3d \leq 1 - dce \Leftrightarrow d(ce - 3) \leq 1$$
If $ce \leq 3$, the inequality is obviously true. Otherwise, if $ce > 3$, then
$$d^2(ce - 3)(ce - 3) \leq \left(\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,600 |
54. Let $a, b, c, d$ be non-negative real numbers, and satisfy $a+b+c+d=4$. Prove:
$\left(1+a^{4}\right)\left(1+b^{4}\right)\left(1+c^{4}\right)\left(1+d^{4}\right) \geq\left(1+a^{3}\right)\left(1+b^{3}\right)\left(1+c^{3}\right)\left(1+d^{3}\right)$ (Pham Kim Hung) | Proof: Note that when $x \geq 0$, $\left(1+x^{4}\right)(1+x) \geq\left(1+x^{3}\right)\left(1+x^{2}\right)$, therefore
$$\prod_{c y c}\left(1+a^{4}\right) \prod_{c y c}(1+a) \geq \prod_{c y c}\left(1+a^{3}\right) \prod_{c y c}\left(1+a^{2}\right)$$
Thus, it suffices to prove
$$\prod_{c y c}\left(1+a^{2}\right) \geq \pr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,601 |
55. Find the best constant $k$ (minimum value) such that the following inequality holds for non-negative real numbers $a, b, c$ satisfying $a+b+c=3$.
$$a^{k}+b^{k}+c^{k} \geq a b+b c+c a$$
(Genralization Of Russia M0 2000) | Solution: In Example 1.1.1 of Chapter 1, the inequality has already been proven when $k=\frac{1}{2}$. Therefore, for $k \geq \frac{1}{2}$, it holds. Now consider the case when $k \leq \frac{1}{2}$.
Lemma: Let $a, b \geq 0, a+b=2 t \geq 1$, then we have $a^{k}+b^{k}-a b \geq \min \left\{(2 t)^{k}, 2 t^{k}-t^{2}\right\}... | k=\frac{2 \ln 3-3 \ln 2}{\ln 3-\ln 2} \approx 0.2905 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,602 |
56. Let $a, b, c$ be positive real numbers, prove: $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \sqrt{\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}}$ (Vo Quoc Ba Can) | Proof: Note that, if $a \geq b \geq c$, then
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)=\frac{(a-b)(a-c)(c-b)}{a b c} \leq 0$$
Therefore, we only need to consider the case $a \geq b \geq c$. Squaring both sides, we have
$$\sum_{c y c} \frac{a^{2}}{b^{2}}+\sum_{c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,603 |
57. Let $a, b, c, d$ be positive real numbers, and satisfy $a^{2}+b^{2}+c^{2}+d^{2}=4$, prove: $\frac{1}{3-a b c}+\frac{1}{3-b c d}+\frac{1}{3-c d a}+\frac{1}{3-d a b} \leq 2$ (Pham Kim Hung) | Proof: Let $x=abc, y=bcd, z=acd, t=bcd$, then the inequality becomes
$$\sum_{\text {cyc }} \frac{1}{3-x} \leq 2 \Leftrightarrow \sum_{c y c} \frac{1-x}{3-x} \geq 0$$
According to the AM-GM inequality, we have
$$x+y=ab(c+d) \leq \frac{1}{2}\left(a^{2}+b^{2}\right) \sqrt{2\left(c^{2}+d^{2}\right)} \leq\left(\frac{4}{3}\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,605 |
Example 1.2.6 Let $a, b, c>0, a+b+c=3$, prove: $\frac{a+1}{b^{2}+1}+\frac{b+1}{c^{2}+1}+\frac{c+1}{a^{2}+1} \geq 3$ | Prove: We use the following estimation
$$\frac{a+1}{b^{2}+1}=a+1-\frac{b^{2}(a+1)}{b^{2}+1} \geq a+1-\frac{b^{2}(a+1)}{2 b}=a+1-\frac{a b+b}{2}$$
For similar results about $a, b, c$, adding them up, we have
$$\sum_{c y c} \frac{a+1}{b^{2}+1} \geq 3+\frac{1}{2} \sum_{c y c} a-\frac{1}{2} \sum_{c y c} a b \geq 3$$
Note... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,606 |
58. Let $n$ positive real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy the condition $\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+\frac{1}{1+x_{n}}=\frac{n}{2}$. Prove: $\sum_{i, j=1}^{n} \frac{1}{x_{i}+x_{j}} \geq \frac{n^{2}}{2}$ (Titu Andreescu, Gabriel Dospinescu) | Prove: For $i \in\{1,2, \cdots, n\}$, we denote $a_{i}=\frac{1-x_{i}}{1+x_{i}}$. Then $a_{1}+a_{2}+\cdots+a_{n}=0$ and $a_{i} \in[-1,1]$.
Consider the expression
$$S=\sum_{i, j=1}^{n} \frac{1}{x_{i}+x_{j}} \Rightarrow 2 S=\sum_{i, j=1}^{n} \frac{\left(1+a_{i}\right)\left(1+a_{j}\right)}{1-a_{i} a_{j}}$$
We have
$$P=\s... | S \geq \frac{n^{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 731,607 |
60. Assume $n$ is an integer greater than 2. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and satisfy $a_{1} a_{2} \cdots a_{n}=1$. Prove: $\frac{a_{1}+3}{\left(a_{1}+1\right)^{2}}+\frac{a_{2}+3}{\left(a_{2}+1\right)^{2}}+\cdots+\frac{a_{n}+3}{\left(a_{n}+1\right)^{2}} \geq 3$ (United Kingdom TST 2005) | Proof: First, note that it is sufficient to prove the inequality for the case $n=3$. For $n \geq 4$, we can select three smaller elements from the set $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, say $a_{1}, a_{2}, a_{3}$. Since $a_{1} a_{2} a_{3} \leq 1$, there exists a positive number $k$ such that $\frac{a}{a_{1}}=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,609 |
61. Let $a, b, c$ be non-negative real numbers, and satisfy $a+b+c=2$. Prove:
$\sqrt{a+b-2ab}+\sqrt{b+c-2bc}+\sqrt{c+a-2ca} \geq 2$ (Pham Kim Hung) | Proof: Without loss of generality, we assume $a \geq b \geq c$. Let $x=a+b-2 a b, y=b+c-2 b c, z=c+a-2 c a$, then the inequality is equivalent to (after squaring)
$$2 \sum_{c y c} \sqrt{x y} \geq 2 \sum_{c y c} a b$$
Notice that $2 x=c(a+b)+(a-b)^{2}, 2 y=a(b+c)+(b-c)^{2}$, so by the Cauchy-Schwarz inequality, we have... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,610 |
62. Let $a, b, c$ be non-negative real numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove:
$$\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} \leq 1 \text { (Vasile Cirtoaje) }$$ | Prove: By eliminating the denominator and rearranging, the inequality is equivalent to
$$a b^{2}+b c^{2}+c a^{2} \leq 2+a b c$$
Without loss of generality, assume $b$ is the second largest number in the set $\{a, b, c\}$, then
$$a(b-a)(b-c) \leq 0 \Leftrightarrow a^{2} b+a b c \geq a b^{2}+c a^{2}$$
Therefore, we onl... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,611 |
63. Let $a, b, c$ be non-negative real numbers, prove:
$\frac{a(b+c)}{a^{2}+b c}+\frac{b(c+a)}{b^{2}+c a}+\frac{c(a+b)}{c^{2}+a b} \geq 2$ (Pham Kim Hung) | Prove that the inequality is equivalent to
$$\begin{array}{l}
\sum_{c y c} a(b+c)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \geq 2\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right) \\
\Leftrightarrow \sum_{c y c} a^{4}\left(b^{2}+c^{2}\right)+3 a b c \sum_{c y c} a^{2}(b+c) \geq 4 a^{2} b^{2} c^{2}+2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,612 |
64. Let $a, b, c$ be the three sides of a triangle, prove:
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \leq \frac{5}{2}$$ | Prove: Using the following identities
$$\begin{array}{l}
-3+\sum_{c y c} \frac{2 a}{b+c}=\sum_{\text {cyc }} \frac{(a-b)^{2}}{(a+c)(b+c)} \\
2-\frac{2(a b+b c+c a)}{a^{2}+b^{2}+c^{2}}=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{a^{2}+b^{2}+c^{2}}
\end{array}$$
We can write the inequality in the form: \( S_{a}(b-c)^{2}+S_{b}(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,613 |
66. Let $a, b, c$ be positive real numbers, prove:
$$\frac{a^{3}}{2 a^{2}-a b+2 b^{2}}+\frac{b^{3}}{2 b^{2}-b c+2 c^{2}}+\frac{c^{3}}{2 c^{2}-c a+2 a^{2}} \geq \frac{a+b+c}{3} \text { (Nguyen Vient Anh ) }$$ | Prove that the inequality is equivalent to
$$\begin{array}{l}
\sum_{c y c} \frac{a^{3}}{2 a^{2}-a b+2 b^{2}}-\frac{1}{3} \sum_{c y c} a=\sum_{c y c} \frac{a\left(a^{2}+a b-2 b^{2}\right)}{3\left(2 a^{2}-a b+2 b^{2}\right)}=\sum_{c y c}(a-b)\left[\frac{a(2 a+b)}{3\left(2 a^{2}-a b+2 b^{2}\right)}-\frac{1}{3}\right] \\
=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,615 |
67. Proof: For all positive real numbers $a, b, c, \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geq 3 \sqrt[4]{\frac{a^{4}+b^{4}+c^{4}}{3}}$ | Prove: By applying the Hölder's inequality, we have
$$\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \geq\left(a^{2}+b^{2}+c^{2}\right)^{3}$$
Let $x=a^{2}, y=b^{2}, z=c^{2}$, then we only need to p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,616 |
Example 1.2.7 Let $a, b, c>0, a+b+c=3$, prove: $\frac{1}{1+2 b^{2} c}+\frac{1}{1+2 c^{2} a}+\frac{1}{1+2 a^{2} b} \geq 1$ | Prove: We use the following estimation
$$\frac{1}{1+2 b^{2} c}=1-\frac{2 b^{2} c}{1+2 b^{2} c} \geq 1-\frac{2 \sqrt[3]{b^{2} c}}{3} \geq 1-\frac{2(2 b+c)}{9}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,617 |
68. Let non-negative real numbers $a, b, c$ satisfy the condition $a+b+c=3$. Prove:
$\left(a+b^{2}\right)\left(b+c^{2}\right)\left(c+a^{2}\right) \leq 13+a b c$
(Pham Kim Hung) | Proof: We first prove that if $a \geq b \geq c$, then
$$\left(a+b^{2}\right)\left(b+c^{2}\right)\left(c+a^{2}\right) \geq\left(a^{2}+b\right)\left(b^{2}+c\right)\left(c^{2}+a\right)$$
In fact, note that
$$\begin{array}{l}
\sum_{c y c} a^{3} b-\sum_{c y c} a b^{3}=(a+b+c)(a-b)(b-c)(a-c) \\
\sum_{c y c} a^{2} b^{3}-\sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,618 |
69. Let $a, b, c$ be positive real numbers. Prove:
$\frac{(a+b)^{2}}{c^{2}+a b}+\frac{(b+c)^{2}}{a^{2}+b c}+\frac{(c+a)^{2}}{b^{2}+a c} \geq 6$ (Peter Scholze, Darij Grinberg) | Prove: We have $(a+b)^{2}-2\left(c^{2}+a b\right)=\left(a^{2}-c^{2}\right)+\left(b^{2}-c^{2}\right)$, therefore
$$\begin{array}{l}
\frac{(a+b)^{2}}{c^{2}+a b}+\frac{(b+c)^{2}}{a^{2}+b c}+\frac{(c+a)^{2}}{b^{2}+a c}-6=\sum_{c y c} \frac{\left(a^{2}-b^{2}\right)+\left(a^{2}-c^{2}\right)}{a^{2}+b c} \\
=\sum_{c y c}\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,619 |
70. Find the maximum value of the expression $k=k(n)$, such that for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, the following inequality always holds. $x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \geq k\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)$ (Le Hong Quy) | Proof: Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, then
$$\begin{array}{l}
a_{1} y_{1}^{2}+\frac{1}{a_{1}} \cdot y_{2}^{2}+2 y_{1} y_{2} \geq 0 \\
a_{2} y_{2}^{2}+\frac{1}{a_{2}} \cdot y_{3}^{2}+2 y_{2} y_{3} \geq 0 \\
\ldots \ldots \ldots \ldots \ldots \ldots \\
a_{n-1} y_{n-1}^{2}+\frac{1}{a_{n}} \cdo... | \frac{1}{4 \cos ^{2} \frac{\pi}{2 n+1}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,620 |
71. Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive real numbers, and $a_{1}+a_{2}+\cdots+a_{n}=n$, prove:
$$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots \frac{1}{a_{n}}-n \geq \frac{8(n-1)\left(1-a_{1} a_{2} \cdots a_{n}\right)}{n^{2}} \text { (Pham Kim Hung) }$$ | Proof: We use induction to prove this inequality. If $n=2$, the inequality becomes
$$\frac{1}{a_{1}}+\frac{1}{a_{2}}-2 \geq 2\left(1-a_{1} a_{2}\right) \Leftrightarrow\left(1-a_{1} a_{2}\right)^{2} \geq 0 \text { . }$$
Assume the inequality holds for $n$, we will prove that it also holds for $n+1$. Assume $a_{1} \leq ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,621 |
72. Let non-negative real numbers $x, y, z$ satisfy the condition $x+y+z=1$. Prove:
$\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}} \geq 2$ (Phan Thanh Nam) | Proof: Note that, if $a, b, c, d$ are non-negative real numbers, and satisfy $a+b=c+d,|a-b|=|c-d|$, then we have
$$\sqrt{a}+\sqrt{b} \geq \sqrt{c}+\sqrt{d}$$
In fact, because $(a+b)^{2}-(a-b)^{2} \geq(c+d)^{2}-(c-d)^{2}$, we have $a b \geq c d$. Therefore,
$$a+b+2 \sqrt{a b} \geq c+d+2 \sqrt{c d}$$
According to $(*)$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,622 |
73. Let positive real numbers $a, b, c$ satisfy the condition $a+b+c=3$, prove:
$\frac{1}{2+a^{2} b^{2}}+\frac{1}{2+b^{2} c^{2}}+\frac{1}{2+c^{2} a^{2}} \geq 1$ (Pham Kim Hung) | Prove: According to the AM-GM inequality, we have
$$\frac{1}{2+a^{2} b^{2}}=\frac{1}{2}-\frac{a^{2} b^{2}}{2\left(2+a^{2} b^{2}\right)} \geq \frac{1}{2}-\frac{a^{2} b^{2}}{6 \sqrt[3]{a^{2} b^{2}}}=\frac{1}{2}-\frac{a^{\frac{4}{3}} b^{\frac{4}{3}}}{6}$$
We have
$$\sum_{c y c} \frac{1}{2+a^{2} b^{2}} \geq \frac{3}{2}-\f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,623 |
74. Consider positive real constants $m, n$, satisfying $3 n^{2}>m^{2}$. Real numbers $a, b, c$ satisfy the conditions: $a+b+c=m, a^{2}+b^{2}+c^{2}=n^{2}$. Find the maximum and minimum values of the expression $P$. $P=a^{2} b+b^{2} c+c^{2} a \quad$ (Le Trung Kien, Vo Quoc Ba Can) | Solution: Let $a=x+\frac{m}{3}, b=y+\frac{m}{3}, c=z+\frac{m}{3}$. From the given conditions, we get $x+y+z=0$ and $x^{2}+y^{2}+z^{2}=\frac{3 n^{2}-m^{2}}{3}$, then the expression P becomes
$$P=x^{2} y+y^{2} z+z^{2} x+\frac{m^{3}}{9}$$
Notice that
$$\begin{array}{l}
\sum_{c y c}\left(3 x \sqrt{\frac{2}{3 n^{2}-m^{2}}}... | \max P=\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{\frac{3}{2}}+\frac{m^{3}}{9}, \min P=-\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{\frac{3}{2}}-\frac{m^{3}}{9} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,624 |
75. Let positive real numbers $a, b, c$ satisfy the condition $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=13$. Find the maximum and minimum values of the following expression.
$$P=\frac{a^{3}+b^{3}+c^{3}}{a b c} \text { (Pham Kim Hung) }$$ | Let $x=\sum_{c y c} \frac{a}{b} ; y=\sum_{c y c} \frac{b}{a} ; m=\sum_{c y c} \frac{a^{2}}{b c} ; n=\sum_{c y c} \frac{b c}{a^{2}}$; then we have $x+y=10$ and
$$x^{3}=3(m+n)+6+\sum_{c y c} \frac{a^{3}}{b^{3}} ; \quad y^{3}=3(m+n)+6+\sum_{c y c} \frac{b^{3}}{a^{3}}$$
From the above identities, we have
$$x^{3}+y^{3}=\le... | 11 \pm 2\sqrt{3} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,625 |
76. Prove for all positive real numbers $a, b, c, d, e$,
$$\frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+d}{2} \cdot \frac{d+e}{2} \cdot \frac{e+a}{3} \leq \frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \frac{c+d+e}{3} \cdot \frac{d+e+a}{3} \cdot \frac{e+a+b}{3} .$$ | Proof: We first prove that for any $a, b>0, a+b \leq 1$
$$\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right) \geq\left(\frac{2}{a+b}-1\right)^{2}$$
In fact, this result can be expressed in the following form
$$\begin{array}{l}
\frac{1}{a b}-\frac{1}{a}-\frac{1}{b} \geq \frac{4}{(a+b)^{2}}-\frac{4}{a+b} \Leftrightarr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,626 |
Example 1.2.8 Let $a, b, c, d \geq 0, a+b+c+d=4$, prove:
$\frac{1+a b}{1+b^{2} c^{2}}+\frac{1+b c}{1+c^{2} d^{2}}+\frac{1+c d}{1+d^{2} a^{2}}+\frac{1+d a}{1+a^{2} b^{2}} \geq 4 \quad$ (Pham Kim Hung) | Prove: By applying the AM-GM inequality, we have
$$\frac{1+a b}{1+b^{2} c^{2}}=(1+a b)-\frac{(1+a b) b^{2} c^{2}}{1+b^{2} c^{2}} \geq 1+a b-\frac{1}{2}(1+a b) b c$$
By adding similar results, we have
$$\sum_{c y c} \frac{1+a b}{1+b^{2} c^{2}} \geq 4+\sum_{c y c} a b-\frac{1}{2} \sum_{c y c} b c(1+a b)=4+\frac{1}{2}\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,628 |
78. Let $a, b, c > 0$, prove:
$\sqrt{\frac{a^{3}}{a^{2}+a b+b^{2}}}+\sqrt{\frac{b^{3}}{b^{2}+b c+c^{2}}}+\sqrt{\frac{c^{3}}{c^{2}+c a+a^{2}}} \geq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{3}}$ (Le Trung Kien) | Proof: Let $x^{2}=a, y^{2}=b, z^{2}=c$, then the inequality becomes
$$\frac{x^{3}}{\sqrt{x^{4}+x^{2} y^{2}+y^{4}}}+\frac{y^{3}}{\sqrt{y^{4}+y^{2} z^{2}+z^{4}}}+\frac{z^{3}}{\sqrt{z^{4}+z^{2} x^{2}+x^{4}}} \geq \frac{x+y+z}{\sqrt{3}}$$
Squaring both sides, we get the equivalent form
$$\sum_{c y c} \frac{x^{6}}{x^{4}+x^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,629 |
79. Let $a, b, c \geq 0, a+b+c=2$, prove:
$$\frac{a b}{1+c^{2}}+\frac{b c}{1+a^{2}}+\frac{c a}{1+b^{2}} \leq 1 \quad \text { (Pham Kim Hung) }$$ | Proof: Let $x=ab+bc+ca, p=abc$. According to the identity
$$\begin{array}{l}
A=(a-b)^{2}(b-c)^{2}(c-a)^{2}=4 x^{2}(1-x)+4(9 x-8) p-27 p^{2}, \\
B=\sum_{cyc} a^{2}(a-b)(a-c)=12 p+4(1-x)(4-x),
\end{array}$$
our inequality is transformed into
$$\begin{array}{l}
(1-x)\left(5-2 x+x^{2}\right)+(6 x-2) p-2 p^{2} \geq 0 \\
\L... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,630 |
80. Let $a_{1}, a_{2}, \cdots, a_{n}$ be non-negative real numbers, and satisfy $a_{1}+a_{2}+\cdots+a_{n}=n$. Find the minimum value of the expression $S$. $S=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}+a_{1} a_{2} \cdots a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots \frac{1}{a_{n}}\right)$ (Pham kim Hung) | Consider the following function
$$F=f\left(a_{1}, a_{2}, \cdots, a_{n}\right)=a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}+a_{1} a_{2} \cdots a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots \frac{1}{a_{n}}\right)$$
We have the following interesting identity:
$$\begin{array}{l}
f\left(a_{1}, a_{2}, \cdots, a_{n}\right)-f\lef... | \min (2 n, \frac{n^{2}}{n-2}, \frac{n^{2}}{n-1}+\left(\frac{n}{n-1}\right)^{n-1}) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,631 |
81. Let positive real numbers $x, y, z$ satisfy the condition $2 x y z=3 x^{2}+4 y^{2}+5 z^{2}$, find the minimum value of the expression $P=3 x+2 y+z$.
| Solution: Let $a=3 x, b=2 y, c=z$, then we have
$$a+b+c=3 x+2 y+z, \quad a^{2}+3 b^{2}+15 c^{2}=a b c$$
According to the weighted AM-GM inequality, we have
$$\begin{array}{l}
a+b+c \geq(2 a)^{\frac{1}{2}}(3 b)^{\frac{1}{3}}(6 c)^{\frac{1}{6}} \\
a^{2}+3 b^{2}+15 c^{2} \geq\left(4 a^{2}\right)^{\frac{1}{4}}\left(9 b^{2... | 36 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 731,632 |
Let $a, b, c, x, y, z$ be positive real numbers, and satisfy the condition $a x^{2}+b y^{2}+c z^{2}=x y z$
(a) Prove that there exists a positive real number $k$ such that
$$\frac{1}{2 \sqrt{k}}=\frac{1}{\sqrt{k}+\sqrt{k+a}}+\frac{1}{\sqrt{k}+\sqrt{k+b}}+\frac{1}{\sqrt{k}+\sqrt{k+c}}$$
(b) Using this $k$ value, prove t... | (a) This part is quite simple. Consider the following function
$$f(k)=\frac{\sqrt{k}}{\sqrt{k}+\sqrt{k+a}}+\frac{\sqrt{k}}{\sqrt{k}+\sqrt{k+b}}+\frac{\sqrt{k}}{\sqrt{k}+\sqrt{k+c}}-\frac{1}{2}$$
Since $f(k)$ is an increasing function of $k$, and $f(0)=-\frac{1}{2}, \lim _{k \rightarrow \infty} f(k)=1$, by the continui... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,633 |
82. Let $a, b, c$ be non-negative real numbers, prove:
$\frac{1}{\sqrt{a^{2}+b c}}+\frac{1}{\sqrt{b^{2}+c a}}+\frac{1}{\sqrt{c^{2}+a b}} \geq \frac{2 \sqrt{2}}{\sqrt{a b+b c+c a}}$ (Pham Kim Hung) | Proof: First, we assume $a \geq b \geq c$. Notice that
$$\frac{1}{\sqrt{b^{2}+c a}}+\frac{1}{\sqrt{c^{2}+a b}} \geq \frac{2 \sqrt{2}}{\sqrt{b^{2}+c^{2}+a b+a c}}$$
Thus, it suffices to prove
$$\frac{1}{\sqrt{a^{2}+b c}}+\frac{2 \sqrt{2}}{\sqrt{b^{2}+c^{2}+a b+a c}} \geq \frac{2 \sqrt{2}}{\sqrt{a b+b c+c a}}$$
Let $M=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,634 |
85. Let $a, b, c$ be positive real numbers. Consider the following inequality:
$$\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b} \geq 3 \sqrt[k]{\frac{a^{k}+b^{k}+c^{k}}{3}}$$
(a) Prove (*) for $k=2$.
(b) Prove that (*) does not hold for $k=3$, but it does hold for $k=3$ and $a, b, c$ satisfying the condition $a^{3} b^{3}+b^... | Prove: (a) For $k=2$, without loss of generality, we assume $\sum_{c y c} a^{2}=3$. Then the inequality $\sum_{c y c} \frac{a b}{c} \geq 3$ is equivalent to
$$\left(\sum_{c y c} \frac{a b}{c}\right)^{2} \geq 9 \Leftrightarrow \sum_{c y c} \frac{a^{2} b^{2}}{c^{2}} \geq 3 \Leftrightarrow \sum_{c y c} a^{2}\left(\frac{b^... | k \leq \frac{\ln 3}{\ln 3-\ln 2} \approx 2.709511 \cdots | Inequalities | proof | Yes | Yes | inequalities | false | 731,637 |
Example 1.2.9 Let $a, b, c>0, a^{2}+b^{2}+c^{2}=3$, prove: $\frac{1}{a^{3}+2}+\frac{1}{b^{3}+2}+\frac{1}{c^{3}+2} \geq 1$ | Prove: According to the AM-GM inequality, we have
$$\sum_{c y c} \frac{1}{a^{3}+2}=\frac{3}{2}-\frac{1}{2} \sum_{c y c} \frac{a^{3}}{a^{3}+1+1} \geq \frac{3}{2}-\frac{1}{2} \sum_{c y c} \frac{a^{3}}{3 a}=1$$ | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 731,638 |
86. Let $a, b, c$ be non-negative real numbers, prove that
$$\frac{1}{\sqrt{4 a^{2}+b c}}+\frac{1}{\sqrt{4 b^{2}+c a}}+\frac{1}{\sqrt{4 c^{2}+a b}} \geq \frac{4}{a+b+c} \text{ (Pham Kim Hung) }$$ | Prove: (1) We denote
$$S=\sum_{c y c} \frac{1}{\sqrt{4 a^{2}+b c}} ; \quad P=\sum_{c y c}(b+c)^{3}\left(4 a^{2}+b c\right)$$
According to the Hölder inequality, we have
$$S \cdot S \cdot P \geq(a+b+c)^{3}$$
Therefore, it suffices to prove
$(a+b+c)^{5} \geq 2 P$. Since
$$\begin{array}{l}
P=\sum_{c y c} a^{4}(b+c)+7 \s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,639 |
87. Let $a, b, c$ be non-negative real numbers, prove:
$\sqrt{\frac{a b}{4 a^{2}+b^{2}+4 c^{2}}}+\sqrt{\frac{b c}{4 b^{2}+c^{2}+4 a^{2}}}+\sqrt{\frac{c a}{4 c^{2}+a^{2}+4 b^{2}}} \leq 1$ (Pham Kim Hung) | Proof: Without loss of generality, assume $a^{2}+b^{2}+c^{2}=3$. By the weighted Jensen's inequality, we have
$$\begin{array}{l}
\sum_{c y c} \sqrt{\frac{a b}{4 a^{2}+b^{2}+4 c^{2}}}=\sum_{c y c} \frac{a^{2}+4 b^{2}+4 c^{2}}{27} \cdot \sqrt{\frac{27^{2} \cdot a b}{\left(4 a^{2}+b^{2}+4 c^{2}\right)\left(a^{2}+4 b^{2}+4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,640 |
88. Let $n$ be a positive integer, and $\left(x_{1}, x_{2}, \cdots, x_{n}\right) ;\left(y_{1}, y_{2}, \cdots, y_{n}\right)$ be two sequences of positive numbers. Let $\left(z_{1}, z_{2}, \cdots, z_{n}\right)$ be a sequence of positive numbers such that $z_{i+j}^{2} \geq x_{i} y_{j} \quad(1 \leq i, j \leq n)$, and let $... | Proof: Let \( X = \max \left\{x_{1}, x_{2}, \cdots, x_{n}\right\}, Y = \max \left\{y_{1}, y_{2}, \cdots, y_{n}\right\} \). Without loss of generality, we can assume \( X = Y = 1 \) (otherwise, replace \( x_{i} \) with \( \frac{x_{i}}{X} \), \( y_{i} \) with \( \frac{y_{i}}{Y} \), and \( z_{i} \) with \( \frac{z_{i}}{\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,641 |
89、( a ) Let $a, b, c$ be three real numbers. Prove:
$$a^{4}+b^{4}+c^{4}+a b^{3}+b c^{3}+c a^{3} \geq 2\left(a^{3} b+b^{3} c+c^{3} a\right)$$
( b ) Let $a, b, c$ be three real numbers, and satisfy $a^{2}+b^{2}+c^{2}+a b+b c+c a=6$. Prove:
$$a^{3} b+b^{3} c+c^{3} a+a b c(a+b+c) \leq 6$$
(c) Find the best (largest) const... | Prove: For all real numbers $a, b, c$, we have
$$\left(a^{2}-k a b+k a c-c^{2}\right)^{2}+\left(b^{2}-k b c+k b a-a^{2}\right)^{2}+\left(c^{2}-k c a+k c b-b^{2}\right)^{2} \geq 0$$
Expanding this inequality, it becomes
$$\sum_{c y c} a^{4}+\left(k^{2}-1\right) \sum_{c y c} a^{2} b^{2}+k \sum_{c y c} a b^{3} \geq 2 k \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,642 |
91. Let $a, b, c$ be positive real numbers, and satisfy $a^{2}+b^{2}+c^{2}=3$, prove:
$$a^{3} b^{2}+b^{3} c^{2}+c^{3} a^{2} \leq 3$$ | Prove that by the Cauchy-Schwarz inequality, we have
$$\left(a^{3} b^{2}+b^{3} c^{2}+c^{3} a^{2}\right)^{2} \leq\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)\left(a^{4} b^{2}+b^{4} c^{2}+c^{4} a^{2}\right)$$
For the remaining proof, if \(x + y + z = 3\), then
$$(x y + y z + z x)\left(x^{2} y + y^{2} z + z^{2} x\rig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,644 |
92. Let $a, b, c$ be any positive real numbers, prove that $\left(a+\frac{b^{2}}{c}\right)^{2}+\left(b+\frac{c^{2}}{a}\right)^{2}+\left(c+\frac{a^{2}}{b}\right)^{2} \geq \frac{12\left(a^{3}+b^{3}+c^{3}\right)}{a+b+c}$ (Pham Kim Hung) | Prove that the inequality is equivalent to
$$a^{2}+b^{2}+c^{2}+\frac{2 a b^{2}}{c}+\frac{2 b c^{2}}{a}+\frac{2 c a^{2}}{b}+\frac{b^{4}}{c^{2}}+\frac{c^{4}}{a^{2}}+\frac{a^{4}}{b^{2}} \geq \frac{12\left(a^{3}+b^{3}+c^{3}\right)}{a+b+c}$$
Using the following identities:
$$\begin{array}{l}
\sum_{c y c} \frac{b^{4}}{c^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,645 |
93. Assume $n$ is a positive integer greater than 2, and $a_{1}, a_{2}, \cdots, a_{n}$ are $n$ real numbers. Prove that for any non-empty subset $S$ of the set $\{1,2, \cdots, n\}$, we have
$$\left(\sum_{i \in S} a_{i}\right)^{2} \leq \sum_{1 \leq i \leq j \leq n}\left(a_{i}+\cdots+a_{j}\right)^{2} \quad(\text{Gabriel ... | Proof: We first prove a lemma.
Lemma: For all real numbers $x_{1}, x_{2}, \cdots, x_{2 k+1}$, we have
$$\left(\sum_{0 \leq i \leq k} x_{2 i+1}\right)^{2} \leq \sum_{1 \leq i \leq j \leq 2 k+1}\left(x_{i}+\cdots+x_{j}\right)^{2}$$
Proof: Let $s_{i}=x_{1}+x_{2}+\cdots+x_{i} \quad i \in\{1,2, \cdots, k\}$, then
$$\sum_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,646 |
94. Let $a, b, c$ be non-negative real numbers, prove:
$\frac{a^{3}}{(a+b)^{3}}+\frac{b^{3}}{(b+c)^{3}}+\frac{c^{3}}{(c+a)^{3}}+\frac{5 a b c}{(a+b)(b+c)(c+a)} \geq 1 \quad$ (Pham Kim Hung) | Prove: The inequality can be rewritten in the following form
$$\frac{1}{(1+x)^{3}}+\frac{1}{(1+y)^{3}}+\frac{1}{(1+z)^{3}}+\frac{5}{(1+x)(1+y)(1+z)} \geq 1$$
where \( x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c} \Rightarrow x y z=1 \), we denote
$$m=1-\frac{2}{1+x}, n=1-\frac{2}{1+y}, p=1-\frac{2}{1+z} \quad m, n, p \i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,647 |
Let $a, b, c$ be non-negative real numbers, prove:
$$\frac{a^{2}}{(a+b)^{2}}+\frac{b^{2}}{(b+c)^{2}}+\frac{c^{2}}{(c+a)^{2}}+\frac{2 a b c}{(a+b)(b+c)(c+a)} \geq 1$$ | We use the Cauchy-Schwarz inequality to prove this problem. In fact, the inequality is equivalent to (after substitution)
$$\sum_{c y c} \frac{x^{4}}{\left(x^{2}+y z\right)^{2}}+\frac{2 x^{2} y^{2} z^{2}}{\left(x^{2}+y z\right)\left(y^{2}+z x\right)\left(z^{2}+x y\right)} \geq 1$$
By the Cauchy-Schwarz inequality, we ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,648 |
95. Assume $a, b, c$ are any positive real numbers, prove:
$\left(a^{3}+b^{3}+c^{3}\right)^{2} \geq 2\left(a^{5} b+b^{5} c+c^{5} a\right)+a b c\left(a^{3}+b^{3}+c^{3}\right) \quad$ (Pham kim hung, Le huu Dien Khue) | Prove: The inequality can be rewritten in the following form
$$\begin{array}{l}
\sum_{c y c} a^{6}+2 \sum_{c y c} a^{3} b^{3} \geq 2 \sum_{c y c} a^{5} b+a b c \sum_{c y c} a^{3} \\
\Leftrightarrow 2\left(\sum_{c y c} a^{6}+\sum_{c y c} a^{4} b^{2}-2 \sum_{c y c} a^{5} b\right)+\left(\sum_{c y c} a^{4} b^{2}+\sum_{c y ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,650 |
96. Let $a, b, c, d$ be non-negative real numbers, and satisfy $(a+b+c+d)^{2}=3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)$, prove the following inequality:
$a^{4}+b^{4}+c^{4}+d^{4} \geq 28 a b c d$ (Pham kim hung) | Proof: Let $m=a+b, n=c+d, x=ab, y=cd$, then $(m+n)^{2}=3\left(m^{2}+n^{2}-2x-2y\right)$ or $3(x+y)=m^{2}+n^{2}-mn$. Therefore, the inequality becomes
$$F=2\left(x^{2}+y^{2}\right)-4\left(m^{2}x+n^{2}y\right)-28xy+m^{4}+n^{4} \geq 0$$
Now, we fix $m, n$ (as constants), and let $x, y$ be variables satisfying the conditi... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,651 |
97. Let positive real numbers $a, b, c$ satisfy the condition $a+b+c=3$. Prove:
$$\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b} \geq \frac{3}{2} \text{ (Pham kim hung) }$$ | Prove: After expansion, the inequality is equivalent to
$$2 \sum_{c y c} a^{4}+2 \sum_{c y c} a^{2} b+3 a b c \geq 3 a^{2} b^{2} c^{2}+\sum_{c y c} a^{3} b^{2}+3 \sum_{c y c} a b^{3}$$
Let $M=a b+b c+c a, S=(a-b)(b-c)(a-c)$. According to the identity
$$\begin{array}{l}
2 \sum_{c y c} a^{2} b=S+3 M-3 a b c \\
2 \sum_{c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,652 |
98. Let $a, b, c$ be positive real numbers, prove:
$$\frac{1}{(2 a+b)^{2}}+\frac{1}{(2 b+c)^{2}}+\frac{1}{(2 c+a)^{2}} \geq \frac{1}{a b+b c+c a} \text{ (Pham Kim Hung) }$$ | Prove that after expansion, the inequality becomes
$$\sum_{c y c}\left(4 a^{5} b+4 a^{5} c-12 a^{4} b^{2}+12 a^{4} c^{2}+5 a^{3} b^{3}+8 a^{4} b c-19 a^{3} b^{2} c+5 a^{3} c^{2} b-7 a^{2} b^{2} c^{2}\right) \geq 0$$
or
$$6 \sum_{c y c} a b\left(a^{2}-b^{2}-2 a b+2 a c\right)^{2}+\sum_{s y m}\left(2 a^{5} b-a^{3} b^{3}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,653 |
99. Let $x_{1} \geq x_{2} \geq \cdots \geq x_{2 n-1} \geq x_{2 n} \geq 0$ be real numbers, and satisfy $x_{1}+x_{2}+\cdots+x_{2 n}=2 n-1$. Find the maximum value of the following expression.
$$P=\left(x_{1}^{2}+x_{2}^{2}\right)\left(x_{3}^{2}+x_{4}^{2}\right) \cdots\left(x_{2 n-1}^{2}+x_{2 n}^{2}\right) \text { (Pham K... | Solution: Although solving this problem directly is very difficult, we unexpectedly found that solving the general case of this problem is relatively simple. In fact, our problem is a direct corollary of the following general result.
Assume $\varepsilon \leq \frac{k}{2 n}$ is a positive constant, $x_{1} \geq x_{2} \geq... | not found | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 731,654 |
Lemma 1. Let $x \geq y \geq z \geq t \geq 0, y+z=2 \alpha$, then
$$\left(x^{2}+y^{2}\right)\left(z^{2}+t^{2}\right) \leq\left(x^{2}+\alpha^{2}\right)\left(\alpha^{2}+t^{2}\right)$$ | Proof: Let $y=\alpha+\beta, z=\alpha-\beta, \beta \geq 0$, then $x \geq \alpha+\beta \geq \alpha-\beta \geq t$. Let
$$f(\beta)=\left[x^{2}+(\alpha+\beta)^{2}\right]\left[(\alpha-\beta)^{2}+t^{2}\right]$$
Then, it is sufficient to prove that $f'(\beta) \leq 0$, which is clearly true, because
$$f'(\beta)=-2 x^{2}(\alpha... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,655 |
Lemma 2: Let $x \geq y \geq z \geq 0,(2 n-1) x+2 y=(2 n+1) \gamma(n \in N, n \geq 2)$, then
$$x^{2 n-2}\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right) \leq 2 \gamma^{2 n}\left(\gamma^{2}+z^{2}\right)$$ | Prove: There exists a real number $\beta \geq 0$, satisfying the conditions $x=\gamma+2 \beta, y=\gamma-(2 n-1) \beta$. Therefore, we must have $\gamma-(2 n-1) \beta \geq z$. Let
$$\begin{array}{l}
g(\beta)=(\gamma+2 \beta)^{2 n}(\gamma-(2 n-1) \beta)^{2}+(\gamma+2 \beta)^{2 n-2}(\gamma-(2 n-1) \beta)^{4}+(\gamma+2 \be... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,656 |
For any two sequences of real numbers $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$, we always have
$$\sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{2}}+\cdots+\sqrt{a_{n}^{2}+b_{n}^{2}} \geq \sqrt{\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2}+\left(b_{1}+b_{2}+\cdots+b_{n}\ri... | Proof: By a simple induction, it suffices to prove the case when $n=2$. In this case, the inequality becomes
$$\sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{2}} \geq \sqrt{\left(a_{1}+a_{2}\right)^{2}+\left(b_{1}+b_{2}\right)^{2}}$$
Squaring both sides and eliminating the common terms, it reduces to the Cauchy-Sch... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,658 |
Example 1.1.2 Let $x, y, z>0$ and $xyz=1$, prove: $\frac{x^{3}}{(1+y)(1+z)}+\frac{y^{3}}{(1+z)(1+x)}+\frac{z^{3}}{(1+x)(1+y)} \geq \frac{3}{4}$ | Prove: Using the AM-GM inequality, we have
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8} \geq \frac{3 x}{4}$$
Therefore,
$$\sum_{c y c} \frac{x^{3}}{(1+y)(1+z)}+\frac{1}{4} \sum_{c y c}(1+x) \geq \sum_{c y c} \frac{3 x}{4} \Rightarrow \sum_{c y c} \frac{x^{3}}{(1+y)(1+z)} \geq \frac{1}{4} \sum_{c y c}(2 x-1) ... | \frac{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 731,659 |
Inference 3: For any real number sequence $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$, we have
$$\left(a_{1}+a_{2}+\cdots+a_{n}\right)^{2} \leq n\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right)$$ | Prove: Applying the Cauchy-Schwarz inequality to the following two sequences, we get
$$\left(a_{1}, a_{2}, \cdots, a_{n}\right),(1,1, \cdots, 1)$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,660 |
Example 2.1.1 Let $a, b, c \geq 0$, prove: $\frac{a^{2}-b c}{2 a^{2}+b^{2}+c^{2}}+\frac{b^{2}-c a}{2 b^{2}+c^{2}+a^{2}}+\frac{c^{2}-a b}{2 c^{2}+a^{2}+b^{2}} \geq 0$ | Prove: The inequality is equivalent to
$$\sum_{c y c} \frac{(a+b)^{2}}{a^{2}+b^{2}+2 c^{2}} \leq 3$$
According to the Cauchy-Schwarz inequality, we have
$$\frac{(a+b)^{2}}{a^{2}+b^{2}+2 c^{2}} \leq \frac{a^{2}}{c^{2}+a^{2}}+\frac{b^{2}}{b^{2}+c^{2}}$$
Thus, we have
$$\sum_{c y c} \frac{(a+b)^{2}}{a^{2}+b^{2}+2 c^{2}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,661 |
Example 2.1.3 Let $a, b, c>0$, prove: $\frac{a^{3}}{a^{3}+b^{3}+a b c}+\frac{b^{3}}{b^{3}+c^{3}+a b c}+\frac{c^{3}}{c^{3}+a^{3}+a b c} \geq 1$ | Proof: Let \( x = \frac{b}{a}, y = \frac{c}{b}, z = \frac{a}{c} \), then we have
\[
\frac{a^{3}}{a^{3}+b^{3}+a b c} = \frac{1}{1+x^{3}+\frac{x}{z}} = \frac{1}{1+x^{3}+x^{2} z} = \frac{y z}{y z+x^{2}+x z}
\]
By the Cauchy-Schwarz inequality, we have
\[
\sum_{c y c} \frac{y z}{y z+x^{2}+x z} \geq \frac{(x y+y z+z x)^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,663 |
Example 2.1.4 Let $a, b, c$ be any three real numbers, and let $x=\sqrt{b^{2}-b c+c^{2}}, y=\sqrt{c^{2}-c a+a^{2}}, z=\sqrt{a^{2}-a b+b^{2}}$. Prove: $x y+y z+z x \geq a^{2}+b^{2}+c^{2} \quad$ (Nguyen Anh Tuan, VMEO 2006) | Prove: Rewrite $x, y$ in the following form
$$x=\sqrt{\frac{3 c^{2}}{4}+\left(b-\frac{c}{2}\right)^{2}}, \quad y=\sqrt{\frac{3 c^{2}}{4}+\left(a-\frac{c}{2}\right)^{2}}$$
According to the Cauchy-Schwarz inequality, we have
$$x y \geq \frac{3 c^{2}}{4}+\frac{1}{4}(2 b-c)(2 a-c)$$
Thus, we have
$$\sum_{c y c} x y \geq ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,664 |
Example 2.1.5 Let $a, b, c, d \geq 0$, prove: [Pham Kim Hung]
$$\frac{a}{b^{2}+c^{2}+d^{2}}+\frac{b}{a^{2}+c^{2}+d^{2}}+\frac{c}{a^{2}+b^{2}+d^{2}}+\frac{d}{a^{2}+b^{2}+c^{2}} \geq \frac{4}{a+b+c+d}$$ | Prove: According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(\frac{a}{b^{2}+c^{2}+d^{2}}+\frac{b}{a^{2}+c^{2}+d^{2}}+\frac{c}{a^{2}+b^{2}+d^{2}}+\frac{d}{a^{2}+b^{2}+c^{2}}\right)(a+b+c+d) \\
\geq\left(\sqrt{\frac{a^{2}}{b^{2}+c^{2}+d^{2}}}+\sqrt{\frac{b^{2}}{a^{2}+c^{2}+d^{2}}}+\sqrt{\frac{c^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,665 |
Example 2.1.6 Prove: For all positive real numbers $a, b, c, d, e, f$, we always have
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+f}+\frac{e}{f+a}+\frac{f}{a+b} \geq 3$ (Six-variable Nesbitt's inequality) | Prove: According to the Cauchy-Schwarz inequality, we have
$$\sum_{c y c} \frac{a}{b+c}=\sum_{c y c} \frac{a^{2}}{a b+a c} \geq \frac{(a+b+c+d+e+f)^{2}}{a b+b c+c d+d e+e f+f a+a c+c e+e a+b d+d f+f b}$$
Let the denominator on the right side of the above inequality be S. Of course,
$$2 S=(a+b+c+d+e+f)^{2}-(a+d)^{2}-(b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,666 |
Example 2.1.7 Two real number sequences $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ satisfy
$$a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1$$
Prove the following inequality
$$\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2} \leq 2\left|a_{1} b_{1}+a_{... | Prove: Using the Cauchy-Schwarz inequality, given the conditions \(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=1\), we can obtain
$$1 \geq a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n} \geq-1$$
According to the Cauchy-Schwarz expansion, we have
$$\begin{array}{l}
\left(a_{1}^{2}+a_{2}^{2}+\c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,667 |
Example 2.1.8 Let $a, b, c>0$, and $a+b+c=3$, prove:
$$\sqrt{a+\sqrt{b^{2}+c^{2}}}+\sqrt{b+\sqrt{c^{2}+a^{2}}}+\sqrt{c+\sqrt{a^{2}+b^{2}}} \geq 3 \sqrt{\sqrt{2}+1}$$ | Prove: Squaring both sides of the inequality, we get
$$\sum_{c y c} \sqrt{b^{2}+c^{2}}+2 \sum_{c y c} \sqrt{\left(a+\sqrt{b^{2}+c^{2}}\right)\left(b+\sqrt{c^{2}+a^{2}}\right)} \geq 9 \sqrt{2}+6$$
According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\sum_{c y c} \sqrt{\left(a+\sqrt{b^{2}+c^{2}}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,668 |
Example 2.1.9 Let $a, b, c>0$, and $abc=1$, prove: $\frac{1}{a^{2}+a+1}+\frac{1}{b^{2}+b+1}+\frac{1}{c^{2}+c+1} \geq 1$ | Proof: Given the conditions, there exist three positive real numbers $x, y, z$, satisfying
$$a=\frac{y z}{x^{2}}, b=\frac{x z}{y^{2}}, c=\frac{x y}{z^{2}}$$
Thus, the inequality becomes
$$\sum_{c y c} \frac{x^{4}}{x^{4}+x^{2} y z+y^{2} z^{2}} \geq 1$$
According to the Cauchy-Schwarz inequality, we have
$$L H S \geq \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,669 |
Example 1.1.3 Let $x, y, z>0$, prove: $\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right) \geq 2+\frac{2(x+y+z)}{\sqrt[3]{x y z}}$ | Prove: After rearranging the inequality, it is equivalent to
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{x y z}}$$
By the AM-GM inequality, we have
$$3\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)=\left(\frac{2 x}{y}+\frac{y}{z}\right)+\left(\frac{2 y}{z}+\frac{z}{x}\right)+\left(\frac{2 z}{x}+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,670 |
Example 2.1.10 Let $a, b, c$ be the lengths of the sides of a triangle, prove: $\frac{a}{3 a-b+c}+\frac{b}{3 b-c+a}+\frac{c}{3 c-a+b} \geq 1$ | Prove: $4 \sum_{c y c} \frac{a}{3 a-b+c}=\sum_{c y c} \frac{4 a}{3 a-b+c}=3+\sum_{c y c} \frac{a+b-c}{3 a-b+c} \geq 3+\frac{(a+b+c)^{2}}{\sum_{c y c}(a+b-c)(3 a-b+c)}=4$ equality holds when $a=b=c$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translat... | null | Inequalities | proof | Yes | Yes | inequalities | false | 731,671 |
Example 2.1.11 Let $a, b, c > 0$, and satisfy the conditions $a \leq b \leq c, a+b+c=3$, prove that
$\sqrt{3 a^{2}+1}+\sqrt{5 a^{2}+3 b^{2}+1}+\sqrt{7 a^{2}+5 b^{2}+3 c^{2}+1} \leq 9$
(Pham Kim Hung) | Proof: According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
\left(\sqrt{3 a^{2}+1}+\sqrt{5 a^{2}+3 b^{2}+1}+\sqrt{7 a^{2}+5 b^{2}+3 c^{2}+1}\right)^{2} \\
=\left(\frac{1}{\sqrt{6}} \sqrt{6\left(3 a^{2}+1\right)}+\frac{1}{\sqrt{4}} \sqrt{4\left(5 a^{2}+3 b^{2}+1\right)}+\frac{1}{\sqrt{3}} \sqrt{3\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,672 |
Example 2.1.12 Let $a, b, c>0$, and $a+b+c=1$, prove:
$\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c} \leq \frac{2a}{b}+\frac{2b}{c}+\frac{2c}{a}$
(Japan TST 2004) | Prove: The inequality can be transformed as follows
$$\frac{3}{2}+\sum_{c y c} \frac{a}{b+c} \leq \sum_{c y c} \frac{a}{b} \Leftrightarrow \sum_{c y c}\left(\frac{a}{b}-\frac{a}{b+c}\right) \geq \frac{3}{2} \Leftrightarrow \sum_{c y c} \frac{a c}{b(b+c)} \geq \frac{3}{2}$$
According to the Cauchy-Schwarz inequality, w... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,673 |
Example 2.1.13 (i) Prove that for all non-negative real numbers $x, y, z$
$$6(x+y-z)\left(x^{2}+y^{2}+z^{2}\right)+27 x y z \leq 10\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}$$
(ii) Prove that for all real numbers $x, y, z$, we have
$$6(x+y+z)\left(x^{2}+y^{2}+z^{2}\right) \leq 27 x y z+10\left(x^{2}+y^{2}+z^{2}\right)^{3 /... | Prove: (i) For this part, we must carefully handle the equality conditions $x=y=2z$ or their permutations. This requires us to estimate the original expression. In fact, by the Cauchy-Schwarz inequality, we have
$$\begin{array}{l}
10\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}-6(x+y-z)\left(x^{2}+y^{2}+z^{2}\right) \\
=\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,674 |
Example 2.1.14 Let $a, b, c, d$ be four positive real numbers, and satisfy $r^{4}=a b c d$. Prove the following inequality: $\frac{a b+1}{a+1}+\frac{b c+1}{b+c}+\frac{c d+1}{c+1}+\frac{d a+1}{d+1} \geq \frac{4\left(1+r^{2}\right)}{1+r} \quad$ (Vasile Cirtoaje, Crux ) | To prove: Given the conditions, there exist four positive real numbers $x, y, z, t$ satisfying
$$a=\frac{r y}{x}, b=\frac{r z}{y}, c=\frac{r t}{z}, d=\frac{r x}{t}$$
The inequality then becomes the following form
We need to prove $A+\left(r^{2}-1\right) B \geq \frac{4\left(r^{2}+1\right)}{r+1}$, where
$$A=\sum_{c y c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,675 |
Example 2.1.15 Let $a, b, c \geq 0$, prove: $\frac{a^{2}}{a^{2}+2(a+b)^{2}}+\frac{b^{2}}{b^{2}+2(b+c)^{2}}+\frac{c^{2}}{c^{2}+2(c+a)^{2}} \geq \frac{1}{3}$ | Proof: Let $x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}$. Then the inequality becomes
$$\sum_{c y c} \frac{1}{1+2(x+1)^{2}} \geq \frac{1}{3}$$
Since $x y z=1$, there exist three positive real numbers $m, n, p$ such that
$$x=\frac{n p}{m^{2}}, y=\frac{m p}{n^{2}}, z=\frac{m n}{p^{2}}$$
Therefore, we only need to prove... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,676 |
Example 2.1.16 Let $a, b, c \geq 0$, prove: $\frac{a}{b^{2}+c^{2}}+\frac{b}{c^{2}+a^{2}}+\frac{c}{a^{2}+b^{2}} \geq \frac{4}{5}\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)$ | Prove: By applying the Cauchy-Schwarz inequality, we have
$$\left(\sum_{c y c} \frac{a}{b^{2}+c^{2}}\right)\left(\sum_{c y c} a\left(b^{2}+c^{2}\right)\right) \geq(a+b+c)^{2}$$
Thus, it suffices to prove that
$$\frac{(a+b+c)^{2}}{a b(a+b)+b c(b+c)+c a(c+a)} \geq \frac{4}{5} \cdot \frac{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,677 |
Inference 1, Let $a, b, c, x, y, z, t, u, v$ be positive real numbers. We have
$$\left(a^{3}+b^{3}+c^{3}\right)\left(x^{3}+y^{3}+z^{3}\right)\left(t^{3}+u^{3}+v^{3}\right) \geq(a x t+b y u+c z v)^{3}$$ | Prove: This is a direct corollary of the Hölder's inequality $(m=n=3)$. I choose this special case of the Hölder's inequality because it can illustrate the details of the proof of the Hölder's inequality. According to the AM-GM inequality, we have
$$\begin{array}{l}
3=\sum_{\text {cyc }} \frac{a^{3}}{a^{3}+b^{3}+c^{3}}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,679 |
Example 1.1.4 Let $a, b, c, d>0$, prove: $16(a b c+b c d+c d a+d a b) \leq(a+b+c+d)^{3}$ | $$\begin{array}{l}
16(a b c+b c d+c d a+d a b)=16 a b(c+d)+16 c d(a+b) \\
\leq 4(a+b)^{2}(c+d)+4(c+d)^{2}(a+b)=4(a+b+c+d)(a+b)(c+d) \\
\leq(a+b+c+d)^{3}
\end{array}$$
Equality holds if and only if $a=b=c=d$.
Proof: Applying the AM-GM inequality, we get the above result. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,681 |
Example 2.2.1 Let $a, b, c$ be positive real numbers, prove: $\frac{a}{\sqrt{a^{2}+8 b c}}+\frac{b}{\sqrt{b^{2}+8 a c}}+\frac{c}{\sqrt{c^{2}+8 a b}} \geq 1$ | Prove: Applying the Hölder's inequality to three sequences (each sequence has three terms which are actually Corollary 1), we have
$$\left(\sum_{\text {cyc }} \frac{a}{\sqrt{a^{2}+8 b c}}\right)\left(\sum_{c y c} \frac{a}{\sqrt{a^{2}+8 b c}}\right)\left(\sum_{c y c} a\left(a^{2}+8 b c\right)\right) \geq(a+b+c)^{3}$$
T... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,682 |
Example 2.2.2 Let $a, b, c > 0$, and $abc = 1$, prove:
$$\begin{array}{l}
\frac{a}{\sqrt{7+b+c}}+\frac{b}{\sqrt{7+c+a}}+\frac{c}{\sqrt{7+a+b}} \geq 1 \\
\frac{a}{\sqrt{7+b^{2}+c^{2}}}+\frac{b}{\sqrt{7+c^{2}+a^{2}}}+\frac{c}{\sqrt{7+a^{2}+b^{2}}} \geq 1
\end{array}$$
Using the same method, determine the truth of the fo... | Prove: Using the Hölder inequality for the first inequality, we have
$$\left(\sum_{c y c} \frac{a}{\sqrt{7+b+c}}\right)\left(\sum_{c y c} \frac{a}{\sqrt{7+b+c}}\right)\left(\sum_{c y c} a(7+b+c)\right) \geq(a+b+c)^{3}$$
Thus, it suffices to prove
$$(a+b+c)^{3} \geq 7(a+b+c)+2(a b+b c+c a)$$
Since \(a+b+c \geq 3 \sqrt... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,683 |
Example 2.2.3 Let $a, b, c$ be positive real numbers, and $k \geq 1$ be a natural number, prove that
$$\frac{a^{k+1}}{b^{k}}+\frac{b^{k+1}}{c^{k}}+\frac{c^{k+1}}{a^{k}} \geq \frac{a^{k}}{b^{k-1}}+\frac{b^{k}}{c^{k-1}}+\frac{c^{k}}{a^{k-1}}$$ | Prove: According to Holder's inequality, we have
$$\left(\frac{a^{k+1}}{b^{k}}+\frac{b^{k+1}}{c^{k}}+\frac{c^{k+1}}{a^{k}}\right)^{k-1}(a+b+c) \geq\left(\frac{a^{k}}{b^{k-1}}+\frac{b^{k}}{c^{k-1}}+\frac{c^{k}}{a^{k-1}}\right)^{k}$$
Thus, it suffices to prove $\square$
$$\frac{a^{k}}{b^{k-1}}+\frac{b^{k}}{c^{k-1}}+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,684 |
Example 2.2.4 Let $a, b, c>0$, prove: $\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right) \geq(a+b+c)^{3}$ | $$\begin{array}{l}
\prod_{c y c}\left(a^{5}-a^{2}+3\right)=\prod_{c y c}\left(a^{3}+2+\left(a^{3}-1\right)\left(a^{2}-1\right)\right) \geq \prod_{c y c}\left(a^{3}+2\right) \\
=\left(a^{3}+1+1\right)\left(b^{3}+1+1\right)\left(c^{3}+1+1\right) \geq(a+b+c)^{3}
\end{array}$$
According to Hölder's inequality, we have:
$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,685 |
Example 2.2.6 Let $a, b, c>0$, and $a+b+c=1$, prove: $\frac{a}{\sqrt[3]{a+2 b}}+\frac{b}{\sqrt[3]{b+2 c}}+\frac{c}{\sqrt[3]{c+2 a}} \geq 1$ | Prove: This inequality can be directly derived from the Holder inequality
$$\left(\sum_{c y c} \frac{a}{\sqrt[3]{a+2 b}}\right)\left(\sum_{c y c} \frac{a}{\sqrt[3]{a+2 b}}\right)\left(\sum_{c y c} \frac{a}{\sqrt[3]{a+2 b}}\right)\left(\sum_{c y c} a(a+2 b)\right) \geq\left(\sum_{c y c} a\right)^{4}=1$$
Because
$$\sum_... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,687 |
Example 2.2.7 Let $a, b, c>0$, prove:
$$a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \geq(a b+b c+c a) \sqrt[3]{(a+b)(b+c)(c+a)}$$ | Prove: Notice that the following expressions are equal
$$\begin{array}{l}
a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b) \\
b^{2}(c+a)+c^{2}(a+b)+a^{2}(b+c) \\
a b(a+b)+b c(b+c)+c a(c+a)
\end{array}$$
According to Holder's inequality, we have
$$\left(\sum_{c y c} a^{2}(b+c)\right)^{3} \geq\left(\sum_{c y c} a b \sqrt[3]{(a+b)(b+c)(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,688 |
Example 2.2.8 Let $a, b, c, d > 0$, and $abcd = 1$, prove that
$$4^{4}\left(a^{4}+1\right)\left(b^{4}+1\right)\left(c^{4}+1\right)\left(d^{4}+1\right) \geq\left(a+b+c+d+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^{4}$$ | $$\begin{array}{l}
\left(a^{4}+1\right)\left(b^{4}+1\right)\left(c^{4}+1\right)\left(d^{4}+1\right) \geq(a+b c d)^{4}=\left(a+\frac{1}{a}\right)^{4} \\
\Rightarrow \sqrt[4]{\left(a^{4}+1\right)\left(b^{4}+1\right)\left(c^{4}+1\right)\left(d^{4}+1\right)} \geq a+\frac{1}{a} \\
\Rightarrow 4 \sqrt[4]{\left(a^{4}+1\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,689 |
Example 2.2.9 Let $a, b, c>0$, prove: $\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geq(a b+b c+c a)^{3}$ | $$\begin{array}{l}
\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right)=\left(a b+a^{2}+b^{2}\right)\left(a^{2}+c a+c^{2}\right)\left(b^{2}+c^{2}+b c\right) \\
\geq(a b+b c+c a)^{3}
\end{array}$$
Prove: By applying the Hölder's inequality, we have
$$\begin{array}{l}
\left(a^{2}+a b+b^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 731,690 |
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