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Problem 25. Let $a, b, c, d$ be positive real numbers such that $a b c d=1$. Prove that
$$\frac{1}{(1+a)^{2}}+\frac{1}{(1+b)^{2}}+\frac{1}{(1+c)^{2}}+\frac{1}{(1+d)^{2}} \geq 1$$ | Solution. First, notice that for any non-negative real numbers $x, y$
$$\frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}} \geq \frac{1}{1+x y}$$
By expanding, the above inequality becomes
$$\begin{aligned}
(2+2 x+2 y+ & \left.x^{2}+y^{2}\right)(1+x y) \geq\left(1+2 x+x^{2}\right)\left(1+2 y+y^{2}\right) \\
& \Leftrightarrow x y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,391 |
Problem 26. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a+2 b}{c+2 b}+\frac{b+2 c}{a+2 c}+\frac{c+2 a}{b+2 a} \geq 3$$ | Solution. By expanding, we can rewrite the inequality to
$$\begin{aligned}
& \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)+3 a b c \geq 3\left(a^{2} b+b^{2} c+c^{2} a\right)
\end{aligned}$$
which is a combination of the third degree-Schur inequality a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,392 |
Problem 27. Let $a, b, c$ be three positive real numbers. Prove that
$$\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}$$ | Solution. Notice that
$$\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 .$$
Therefore, the inequality can be rewritten in the following form
$$\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... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,393 |
Problem 30. Let $a, b, c, d$ be non-negative real numbers. Prove that
$$\left(1+\frac{2 a}{b+c}\right)\left(1+\frac{2 b}{c+d}\right)\left(1+\frac{2 c}{d+a}\right)\left(1+\frac{2 d}{a+b}\right) \geq 9$$ | Solution. Rewrite it 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's easy to see that
$$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \geq\left(1+\frac{2}{x+y}\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,395 |
Problem 31. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}}+\frac{8}{a^{2}+b^{2}+c^{2}} \geq \frac{6}{a b+b c+c a} .$$ | Solution. WLOG, assume that $a \geq b \geq c$. Denote $t=\sqrt{b^{2}}+c^{2}$ and
$$f(a, b, c)=\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}}+\frac{8}{a^{2}+b^{2}+c^{2}}-\frac{6}{a b+b c+c a}$$
We have
$$\begin{aligned}
f(a, b, c)-f(a, t, 0) & =\frac{c^{2}}{\left(a^{2}+b^{2}\right)\left(a^{2}+t^{2}\r... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,396 |
Problem 32. Let $a, b, c, k$ be positive real numbers and $k \geq \frac{3}{2}$. Prove that
$$\frac{a^{k}}{a+b}+\frac{b^{k}}{b+c}+\frac{c^{k}}{c+a} \geq \frac{1}{2}\left(a^{k-1}+b^{k-1}+c^{k-1}\right)$$
(Vasile Cirtoaje and Pham Kim Hung)
Solution. I have two solutions to this problem. | First Solution. (Cauchy reverse) Rewrite the inequality in the form
$$\sum_{c y c}\left(a^{k-1}-\frac{a^{k}}{a+b}\right) \leq \frac{1}{2}\left(\sum_{c y c} a^{k-1}\right) \Leftrightarrow \sum_{c y c} \frac{a^{k-1} b}{a+b} \leq \frac{1}{2}\left(\sum_{c y c} a^{k-1}\right)$$
Notice that
$$\sum_{c y c} \frac{a^{k-1} b}{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,397 |
Problem 33. Let $a, b, c$ be non-negative real numbers and $a+b+c=3$. Prove that
$$a \sqrt{1+b^{3}}+b \sqrt{1+c^{3}}+c \sqrt{1+a^{3}} \leq 5$$ | Solution. By AM-GM inequality, we deduce that
$$\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)$$
It remains to prove that
$$a b^{2}+b c^{2}+c a^{2} \leq 4$$
WLOG, we may suppose that $b$ is the middle number between $a, b, c$. That m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,398 |
Example 1.2.2. Suppose that $a, b, c, d$ are four positive real numbers with sum 4. Prove that
$$\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$$ | Solution. According to AM-GM, we deduce that
$$\begin{aligned}
\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}
\end{aligned}$$
According to this estimation,
$$\sum_{c y c} \frac{a}{1+b^{... | 2 | Inequalities | proof | Yes | Yes | inequalities | false | 737,399 |
Problem 36. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{b+c}{\sqrt{a^{2}+b c}}+\frac{c+a}{\sqrt{b^{2}+c a}}+\frac{a+b}{\sqrt{c^{2}+a b}} \geq 4 .$$ | Solution. Applying Hölder inequality, we obtain
$$\left(\sum_{c y c} \frac{b+c}{\sqrt{a^{2}+b c}}\right)^{2}\left(\sum_{c y c}(b+c)\left(a^{2}+b c\right)\right) \geq 8\left(\sum_{c y c} a\right)^{3}$$
Therefore, it's enough to prove that
$$(a+b+c)^{3} \geq 4 \sum_{c y c} a^{2}(b+c) \Leftrightarrow 6 a b c+\sum_{c y c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,402 |
Problem 37. Let $a, b, c$ be non-negative real numbers satisfying $a^{2}+b^{2}+c^{2}+a b c=4$. Prove that $2+a b c \geq a b+b c+c a \geq a b c$. | SOLUTION. To prove the right hand inequality, just notice that at least one of $a, b, c$, say $a$, is not bigger than 1. Therefore, we have $a b+b c+c a \geq b c \geq a b c$. Equality holds for $(a, b, c)=(0,0,2)$ up to permutation.
To prove the right hand inequality, notice that two numbers among $a, b, c$, say $a$ a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,403 |
Problem 38. Let $a, b, c$ be non-negative real numbers. Prove that
$$\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$$ | Solution. WLOG, assume that $a \geq b \geq c$. First, we will prove that
$$\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}}$$
Indeed, this condition is equivalent to
$$\begin{aligned}
& \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 | 737,404 |
Problem 39. Let $a, b, c$ be three distinct positive real numbers. Prove that
$$\frac{1}{\left|a^{2}-b^{2}\right|}+\frac{1}{\left|b^{2}-c^{2}\right|}+\frac{1}{\left|c^{2}-a^{2}\right|}+\frac{8}{a^{2}+b^{2}+c^{2}} \geq \frac{28}{(a+b+c)^{2}}$$ | Solution. WLOG, we may assume that $a>b>c$. Notice that
$$\begin{aligned}
\frac{1}{a^{2}-b^{2}} & +\frac{1}{b^{2}-c^{2}}+\frac{1}{a^{2}-c^{2}}+\frac{8}{a^{2}+b^{2}+c^{2}}-\left(\frac{1}{a^{2}-b^{2}}+\frac{1}{b^{2}}+\frac{1}{a^{2}}+\frac{8}{a^{2}+b^{2}}\right) \\
& =c^{2}\left(\frac{1}{a^{2}\left(a^{2}-c^{2}\right)}+\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,405 |
Problem 40. Find the best positive real constant $k$ such that the following inequality holds for all positive real numbers $a, b$ and $c$
$$\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$$ | Solution. We clearly have
$$\begin{aligned}
\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{aligned}$$
So we need to find a positive number $k$ satisfying the condi... | 4 \sqrt{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,406 |
Problem 41. Suppose $a, b, c$ are positive real numbers satisfying the condition $a+$ $b+c+a b c=4$. Prove that
$$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \geq \frac{a+b+c}{\sqrt{2}}$$ | Solution. First we will prove that \(a + b + c \geq ab + bc + ca\). Indeed, we may suppose that \(c \geq b \geq a\) without loss of generality. We need to prove that
\[
a + b - ab \geq \frac{4 - a - b}{ab + 1}(a + b - 1) \Leftrightarrow (a + b - 2)^2 \geq ab(a - 1)(b - 1)
\]
Applying the AM-GM inequality, we are done ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,407 |
Problem 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). With the same condition, prove that
\[
\sqrt{\frac{a^{2}+2 b c}{2 a^{2}+b c}}+\sqrt{\frac{b^... | Solution. (i). Since the inequality is homogeneous, we may assume that $a b c=1$. The problem 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 $x=a^{3}, y=b^{3}, z=c^{2}, x y z=1$. WLOG, suppose that $x \geq y \geq z$. Let $t=\sqrt{y z}$, then $t \leq 1$. Firs... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,408 |
Problem 43. Let \(x, y, z\) be non-negative real numbers with sum 1. Prove that
\[
\sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3}
\] | Solution. Suppose \( z = \min \{x, y, z\} \). First we will prove that if \( u = y - z \), \( v = x - z \), and \( k = \frac{1}{12} \), then
\[
\sqrt{x + k u^2} + \sqrt{y + k v^2} \leq \sqrt{2(x + y) + k(u + v)^2}
\]
Indeed, this one is equivalent to
\[
\begin{aligned}
& 2 \sqrt{(x + k u^2)(y + k v^2)} \leq x + y + 2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,409 |
Example 1.2.3. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a^{3}}{a^{2}+b^{2}}+\frac{b^{3}}{b^{2}+c^{2}}+\frac{c^{3}}{c^{2}+d^{2}}+\frac{d^{3}}{d^{2}+a^{2}} \geq \frac{a+b+c+d}{2} .
\] | 1.0. AM-GM inequality
Solution. We use the following estimation
$$\frac{a^{3}}{a^{2}+b^{2}}=a-\frac{a b^{2}}{a^{2}+b^{2}} \geq a-\frac{a b^{2}}{2 a b}=a-\frac{b}{2}.$$
Comment. Here is a similar result for four variables
$$\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}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,410 |
Problem 44. Let \(x, y, z\) be three non-negative real numbers satisfying the condition \(xy + yz + zx = 1\). Prove that
\[
\frac{1}{\sqrt{x+y}} + \frac{1}{\sqrt{y+z}} + \frac{1}{\sqrt{z+x}} \geq 2 + \frac{1}{\sqrt{2}}
\] | Solution. WLOG, we may assume that \( x = \max (x, y, z) \). Denote \( a = y + z > 0 \), then obviously, \( a x = 1 - y z \leq 1 \). Consider the function
\[
\begin{aligned}
f(x) & = \frac{1}{\sqrt{x+y}} + \frac{1}{\sqrt{y+z}} + \frac{1}{\sqrt{z+x}} \\
& = \frac{1}{\sqrt{y+z}} + \sqrt{\frac{2 x + y + z + 2 \sqrt{x^2 + ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,411 |
Problem 45. Let $a, b, c$ be positive real numbers. Prove that
$$\left(2+\frac{a}{b}\right)^{2}+\left(2+\frac{b}{c}\right)^{2}+\left(2+\frac{c}{a}\right)^{2} \geq \frac{9(a+b+c)^{2}}{a b+b c+c a}$$ | Solution. We can rewrite the inequality in the following form
$$\sum_{c y c} \frac{a^{2}}{b^{2}}+4 \sum_{c y c} \frac{a}{b} \geq \frac{9\left(a^{2}+b^{2}+c^{2}\right)}{a b+b c+c a}+6$$
Taking into account the following identities
$$\begin{array}{c}
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3=\frac{(a-b)^{2}}{a b}+\frac{(c-a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,412 |
Problem 46. Let $a, b, c$ be non-negative real numbers and $a^{2}+b^{2}+c^{2}=3$. Prove that
$$\frac{a}{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 .$$ | Solution. Rewrite the inequality in the form
$$\begin{array}{c}
\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
\end{array}$$
Applying Cauchy-Schwarz, we obtain
$$\text { LHS } \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,413 |
Problem 47. Let a, b, c be three positive real numbers. Prove that
$$\frac{1}{a \sqrt{a+b}}+\frac{1}{b \sqrt{b+c}}+\frac{1}{c \sqrt{c+a}} \geq \frac{3}{\sqrt{2 a b c}}$$ | SolutION. Let \( x = \frac{\sqrt{2bc}}{\sqrt{a(a+b)}}, y = \frac{\sqrt{2ca}}{\sqrt{b(b+c)}}, z = \frac{\sqrt{2ab}}{\sqrt{c(c+a)}} \). We need to prove that \( x + y + z \geq 3 \). However, it suffices to prove the following stronger result as follows:
\[ 3 \leq xy + yz + zx = \frac{2c}{\sqrt{(a+b)(b+c)}} + \frac{2a}{\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,414 |
Problem 48. Prove that $a x+b y+c z \geq 0$ if $a, b, c, x, y, z$ are real numbers such that
$$(a+b+c)(x+y+z)=3 ; \quad\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right)=4$$ | SOLUTION. Let $\alpha=\sqrt[4]{\frac{a^{2}+b^{2}+c^{2}}{x^{2}+y^{2}+z^{2}}}$ and $a_{1}=\frac{a}{\alpha}, b_{1}=\frac{b}{\alpha}, c_{1}=\frac{c}{\alpha}, x_{1}=x \alpha, y_{1}=$ $y \alpha, z_{1}=z \alpha$. We infer that
$$\begin{aligned}
a_{1}^{2}+b_{1}^{2}+c_{1}^{2}=\frac{a^{2}+b^{2}+c^{2}}{\alpha^{2}} & =\sqrt{\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,415 |
Problem 49. Let \(a, b, c, d\) be non-negative real numbers with sum 4. Prove that
\[
\sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4
\] | Solution. According to AM-GM inequality, we get
$$\mathrm{LHS} \geq 4 \sqrt[8]{\frac{(a+1)(b+1)(c+1)(b+1)}{(a b+1)(b c+1)(c d+1)(d a+1)}}$$
and it remains to prove that
$$(a+1)(b+1)(c+1)(d+1) \geq(a b+1)(b c+1)(c d+1)(d a+1)$$
After expanding, the inequality becomes
$$\begin{array}{c}
a b c d+\sum_{\text{sym}} a b c+\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,416 |
Problem 50. Prove that for all non-negative real numbers $a, b, c$ then
$$\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}}$$ | SOLUTION. Let $x=\sqrt{a}, y=\sqrt{b}, z=\sqrt{c}$. The inequality becomes
$$\begin{aligned}
& \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}}{\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,417 |
Problem 53. Let \(a, b, c\) be three real numbers satisfying the condition \(a^{2}+b^{2}+c^{2}=9\). Prove that
\[3 \min (a, b, c) \leq 1 + abc\] | Solution. WLOG, we may assume that $c \geq b \geq a$. Consider the following cases:
(i). $a \leq 0$ : Let $d=-a$ and $e=|b|$. We will prove that
$$-3 d \leq 1-d c e \Leftrightarrow d(c e-3) \leq 1$$
If $c e \leq 3$, the conclusion follows immediately. Otherwise, if $c e \geq 3$ then
$$d^{2}(c e-3)(c e-3) \leq\left(\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,420 |
Example 1.1.2. Let \( x, y, z \) be positive real numbers such that \( xyz = 1 \). Prove that
\[
\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+z)(1+x)} + \frac{z^3}{(1+x)(1+y)} \geq \frac{3}{4}
\] | Solution. We use AM-GM in the following form:
$$\frac{x^{3}}{(1+y)(1+z)}+\frac{1+y}{8}+\frac{1+z}{8} \geq \frac{3 x}{4} .$$
We conclude that
$$\begin{array}{l}
\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{3}{4} | Inequalities | proof | Yes | Yes | inequalities | false | 737,421 |
Example 1.2.4. Let \(a, b, c\) be positive real numbers with sum 3. Prove that
\[
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
\] | Solution. We use the following estimation according to AM-GM
$$\frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{2 / 3}}{3}$$
which implies that
$$\sum_{\text {cyc }} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{\text {cyc }} a-\frac{2}{3} \sum_{\text {cyc }}(a b)^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,422 |
Problem 54. Let \(a, b, c, d\) be non-negative real numbers such that \(a+b+c+d=4\).
Prove that
\[
\left(1+a^{4}\right)\left(1+b^{4}\right)\left(1+c^{4}\right)\left(1+d^{4}\right) \geq\left(1+a^{3}\right)\left(1+b^{3}\right)\left(1+c^{3}\right)\left(1+d^{3}\right)
\] | Solution. Notice that for all $x \geq 0,\left(1+x^{4}\right)(1+x) \geq\left(1+x^{3}\right)\left(1+x^{2}\right)$, therefore
$$\prod_{c y c}\left(1+a^{4}\right) \prod_{c y c}(1+a) \geq \prod_{c y c}\left(1+a^{3}\right) \prod_{c y c}\left(1+a^{2}\right)$$
It's enough to prove that $\prod_{c y c}\left(1+a^{2}\right) \geq ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,423 |
Problem 55. Find the best constant $k$ (smallest) for the inequality
$$a^{k}+b^{k}+c^{k} \geq a b+b c+c a$$
to hold for $a, b, c$ are three non-negative real numbers with $a+b+c=3$. | SOLUTION. In example 1.1.1 in chapter I, this inequality is proved for \( k = \frac{1}{2} \) and therefore it's true for every \( k \geq \frac{1}{2} \). Consider the inequality in the case \( k \leq \frac{1}{2} \).
Lemma. Suppose \( a, b \geq 0 \) and \( a + b = 2t \geq 1 \) then we have
\[ a^k + b^k - ab \geq \min \l... | k = \frac{2 \ln 3 - 3 \ln 2}{\ln 3 - \ln 2} \approx 0.2905 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,424 |
Problem 56. Let \(a, b, c\) be positive real numbers. Prove that
\[
\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \sqrt{\frac{a^{2}+b^{2}+c^{2}}{a b+b c+c a}}
\] | Solution. Notice that if $a \geq b \geq c$ then
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)=\frac{(a-b)(a-c)(c-b)}{a b c} \leq 0$$
so it's enough to consider the case $a \geq b \geq c$. By squaring both sides, we get
$$\sum_{c y c} \frac{a^{2}}{b^{2}}+\sum_{c y c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,425 |
Problem 57. Suppose that a, b, c, d are positive real numbers satisfying $a^{2}+b^{2}+c^{2}+$ $d^{2}=4$. Prove that
$$\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 .$$ | SOLUTION. Let $x=a b c, y=a b d_{a} z=a c d, t=b c d$. The problem becomes
$$\sum_{c y c} \frac{1}{3-x} \leq 2 \Leftrightarrow \sum_{c y c} \frac{1-x}{3-x} \geq 0$$
According to the AM-GM inequality, we deduce
$$x+y=a b(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 | 737,426 |
Problem 58. Suppose that $n$ positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$ satisfy the condition
$$\frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\ldots+\frac{1}{1+x_{n}}=\frac{n}{2}$$
Prove that
$$\sum_{i, j=1}^{n} \frac{1}{x_{i}+x_{j}} \geq \frac{n^{2}}{2}$$ | Solution. For each $i \in\{1,2, \ldots, n\}$, we denote $a_{i}=\frac{1-x_{i}}{1+x_{i}}$. Therefore, $a_{1}+a_{2}+\ldots+ a_{n}=0$ and $a_{i} \in[-1,1]$. Consider the expression
$$S=\sum_{i, j=1}^{n} \frac{1}{x_{i}+x_{j}} \Rightarrow 2 S=\sum_{i, j=1}^{n} \frac{\left(1+a_{i}\right)\left(1+a_{j}\right)}{1-a_{i} a_{j}}$$
... | S \geq \frac{n^{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,427 |
Problem 60. Suppose that $n$ is an integer greater than 2. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1} a_{2} \ldots a_{n}=1$. Prove the following inequality
$$\frac{a_{1}+3}{\left(a_{1}+1\right)^{2}}+\frac{a_{2}+3}{\left(a_{2}+1\right)^{2}}+\ldots+\frac{a_{n}+3}{\left(a_{n}+1\right)^{2}}... | Solution. Notice first that it is sufficient to prove the inequality in the case $n=3$. For a bigger value of $n(n \geq 4)$, we only need to choose from the set $\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ the three smallest numbers, say $a_{1}, a_{2}, a_{3}$. Since $a_{1} a_{2} a_{3} \leq 1$, there exists a positive n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,429 |
Problem 61. Let $a, b, c$ be non-negative real numbers with sum 2. Prove that
$$\sqrt{a+b-2 a b}+\sqrt{b+c-2 b c}+\sqrt{c+a-2 c a} \geq 2$$ | SOLUTION. WLOG, we may assume that $a \geq b \geq c$. Let $x=a+b-2 a b, y=b+c-2 b c$ and $z=c+a-2 c a$. The inequality is equivalent to (after squaring)
$$2 \sum_{c y c} \sqrt{x y} \geq 2 \sum_{c y c} a b .$$
Notice that $2 x=c(a+b)+(a-b)^{2}$ and $2 y=a(b+c)+(b-c)^{2}$, so Cauchy-Schwarz inequality gives us that
$$2 ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,430 |
Problem 62. Let \(a, b, c\) be non-negative real numbers such that \(a^{2}+b^{2}+c^{2}=3\).
Prove that
\[
\frac{a}{b+2}+\frac{b}{c+2}+\frac{c}{a+2} \leq 1
\] | Solution. After expanding, the inequality gets a simpler form WLOG, suppose that $b$ is the second greatest 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 suffices to prove that
$$2 \geq a^{2} b+b c^{2} \Leftrightarrow b\left(a^{2}+c^{2}\right)... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,431 |
Problem 63. Let $a, b, c$ be non-negative real numbers. Prove that
$$\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$$ | Solution. The inequality is equivalent to
$$\begin{array}{c}
\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 | 737,432 |
Problem 64. Suppose that $a, b, c$ are the side lengths of a triangle. Prove that
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \leq \frac{5}{2}$$ | Solution. With the following identities
$$\begin{array}{c}
-3+\sum_{c y c} \frac{2 a}{b+c}=\sum_{c y c} \frac{(a-b)^{2}}{(a+c)(b+c)} \\
2-\frac{2(a b+b c+c a)}{a^{2}+b^{2}+c^{2}}=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{a^{2}+b^{2}+c^{2}}
\end{array}$$
we can transform our inequality to \( S_{a}(b-c)^{2}+S_{b}(a-c)^{2}+S_{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,434 |
Problem 65. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{1}{\sqrt{a^{2}+b c}}+\frac{1}{\sqrt{b^{2}+c a}}+\frac{1}{\sqrt{c^{2}+a b}} \geq \frac{6}{a+b+c}$$ | Solution. First solution. Taking into account problem 15, we have
$$\frac{1}{\sqrt{a^{2}+b c}}+\frac{1}{\sqrt{b^{2}+c a}}+\frac{1}{\sqrt{c^{2}+a b}} \geq \frac{9}{\sqrt{a^{2}+b c}+\sqrt{b^{2}+c a}+\sqrt{c^{2}+a b}} \geq \frac{6}{a+b+c}.$$
Second solution. Applying AM-GM inequality, directly we have
$$\frac{1}{\sqrt{a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,435 |
Problem 66. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a^{3}}{2 a^{2}-a b+2 b^{2}}+\frac{b^{3}}{2 b^{2}-b c+2 c^{2}}+\frac{c^{3}}{2 c^{2}-c a+2 a^{2}} \geq \frac{a+b+c}{3}$$ | SolutION. Rewrite the inequality form as follows
$$\begin{aligned}
\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}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,436 |
Problem 67. Prove that for all positive real numbers $a, b, c$
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geq 3 \sqrt[4]{\frac{a^{4}+b^{4}+c^{4}}{3}}$$ | Solution. Applying Hölder inequality, we obtain
$$\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right)\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \geq\left(a^{2}+b^{2}+c^{2}\right)^{3}$$
Let $x=a^{2}, y=b^{2}, z=c^{2}$. It remains to prove that
$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,437 |
Problem 68. Let \(a, b, c\) be non-negative real numbers with sum 3. Prove that
\[
\left(a+b^{2}\right)\left(b+c^{2}\right)\left(c+a^{2}\right) \leq 13+abc
\] | Solution. We will first prove that if $a \geq b \geq c$ then
$$\left(a+b^{2}\right)\left(b+c^{2}\right)\left(c+a^{2}\right) \geq\left(a^{2}+b\right)\left(b^{2}+c\right)\left(c^{2}+a\right)$$
Indeed, just notice that
$$\begin{aligned}
\sum_{c y c} a^{3} b-\sum_{c y c} a b^{3} & =(a+b+c)(a-b)(b-c)(a-c) \\
\sum_{c y c} a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,438 |
Problem 69. Let $a, b, c$ be positive real numbers. Prove that
$$\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$$ | Solution. We have $(a+b)^{2}-2\left(c^{2}+a b\right)=\left(a^{2}-c^{2}\right)+\left(b^{2}-c^{2}\right)$, therefore
$$\begin{array}{c}
\frac{(a+b)^{2}}{c^{2}+a b}+\frac{(b+c)^{2}}{a^{2}+b c}+\frac{(c+a)^{2}}{b^{2}+c a}-6=\sum_{c y c} \frac{\left(a^{2}-b^{2}\right)+\left(a^{2}-c^{2}\right)}{a^{2}+b c} \\
\quad=\sum_{c y ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,439 |
Problem 70. Find the maximum value of $k=k(n)$ for which the following inequality is true for all real numbers $x_{1}, x_{2}, \ldots, x_{n}$
$$x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\ldots+\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2} \geq k\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{2}$$ | SOLUTION. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers, then
$$\begin{array}{c}
a_{1} y_{1}^{2}+\frac{1}{a_{1}} \cdot y_{2}^{2}+2 y_{1} y_{2} \geq 0 \\
a_{2} y_{2}^{2}+\frac{1}{a_{2}} \cdot y_{3}^{2}+2 y_{2} y_{3} \geq 0 \\
\cdots \\
\cdots \\
a_{n-1} y_{n-1}^{2}+\frac{1}{a_{n}} \cdot y_{n}^{2}+2 y_{n-1} ... | \frac{1}{4 \cos ^{2} \frac{\pi}{2 n+1}} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,440 |
Problem 71. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers with sum $n$. Prove that
$$\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}-n \geq \frac{8(n-1)\left(1-a_{1} a_{2} \ldots a_{n}\right)}{n^{2}}$$ | SOLUTION. We prove this inequality by induction. If $n=2$, the problem is obvious:
$$\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$$
Let's consider the problem for $n+1$ numbers with the supposition that it is true for $n$ numbers. We assume tha... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,441 |
Problem 72. Let \(x, y, z\) be non-negative real numbers with sum 1. Prove that
\[
\sqrt{x+y^{2}}+\sqrt{y+z^{2}}+\sqrt{z+x^{2}} \geq 2
\] | Solution. Notice that if $a, b, c, d$ are non-negative real numbers such that $a+b=c+d$ and $|a-b| \leq|c-d|$, then we have
$$\sqrt{a}+\sqrt{b} \geq \sqrt{c}+\sqrt{d}(\star)$$
Indeed, since $(a+b)^{2}-(a-b)^{2} \geq(c+d)^{2}-(c-d)^{2}$, we have $a b \geq c d$, therefore
$$a+b+2 \sqrt{a b} \geq c+d+2 \sqrt{c d} \Righta... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,442 |
Problem 73. Let \(a, b, c\) be positive real numbers with sum 3. Prove that
\[
\frac{1}{2+a^{2} b^{2}}+\frac{1}{2+b^{2} c^{2}}+\frac{1}{2+c^{2} a^{2}} \geq 1
\] | Solution. According to AM-GM inequality, we have
$$\frac{1}{2+a^{2} b^{2}}=\frac{1}{2}-\frac{a^{2} b^{2}}{2\left(2+a^{2} b^{2}\right)} \geq \frac{1}{2}-\frac{a^{2} b^{2}}{6 \sqrt[3]{a^{2} b^{2}}}=\frac{1}{2}-\frac{a^{4 / 3} b^{4 / 3}}{6}$$
We deduce that
$$\sum_{c y c} \frac{1}{2+a^{2} b^{2}} \geq \frac{3}{2}-\frac{1}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,443 |
Example 1.2.6. Let $a, b, c$ be positive real numbers which sum up to 3. Prove that
$$\frac{a+1}{b^{2}+1}+\frac{b+1}{c^{2}+1}+\frac{c+1}{a^{2}+1} \geq 3$$ | Solution. We use the following estimation
$$\frac{a+1}{b^{2}+1}=a+1-\frac{b^{2}(a+1)}{b^{2}+1} \geq a+1-\frac{b^{2}(a+1)}{2b}=a+1-\frac{ab+b}{2}$$
Summing up the similar results for $a, b, c$, we deduce that
$$\sum_{cyc} \frac{a+1}{b^{2}+1} \geq 3+\frac{1}{2} \sum_{cyc} a - \frac{1}{2} \sum_{cyc} ab \geq 3$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,444 |
Problem 74. Consider the positive real constants $m, n$, such that $3 n^{2}>m^{2}$. For real numbers $a, b, c$ such that $a+b+c=m, a^{2}+b^{2}+c^{2}=n^{2}$, find the maximum and minimum of
$$P=a^{2} b+b^{2} c+c^{2} a$$ | SOLUTION. Let \(a=x+\frac{m}{3}, b=y+\frac{m}{3}, c=z+\frac{m}{3}\). From the given conditions, we get that \(x+y+z=0\) and \(x^{2}+y^{2}+z^{2}=\frac{3 n^{2}-m^{2}}{3}\). 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^{... | \max P=\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{3 / 2}+\frac{m^{3}}{9}, \min P=-\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{3 / 2}-\frac{m^{3}}{9} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,445 |
Problem 75. Suppose that \(a, b, c\) are three positive real numbers verifying
\[
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=13
\]
Find the minimum and maximum values of the expression
\[
P=\frac{a^{3}+b^{3}+c^{3}}{a b c} .
\] | Solution. Denote
$$x=\sum_{c y c} \frac{a}{b} ; y=\sum_{c y c} \frac{b}{a} ; m=\sum_{c y c} \frac{a^{2}}{b c} ; n=\sum_{c y c} \frac{b c}{a^{2}}$$
We have \(x+y=10\) and
$$x^{3}=3(m+n)+6+\sum_{c y c} \frac{a^{3}}{b^{3}} ; y^{3}=3(m+n)+6+\sum_{c y c} \frac{b^{3}}{a^{3}}$$
The identities above yield that
$$x^{3}+y^{3}=... | 11-2\sqrt{3} \text{ and } 11+2\sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,446 |
Problem 76. Prove that for all positive real numbers \(a, b, c, d, e\),
\[
\frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+d}{2} \cdot \frac{d+e}{2} \cdot \frac{e+a}{2} \leq \frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \frac{c+d+e}{3} \cdot \frac{d+e+a}{3} \cdot \frac{e+a+b}{3}
\] | Solution. We will first prove that for all $a, b>0$ and $a+b \leq 1$
$$\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right) \geq\left(\frac{2}{a+b}-1\right)^{2}$$
Indeed, this result can be rewritten in the following form
$$\begin{aligned}
\frac{1}{a b}- & \frac{1}{a}-\frac{1}{b} \geq \frac{4}{(a+b)^{2}}-\frac{4}{a+b}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,447 |
Problem 77. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{a^{4}}{a^{3}+b^{3}}+\frac{b^{4}}{b^{3}+c^{3}}+\frac{c^{4}}{c^{3}+a^{3}} \geq \frac{a+b+c}{2}$$ | Solution. Notice that
$$\frac{2 a^{4}}{a^{3}+b^{3}}-a-\frac{3(a-b)}{2}=(a-b)\left(\frac{a\left(a^{2}+a b+b^{2}\right)}{a^{3}+b^{3}}-\frac{3}{2}\right)=\frac{2 b^{2}+a b-b^{2}}{3\left(a^{3}+b^{3}\right)}(a-b)^{2}$$
Therefore the inequality can be transformed to
$$S_{a}(b-c)^{2}+S_{b}(a-c)^{2}+S_{c}(a-b)^{2} \geq 0$$
in... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,448 |
Problem 78. Let $a, b, c$ be positive real numbers. Prove that
$$\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}}$$ | Solution. Let $x^{2}=a, y^{2}=b$ and $z^{2}=c$. The inequality becomes
$$\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}}$$
Squaring both sides, we obtain an equivalent form
$$\sum_{c y c} \frac{x^{6}}{x^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,449 |
Problem 79. Let \(a, b, c\) be non-negative real numbers with sum 2. Prove that
\[
\frac{a b}{1+c^{2}}+\frac{b c}{1+a^{2}}+\frac{c a}{1+b^{2}} \leq 1
\] | Solution. We denote $x=a b+b c+c a$ and $p=a b c$. According to the identities
$$\begin{array}{l}
A=(a-b)^{2}(b-c)^{2}(c-a)^{2}=4 x^{2}(1-x)+4(9 x-8) p-27 p^{2} \\
B=\sum_{c y c} a^{2}(a-b)(a-c)=12 p+4(1-x)(4-x)
\end{array}$$
we can rewrite our inequality as
$$\begin{aligned}
& (1-x)\left(5-2 x+x^{2}\right)+(6 x-2) p-2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,450 |
Problem 80. Suppose that \(a_{1}, a_{2}, \ldots, a_{n}\) are non-negative real numbers which add up to \(n\). Find the minimum of the expression
\[S=a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}+a_{1} a_{2} \ldots a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right)\] | Solution. Consider the following function
$$F=f\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}+a_{1} a_{2} \ldots a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\ldots+\frac{1}{a_{n}}\right)$$
Surprisingly, these exceptional identities will help
$$\begin{aligned}
f\left(a_{1}, a_{2}, \ldots, ... | \min \left(2 n, \frac{n^{2}}{n-2}, \frac{n^{2}}{n-1}+\left(\frac{n}{n-1}\right)^{n-1}\right) | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,451 |
Problem 81. Let \( x, y, z \) be positive real numbers satisfying \( 2xyz = 3x^2 + 4y^2 + 5z^2 \). Find the minimum of the expression \( P = 3x + 2y + z \). | Solution. Let $a=3 x, b=2 y, z=c$ We then obtain
$$a+b+c=3 x+2 y+z, a^{2}+3 b^{2}+15 c^{2}=a b c .$$
According to the weighted AM-GM inequality, we have that
$$\begin{array}{c}
a+b+c \geq(2 a)^{1 / 2}(3 b)^{1 / 3}(6 c)^{1 / 6} \\
a^{2}+3 b^{2}+15 c^{2} \geq\left(4 a^{2}\right)^{1 / 4}\left(9 b^{2}\right)^{3 / 9}\left(... | 36 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,452 |
Problem 82. Let $a, b, c$ be non-negative real numbers. Prove that
$$\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}}$$ | Solution. First we may assume that $a \geq b \geq c$. Notice that
$$\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}}$$
so it suffices to prove that
$$\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}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,453 |
Example 1.2.7. Let $a, b, c$ be positive real numbers with sum 3. Prove that
$$\frac{1}{1+2 b^{2} c}+\frac{1}{1+2 c^{2} a}+\frac{1}{1+2 a^{2} b} \geq 1$$ | Solution. We use the following estimation
$$\frac{d}{1+2 b^{2} c} = 1 - \frac{2 b^{2} c}{1+2 b^{2} c} \geq 1 - \frac{2 \sqrt[3]{b^{2} c}}{3} \geq 1 - \frac{2(2 b + c)}{9}.$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,455 |
Problem 84. Let $a, b, c$ be three arbitrary real numbers. Prove that
$$\frac{1}{(2 a-b)^{2}}+\frac{1}{(2 b-c)^{2}}+\frac{1}{(2 c-a)^{2}} \geq \frac{11}{7\left(a^{2}+b^{2}+c^{2}\right)}$$ | SOLUTION. Denote \( x = 2a - b, y = 2b - c, z = 2c - a \). We get
\[
\begin{aligned}
a &= \frac{4x + 2y + z}{7}, \quad b = \frac{4y + 2z + x}{7}, \quad c = \frac{4z + 2x + y}{7} \\
& \Rightarrow a^2 + b^2 + c^2 = \frac{2(x + y + z)^2 + x^2 + y^2 + z^2}{7}
\end{aligned}
\]
It remains to prove that for all \( x, y, z \i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,456 |
Problem 85. Let \(a, b, c\) be positive real numbers. Consider the following inequality
\[
\frac{a b}{c} + \frac{b c}{a} + \frac{c a}{b} \geq 3 \sqrt[k]{\frac{a^{k} + b^{k} + c^{k}}{3}} (\star)
\]
(a). Prove that ( \(\star\) ) is true for \(k=2\).
(b). Prove that ( \(\star\) ) is not true for \(k=3\) but true for \(k=3... | Solution. (a). For $k=2$, we can assume that $\sum_{cyc} a^{2}=3$ without loss of generality. The inequality $\sum_{cyc} \frac{a b}{c} \geq 3$ is equivalent to
$$\left(\sum_{cyc} \frac{a b}{c}\right)^{2} \geq 9 \Leftrightarrow \sum_{cyc} \frac{a^{2} b^{2}}{c^{2}} \geq 3 \Leftrightarrow \sum_{cyc} a^{2}\left(\frac{b^{2}... | 2.709511 | Inequalities | proof | Yes | Yes | inequalities | false | 737,457 |
Problem 87. Let $a, b, c$ be non-negative real numbers. Prove that
$$\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 .$$ | SOLUTION. WLOG, suppose that $a^{2}+b^{2}+c^{2}=3$. By the weighted Jensen inequality we deduce that
$$\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 c^{2}\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,459 |
Problem 88. Suppose that $n$ is a positive integer and $\left(x_{1}, x_{2}, \ldots, x_{n}\right) ;\left(y_{1}, y_{2}, \ldots, y_{n}\right)$ are two positive real sequences. Let $\left(z_{2}, z_{3}, \ldots, z_{2 n}\right)$ be a positive sequence satisfying
$$z_{i+j}^{2} \geq x_{i} y_{j} \forall 1 \leq i, j \leq n$$
Den... | Solution. Let $X=\max \left\{x_{1}, x_{2}, \ldots, x_{n}\right\}$ and $Y=\max \left\{y_{1}, y_{2}, \ldots, y_{n}\right\}$. WLOG, we can assume that $X=Y=1$ (otherwise, replace $x_{i}$ with $x_{i} / X, y_{i}$ with $y_{i} / Y$ and $z_{i}$ with $z_{i} / \sqrt{X Y}$ ). According to AM-GM inequality, the following result is... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,460 |
Problem 89. (a). Let \(a, b, c\) be three real numbers. Prove that
\[a^{4}+b^{4}+c^{4}+a b^{3}+b c^{3}+c a^{3} \geq 2\left(a^{3} b+b^{3} c+c^{3} a\right)\]
(b). Let \(a, b, c\) be three real numbers and \(a^{2}+b^{2}+c^{2}+a b+b c+c a=6\). Prove that
\[a^{3} b+b^{3} c+c^{3} a+a b c(a+b+c) \leq 6\]
(c). Find the best (g... | Solution. For all real numbers $a, b, c$, we have that
$$\left(a^{2}-k a b+k a c-c^{2}\right)^{2}+\left(b^{2}-k b c+k b a-a^{2}\right)^{2}+\left(c^{2}-k c a+k c b-b^{2}\right)^{2} \geq 0$$
After expanding, this inequality 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}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,461 |
Problem 91. Let \(a, b, c\) be positive real numbers such that \(a^{2}+b^{2}+c^{2}=3\). Prove that
\[
a^{3} b^{2}+b^{3} c^{2}+c^{3} a^{2} \leq 3
\] | Solution. By Cauchy-Schwarz inequality, we have
$$\left(a^{3} b^{2}+b^{3} c^{2}+c^{3} a^{2}\right)^{2} \leq\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)\left(a^{4} b^{2}+b^{4} c^{2}+c^{4} a^{2}\right)$$
It remains to prove that if $x+y+z=3$ then Notice that
$$\begin{array}{c}
3(x y+y z+z x)\left(x^{2} y+y^{2} z+z^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,463 |
Problem 92. Let $a, b, c$ be arbitrary positive real numbers. Prove that
$$\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}$$ | Solution. The inequality is equivalent to
$$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{aligned}
\sum_{c y c} \frac{b^{4}}{c^{2}}-\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,464 |
Problem 93. Suppose that $n$ is an integer greater than 2 and $a_{1}, a_{2}, \ldots, a_{n}$ are $n$ real numbers. Prove that for any non-empty subset $S$ of $\{1,2, \ldots, n\}$, we have
$$\left(\sum_{i \in S} a_{i}\right)^{2} \leq \sum_{1 \leq i \leq j \leq n}\left(a_{i}+\ldots+a_{j}\right)^{2}$$ | Solution. We will first prove the following lemma:
Lemma. For all real numbers $x_{1}, x_{2}, \ldots, x_{2 k+1}$
$$\left(\sum_{0 \leq i \leq k} a_{2 i+1}\right)^{2} \leq \sum_{1 \leq i \leq j \leq 2 k+1}\left(a_{i}+\ldots+a_{j}\right)^{2}(\star)$$
Proof. Let $s_{i}=x_{1}+x_{2}+\ldots+x_{i} \forall i \in\{1,2, \ldots, ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,465 |
Example [1.2.8. Let $, a, b, c, d$ be non-negative real numbers with sum 4. Prove that
$$\frac{1+a b}{1+b^{2} c^{2}}+\frac{1+b c}{1+c^{2} d^{2}}+\frac{1+c d}{1+d^{2} a^{2}}+\frac{1+x a}{1+a^{2} b^{2}} \geq 4$$ | Solution. Applying AM-GM, we have
$$\frac{1+a b}{1+b^{2} c^{2}}=(1+a b)-\frac{(1+a b) b^{2} c^{2}}{1+b^{2} c^{2}} \geq 1+a b-\frac{1}{2}(1+a b) b c$$
Summing up similar results, we get
$$\sum_{c y c} \frac{1+a b}{1+b^{2} c^{2}} \geq 4+\sum_{c y c} a b-\frac{1}{2} \sum_{c y c} b c(1+a b)=4+\frac{1}{2}\left(\sum_{c y c}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,466 |
Problem 94. Let $a, b, c$ be three non-negative real numbers. Prove that
$$\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 .$$ | SOLUTION. The inequality can be changed into an equivalent form as follows:
$$\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$$
where $x=\frac{b}{a}, y=\frac{c}{b}, z=\frac{a}{c}$ and $x y z=1$. Denote
$$m=1-\frac{2}{1+x}, n=1-\frac{2}{1+y}, p=1-\frac{2}{1+z}$$
Certainly $m... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,467 |
Problem 95. Let $a, b, c$ be arbitrary positive real numbers. Prove that
$$\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)$$ | Solution. Rewrite the inequality in the following form
$$\begin{array}{c}
\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 c} a^{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,468 |
Problem 96. Let \(a, b, c, d\) be non-negative real numbers such that
\[
(a+b+c+d)^{2}=3\left(a^{2}+b^{2}+c^{2}+d^{2}\right)
\]
Prove the following inequality
\[
a^{4}+b^{4}+c^{4}+d^{4} \geq 28 a b c d
\] | Solution. Let $m=a+b, n=c+d, x=ab$ and $y=cd$ then $(m+n)^{2}=3\left(m^{2}+n^{2}-2x-2y\right)$ or $3(x+y)=m^{2}+n^{2}-mn$. The problem 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 $x, y$ vary (as variables) such that $x \leq \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,469 |
Problem 97. Let \(a, b, c\) be positive real numbers with sum 3. Prove that
\[
\frac{a}{b^{2}+c}+\frac{b}{c^{2}+a}+\frac{c}{a^{2}+b} \geq \frac{3}{2} .
\] | SOLUTION. After expanding, we can change the inequality into
$$2 \sum_{c y c} a^{4}+2 \sum_{c y c} a^{2} b+3 a b c \geq 3 a^{2} b^{2} c^{2}+\sum_{c y c} a^{3} b^{2}+3 \sum_{c y c} a b^{3}$$
Let $M=a b+b c+c a$ and $S=(a-b)(a-c)(b-c)$. According to the identities
$$\begin{aligned}
2 \sum_{c y c} a^{2} b & =S+3 M-3 a b ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,470 |
Problem 98. Let $a, b, c$ be positive real numbers. Prove that
$$\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}$$ | Solution. After expanding, the inequality becomes
$$\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}-4... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,471 |
Problem 99. Suppose that \(x_{1} \geq x_{2} \geq \ldots \geq x_{2 n-1} \geq x_{2 n} \geq 0\) are real numbers and \(x_{1}+x_{2}+\ldots+x_{2 n}=2 n-1\). Find the maximum of the following expression
\[P=\left(x_{1}^{2}+x_{2}^{2}\right)\left(x_{3}^{2}+x_{4}^{2}\right) \ldots\left(x_{2 n-1}^{2}+x_{2 n}^{2}\right)\] | SOLUTION. Although it's very hard to solve this problem directly, we find out unexpectedly that proving the general problem is simpler. In fact, the proposed problem is a direct corollary of the following general result:
$\star$ Suppose that $\epsilon \leq \frac{k}{2 n}$ is a positive constant and $x_{1} \geq x_{2} \ge... | proof | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,472 |
Example 1.2.9. Let $a, b, c$ be positive real numbers satisfying $a^{2}+b^{2}+c^{2}=3$. Prove that
$$\frac{1}{a^{3}+2}+\frac{1}{b^{3}+2}+\frac{1}{c^{3}+2} \geq 1$$ | Solution. According to AM-GM, we obtain
$$\begin{aligned}
\sum_{c y c} \frac{1}{a^{3}+2} & =\frac{3}{2}-\frac{1}{2} \sum_{c y c} \frac{a^{3}}{a^{3}+1+1} \\
& \geq \frac{3}{2}-\frac{1}{2} \sum_{c y c} \frac{a^{3}}{3 a}=1
\end{aligned}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,473 |
Example 2.1.1. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{a^{2}-b c}{2 a^{2}+b^{2}+c^{2}}+\frac{b^{2}-c a}{2 b^{2}+c^{2}+a^{2}}+\frac{c^{2}-a b}{2 c^{2}+a^{2}+b^{2}} \geq 0$$ | Solution. 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
$$\frac{(a+b)^{2}}{a^{2}+b^{2}+2 c^{2}} \leq \frac{a^{2}}{a^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}}$$
This concludes
$$\sum_{c y c} \frac{(a+b)^{2}}{a^{2}+b^{2}+2 c^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,474 |
Example 2.1.2. Suppose that $x, y, z \geq 1$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove that
$$\sqrt{x+y+z} \geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$ | Solution. By hypothesis, we obtain
$$\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1 .$$
According to Cauchy-Schwarz, we have
$$\sum_{cyc} x=\left(\sum_{cyc} x\right)\left(\sum_{cyc} \frac{x-1}{x}\right) \geq\left(\sum_{cyc} \sqrt{x-1}\right)^{2}$$
which implies
$$\sqrt{x+y+z} \geq \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$$ | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,475 |
Example 2.1.3. Let $a, b, c$ be positive real numbers. Prove that
$$\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$$ | Solution. Let $x=\frac{b}{a}, y=\frac{c}{b}$ and $z=\frac{a}{c}$. Then we have
$$\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}$$
By Cauchy-Schwarz inequality, we deduce that
$$\sum_{c y c} \frac{y z}{y z+x^{2}+x z} \geq \frac{(x y+y z+z x)^{2}}{y z\le... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,476 |
Example 2.1.4. Let $a, b, c$ be three arbitrary real numbers. Denote
$$x=\sqrt{b^{2}-b c+c^{2}}, y=\sqrt{c^{2}-c a+a^{2}}, z=\sqrt{a^{2}-a b+b^{2}}.$$
Prove that
$$x y+y z+z x \geq a^{2}+b^{2}+c^{2}$$ | SolutION. Rewrite $x, y$ in the following forms
$$x=\sqrt{\frac{3 c^{2}}{4}+\left(b-\frac{c}{2}\right)^{2}}, y=\sqrt{\frac{3 c^{2}}{4}+\left(a-\frac{c}{2}\right)^{2}}.$$
According to the Cauchy-Schwarz inequality, we conclude
$$x y \geq \frac{3 c^{2}}{4}+\frac{1}{4}(2 b-c)(2 a-c)$$
which implies
$$\sum_{c y c} x y \ge... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,477 |
Example 1.1.3. Let $a, b, c$ be positive real numbers. Prove that
$$\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}}$$ | 1.0. $A M-G M$ inequality
19
Solution. Certainly, the problem follows the inequality
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \geq \frac{x+y+z}{\sqrt[3]{x y z}}$$
which is true by AM-GM because
$$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... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,478 |
Example 2.1.5. Let $a, b, c, d$ be non-negative real numbers. Prove that
$$\frac{a}{b^{2}+c^{2}+d^{2}}+\frac{b}{a^{2}+c^{2}+d^{2}}+\frac{c}{a^{2}+b^{2}+d^{2}}+\frac{d}{a^{2}+b^{2}+c^{2}} \geq \frac{4}{a+b+c+d}$$ | Solution. According to Cauchy-Schwarz, 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}}{a^{2}+b^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,479 |
Example 2.1.6. Prove that for all positive real numbers \(a, b, c, d, e, f\), we always have
\[
\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+f}+\frac{e}{f+a}+\frac{f}{a+b} \geq 3
\] | Solution. According to Cauchy-Schwarz inequality
$$\sum_{c y c} \frac{a}{b+c}=\sum_{c y c} \frac{a^{2}}{a b+a c} \geq \frac{(a+b+c+d+e+f)^{2}}{a b+b c+c d+d e+e f+f a+a c+c e+e a+b d+d f+f b}$$
Denote the denominator of the right fraction above by \( S \). Certainly,
$$2 S = (a+b+c+d+e+f)^{2} - (a+d)^{2} - (b+e)^{2} -... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,480 |
Example 2.1.7. Two real sequences $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ satisfy
$$a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}=b_{1}^{2}+b_{2}^{2}+\ldots+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_{2} b_{... | SolutION. By Cauchy-Schwarz, the condition $\sum_{i=1}^{n} a_{i}^{2}=\sum_{i=1}^{n} b_{i}^{2}=1$ yields
$$1 \geq a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{n} b_{n} \geq-1$$
According to the expansion of the Cauchy-Schwarz inequality, we have
$$\left(a_{1}^{2}+\ldots+a_{n}^{2}\right)\left(b_{1}^{2}+\ldots+b_{n}^{2}\right)-\lef... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,481 |
Example 2.1.8. Suppose \(a, b, c\) are positive real numbers with sum 3. Prove that
\[
\sqrt{a+\sqrt{b^{2}+c^{2}}}+\sqrt{b+\sqrt{c^{2}+a^{2}}}+\sqrt{c+\sqrt{a^{2}+b^{2}}} \geq 3 \sqrt{\sqrt{2}+1}
\] | Solution. We rewrite the inequality in the following form (after squaring both sides)
$$\sum_{c y c} \sqrt{b^{2}+c^{2}}+2 \sum_{c y c} \sqrt{\left(a+\sqrt{b^{2}+c^{2}}\right)\left(b+\sqrt{c^{2}+a^{2}}\right)} \geq 9 \sqrt{2}+6$$
According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{c}
\sum_{c y c} \sqrt{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,482 |
Example 2.1.9. Suppose \(a, b, c\) are positive real numbers such that \(abc=1\). Prove the following inequality
\[
\frac{1}{a^{2}+a+1}+\frac{1}{b^{2}+b+1}+\frac{1}{c^{2}+c+1} \geq 1
\] | Solution. By hypothesis, there exist three positive real numbers $x, y, z$ for which
$$a=\frac{y z}{x^{2}}, b=\frac{x z}{y^{2}}, c=\frac{x y}{z^{2}}$$
The inequality can be rewritten to
$$\sum_{cyc} \frac{x^{4}}{x^{4}+x^{2} y z+y^{2} z^{2}} \geq 1$$
According to Cauchy-Schwarz, we have
$$\mathrm{LHS} \geq \frac{\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,483 |
Example 2.1.10. Let \(a, b, c\) be the side-lengths of a triangle. Prove that
\[
\frac{a}{3a - b + c} + \frac{b}{3b - c + a} + \frac{c}{3c - a + b} \geq 1.
\] | Solution. By Cauchy-Schwarz, we have
$$\begin{aligned}
4 \sum_{c y c} \frac{a}{3 a-b+c} & =\sum_{c y c} \frac{4 a}{3 a-b+c} \\
& =3+\sum_{c y c} \frac{a+b-c}{3 a-b+c} \\
& \geq 3+\frac{(a+b+c)^{2}}{\sum_{c y c}(a+b-c)(3 a-b+c)} \\
& =4
\end{aligned}$$
Equality holds for \(a=b=c\). | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,484 |
Example 2.1.12. Let \(a, b, c\) be positive real numbers with sum 1. Prove that
\[
\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c} \leq \frac{2a}{b}+\frac{2b}{c}+\frac{2c}{a}
\] | Solution. Rewrite this inequality in the form
$$\frac{3}{2}+\sum_{c y c} \frac{a}{b+c} \leq \sum_{c y c} \frac{a}{b} \Leftrightarrow \sum_{c y c}\left(\frac{a}{b}-\frac{a}{b+c}\right) \geq \frac{3}{2} \Leftrightarrow \sum_{c y c} \frac{a c}{b(b+c)} \geq \frac{3}{2}$$
According to Cauchy-Schwarz,
$$\sum_{c y c} \frac{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,485 |
Example 2.1.14. Let \(a, b, c, d\) be four positive real numbers such that \(r^{4}=a b c d \geq 1\)
Prove the following inequality
\[
\frac{a b+1}{a+1}+\frac{b c+1}{b+1}+\frac{c d+1}{c+1}+\frac{d a+1}{d+1} \geq \frac{4\left(1+r^{2}\right)}{1+r}
\] | Solution. The hypothesis implies the existence of four positive real numbers $x, y, z, t$ such that
$$a=\frac{r y}{x}, \quad b=\frac{r z}{y}, \quad c=\frac{r t}{z}, \quad d=\frac{r x}{t}.$$
The inequality is therefore rewritten in the following form
$$\sum_{c y c} \frac{\frac{r^{2} z}{x}+1}{\frac{r y}{x}+1} \geq \frac... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,487 |
Example 1.1.4. Let \(a, b, c, d\) be positive real numbers. Prove that
\[16(a b c + b c d + c d a + d a b) \leq (a + b + c + d)^{4}\] | Solution. Applying AM-GM for two numbers, we obtain
$$\begin{aligned}
16(a b c+b c d+c d a & +d a b)=16 a b(c+d)+16 c d(a+b) \\
& \leq 4(a+b)^{2}(c+d)+4(c+d)^{2}(a+b) \\
& =4(a+b+c+d)(a+b)(c+d) \\
& \leq(a+b+c+d)^{3} .
\end{aligned}$$
The equality holds for $a=b=c=d$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,488 |
Example 2.1.15. Let \(a, b, c\) be non-negative real numbers. Prove that
\[
\frac{a^{2}}{a^{2}+2(a+b)^{2}}+\frac{b^{2}}{b^{2}+2(b+c)^{2}}+\frac{c^{2}}{c^{2}+2(c+a)^{2}} \geq \frac{1}{3}
\] | SOLUTION. We denote \( x = \frac{b}{a}, y = \frac{c}{b}, z = \frac{a}{c} \). The problem becomes
\[ \sum_{cyc} \frac{1}{1 + 2(x + 1)^2} \geq \frac{1}{3} \]
Because \( xyz = 1 \), there exist three positive real numbers \( m, n, p \) such that
\[ x = \frac{np}{m^2}, y = \frac{mp}{n^2}, z = \frac{mn}{p^2} \]
It remains... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,489 |
Example 2.1.16. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{a}{b^{2}+c^{2}}+\frac{b}{a^{2}+c^{2}}+\frac{c}{a^{2}+b^{2}} \geq \frac{4}{5}\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) .$$ | SolutION. Applying Cauchy-Schwarz, we obtain
$$\left(\sum_{c y c} \frac{a}{b^{2}+c^{2}}\right)\left(\sum_{c y c} a\left(b^{2}+c^{2}\right)\right) \geq(a+b+c)^{2}$$
It remains to prove that
$$\frac{(a+b+c)^{2}}{a b(a+b)+b c(b+c)+c a(c+a)} \geq \frac{4}{5} \cdot \frac{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)}{a b(a+b)+b c(b+c)+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,490 |
Example 2.2.1. Let $a, b, c$ be positive real numbers. Prove that
$$\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$$ | Solution. Applying Hölder inequality for three sequences, each of which has three terms (actually, that's corollary 1), we deduce that
$$\left(\sum_{c y c} \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}$$
So i... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,491 |
Example 2.2.2. Let \(a, b, c\) be positive real numbers such that \(abc = 1\). Prove that
\[
\begin{array}{r}
\frac{a}{\sqrt{7+b+c}}+\frac{b}{\sqrt{7+c+a}}+\frac{c}{\sqrt{7+a+b}} \geq 1 \\
\frac{a}{\sqrt{7+b^{2}+c^{2}}}+\frac{b}{\sqrt{7+c^{2}+a^{2}}}+\frac{c}{\sqrt{7+a^{2}+b^{2}}} \geq 1
\end{array}
\]
With the same c... | Solution. For the first one, apply Hölder inequality in the following form
$$\left(\sum_{c y c} \frac{a}{\sqrt{7+b+c}}\right)\left(\sum_{c y c} \frac{a}{\sqrt{7+b+c}}\right)\left(\sum_{c y c} a(7+b+c)\right) \geq(a+b+c)^{3}$$
It's enough to prove that
$$(a+b+c)^{3} \geq 7(a+b+c)+2(a b+b c+c a)$$
Because $a+b+c \geq 3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,492 |
Example 2.2.3. Let $a, b, c$ be positive real numbers. Prove that for all natural numbers $k,(k \geq 1)$, the following inequality holds
$$\frac{a^{k+1}}{b^{k}}+\frac{b^{k+1}}{c^{k}}+\frac{c^{k+1}}{a^{k}} \geq \frac{a^{k}}{b^{k-1}}+\frac{b^{k}}{c^{k-1}}+\frac{c^{k}}{a^{k-1}}$$ | Solution. According to Hölder inequality, we deduce that
$$\left(\frac{a^{k+1}}{b^{k}}+\frac{b^{k+1}}{c^{k}}+\frac{c^{k+1}}{a^{k}}\right)^{k-1}(a+b+c) \geq\left(\frac{a^{k}}{b^{k-1}}+\frac{b^{k}}{c^{k-1}}+\frac{c^{k}}{a^{k-1}}\right)^{k}$$
It suffices to prove that
$$\frac{a^{k}}{b^{k-1}}+\frac{b^{k}}{c^{k-1}}+\frac{c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,493 |
Example 2.2.4. Let \(a, b, c\) be positive real numbers. Prove that
\[
\left(a^{5}-a^{2}+3\right)\left(b^{5}-b^{2}+3\right)\left(c^{5}-c^{2}+3\right) \geq (a+b+c)^{3}
\] | $$\begin{array}{c}
\prod_{c y c}\left(a^{5}-a^{2}+3\right)=\prod_{c y c}\left(a^{3}+2+\left(a^{3}-1\right)\left(a^{2}-1\right)\right) \geq \prod\left(a^{3}+2\right) \\
\quad=\left(a^{3}+1+1\right)\left(1+b^{3}+1\right)\left(1+1+c^{3}\right) \geq(a+b+c)^{3}
\end{array}$$
Solution. According to Hölder's inequality, we c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,494 |
Example 2.2.6. Let \(a, b, c\) be positive real numbers which sum up to 1. Prove that
$$\frac{a}{\sqrt[3]{a+2 b}}+\frac{b}{\sqrt[3]{b+2 c}}+\frac{c}{\sqrt[3]{c+2 a}} \geq 1$$ | Solution. This inequality is directly obtained from the Hölder inequality:
$$\left(\sum_{c y c} \frac{a}{\sqrt[3]{a+2 b}}\right)\left(\sum_{c y c} \frac{a}{\sqrt[3]{a+2 b}}\right)\left(\sum_{c y c} \frac{a}{\sqrt[3]{a+2 b}}\right)\left(\sum_{c y c} a(a+2 b)\right) \geq\left(\sum a\right)^{4}=1$$
because
$$\sum_{\text {... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,496 |
Example 2.2.7. Let $a, b, c$ be positive real numbers. Prove that
$$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)}$$ | Solution. Notice that the following expressions are equal to each other
$$\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}$$
According to Hölder's inequality, we get that
$$\left(\sum_{c y c} a^{2}(b+c)\right)^{3} \geq\left(\sum_{c y c} a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,497 |
Example 2.2.8. Let \(a, b, c, d\) be positive real numbers such that \(a b c d=1\). Prove that
\[
4^{4}\left(a^{4}+1\right)\left(b^{4}+1\right)\left(c^{4}+1\right)\left(d^{4}+1\right) \geq\left(a+b+c+d+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^{4}
\] | Solution. By Hölder inequality, we get that
$$\begin{array}{c}
\left(a^{4}+1\right)\left(1+b^{4}\right)\left(1+c^{4}\right)\left(1+d^{4}\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} \\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,498 |
Example 1.1.5. Suppose that \(a, b, c\) are three side-lengths of a triangle with perimeter
3. Prove that
\[
\frac{1}{\sqrt{a+b-c}}+\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{c+a-b}} \geq \frac{9}{a b+b c+c a}
\] | SOLUTION. Let $x=\sqrt{b+c-a}, y=\sqrt{c+a-b}, z=\sqrt{a+b-c}$. We get $x^{2}+y^{2}+z^{2}=$
3. The inequality becomes
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{36}{9+x^{2} y^{2}+y^{2} z^{2}+z^{2} x^{2}}$$
Let $m=x y, n=y z, p=z x$. The inequality above is equivalent to
$$(m+n+p)\left(m^{2}+n^{2}+p^{2}+9\right) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,499 |
Example 2.2.9. Let \(a, b, c\) be positive real numbers. Prove that
\[
\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \geq (a b+b c+c a)^{3}.
\] | $$\begin{array}{c}
\left(a^{2}+a b+b^{2}\right)\left(b^{2}+b c+c^{2}\right)\left(c^{2}+c a+a^{2}\right) \\
=\left(a b+a^{2}+b^{2}\right)\left(a^{2}+a c+c^{2}\right)\left(b^{2}+c^{2}+b c\right) \geq(a b+a c+b c)^{3}
\end{array}$$
Applying Hölder inequality, we obtain
$$\begin{array}{c}
\left(a^{2}+a b+b^{2}\right)\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,500 |
Example 2.2.10. Suppose that $a, b, c$ are positive real numbers satisfying the condition $3 \max \left(a^{2}, b^{2}, c^{2}\right) \leq 2\left(a^{2}+b^{2}+c^{2}\right)$. Prove that
$$\frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}+\frac{b}{\sqrt{2 c^{2}+2 a^{2}-b^{2}}}+\frac{c}{\sqrt{2 a^{2}+2 b^{2}-c^{2}}} \geq \sqrt{3}$$ | Solution. By Hölder, we deduce that
$$\left(\sum_{c y c} \frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}\right)\left(\sum_{c y c} \frac{a}{\sqrt{2 b^{2}+2 c^{2}-a^{2}}}\right)\left(\sum_{c y c} a\left(2 b^{2}+2 c^{2}-a^{2}\right)\right) \geq(a+b+c)^{3}$$
It remains to prove that
$$(a+b+c)^{3} \geq 3 \sum_{\text {cyc }} a\left(... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,501 |
Example 3.1.2. Let \(a, b, c, d\) be positive real numbers such that \(a^{2}+b^{2}+c^{2}+d^{2}=4\). Prove the following inequality
\[
\frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b}+\frac{d^{2}}{a+b+c} \geq \frac{4}{3}
\] | Solution. Notice that if $(a, b, c, d)$ is arranged in an increasing order then
$$\frac{1}{b+c+d} \geq \frac{1}{c+d+a} \geq \frac{1}{d+a+b} \geq \frac{1}{a+b+c}$$
Therefore, by Chebyshev inequality, we have
$$\begin{aligned}
4 \mathrm{LHS} & \geq\left(\sum_{c y c} a^{2}\right)\left(\sum_{c y c} \frac{1}{b+c+d}\right) ... | \frac{a^{2}}{b+c+d}+\frac{b^{2}}{c+d+a}+\frac{c^{2}}{d+a+b}+\frac{d^{2}}{a+b+c} \geq \frac{4}{3} | Inequalities | proof | Yes | Yes | inequalities | false | 737,502 |
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