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Example 1.7.7. Let $a, b, c$ be non-negative real numbers. Prove that
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}+\sqrt{\frac{27(ab+bc+ca)}{a^2+b^2+c^2}} \geq \frac{7 \sqrt{2}}{2} .$$ | SOLUTION. Similarly as in the previous problem, we get that if $a \geq b \geq c$ and $t=b+c$ then
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}+\sqrt{\frac{27(a b+b c+c a)}{a^{2}+b^{2}+c^{2}}} \geq \frac{a+t}{\sqrt{a t}}+3 \sqrt{3} \cdot \sqrt{\frac{a t}{a^{2}+t^{2}}}$$
Denote $x=\frac{a+t}{\sqrt{a ... | \frac{7 \sqrt{2}}{2} | Inequalities | proof | Yes | Yes | inequalities | false | 737,610 |
Example 1.7.8. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{8}{a+b+c} \geq \frac{6}{\sqrt{a b+b c+c a}}$$ | SOLUTION. Similarly to the previous proofs, we assume first that $a \geq b \geq c$ and denote
$$f(a, b, c)=\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{8}{a+b+c}-\frac{6}{\sqrt{a b+b c+c a}}$$
Let $t=b+c$, then we have
$$\begin{aligned}
f(a, b, c)-f(a, t, 0) & =\frac{1}{a+b}+\frac{1}{a+c}-\frac{1}{a+b+c}-\frac{1}{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,611 |
Example 1.8.1. Suppose that $a, b, c$ are three positive real numbers satisfying
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=13 .$$
Find the minimum value of
$$P=\left(a^{2}+b^{2}+c^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) .$$ | SOLUTION. Let now
$$x=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}, y=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$$
then we obtain that
$$x^{2}=2 y+\sum_{c y c} \frac{a^{2}}{b^{2}} ; y^{2}=2 x+\sum_{c y c} \frac{b^{2}}{a^{2}}$$
Because $x+y=10$, AM-GM inequality yields that
$$P-3=x^{2}+y^{2}-2(x+y) \geq \frac{1}{2}(x+y)^{2}-2(x+y)=50-... | 33 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,612 |
Example 1.8.3. Suppose that \(a, b, c\) are three positive real numbers satisfying that
\[
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16
\]
Find the minimum and maximum value of
\[
P=\left(a^{4}+b^{4}+c^{4}\right)\left(\frac{1}{a^{4}}+\frac{1}{b^{4}}+\frac{1}{c^{4}}\right)
\] | Solution. We denote $x, y, m, n$ as in the previous problem and also denote
$$p=\sum_{c y c} \frac{a^{2}}{b^{2}} ; q=\sum_{c y c} \frac{b^{2}}{a^{2}}$$
The expression $P$ can be rewritten as
$$P=3+\sum_{c y c} \frac{a^{4}}{b^{4}}=3+p^{2}+q^{2}-2(p+q) \text {. }$$
The hypothesis yields that $x+y=13$. Moreover, we have... | \frac{12777}{8} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,614 |
Example 1.8.4. Let \(a, b, c, d\) be positive real numbers such that
\[
(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)=20.
\]
Prove that
\[
\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}\right) \geq 36.
\] | SOLUTION. For each triple of positive real numbers $(x, y, z)$, we denote
$$F(x, y, z)=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$$
Clearly
$$F(x, y, z)^{2}=F\left(x^{2}, y^{2}, z^{2}\right)+2 F(z, y, x) ; F(z, y, x)^{2}=F\left(z^{2}, y^{2}, x^{2}\right)+2 F(x, y, z)$$
The condition of the problem can be rewritten as
$$\sum... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,615 |
Example 1.10.1. Let $a, b, c$ be non-negative real numbers. Prove that
$$a^{3}+b^{3}+c^{3}+\left(\frac{3}{\sqrt[3]{4}}-1\right) a b c \geq \frac{3}{\sqrt[3]{4}}\left(a^{2} b+b^{2} c+c^{2} a\right)$$ | Solution. Clearly, the above inequality is true if $a=b=c$. According to CD3 theorem, it suffices to consider the inequality in one case $b=1, c=0$. The inequality becomes
$$a^{3}+1 \geq \frac{3}{\sqrt[3]{4}} a^{2}$$
which simply follows from AM-GM inequality
$$a^{3}+1=\frac{a^{3}}{2}+\frac{a^{3}}{2}+1 \geq \frac{3}{\s... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,617 |
Example 1.10.2. Let \(a, b, c\) be non-negative real numbers with sum 3. Prove that
\[a^{2} b + b^{2} c + c^{2} a + a b c \leq 4\] | SOLUTION. The inequality is equivalent to
$$27\left(a^{2} b+b^{2} c+c^{2} a+a b c\right) \leq 4(a+b+c)^{3}$$
Since this is a cyclic inequality and holds for $a=b=c$, we only need to consider the case $c=0$ due to the CD3 theorem. In this case, the inequality becomes
$$24 a^{2} b \leq 4(a+b)^{3}$$
By the AM-GM inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,618 |
Example 1.10.3. Let $a, b, c$ be non-negative real numbers. Prove that
$$4\left(a^{3}+b^{3}+c^{3}\right)+12\left(a^{2} b+b^{2} c+c^{2} a\right) \geq 15\left(a b^{2}+b c^{2}+c a^{2}\right)+3 a b c$$ | SOLUTION. Because the inequality is cyclic and holds for $a=b=c$, according to CD3 theorem, it is enough to consider the case $c=0$. The inequality becomes
$$4\left(a^{3}+b^{3}\right)+12 a^{2} b \geq 15 a b^{2}$$
or
$$(2 a-b)^{2}(a+4 b) \geq 0$$
which is obvious. The equality holds for $a=b=c$ or $(a, b, c) \sim (1,2,0... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,619 |
Example 1.1.5. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a^{2}+2 b c}{(b+c)^{2}}+\frac{b^{2}+2 a c}{(a+c)^{2}}+\frac{c^{2}+2 a b}{(a+b)^{2}} \geq \frac{9}{4}$$ | SOLUTION. The inequality can be rewritten as
$$\sum_{c y c} \frac{(a-b)(a-c)+(a b+b c+c a)}{(b+c)^{2}} \geq \frac{9}{4}$$
or equivalently $A+B \geq \frac{9}{4}$, where
$$A=\frac{(a-b)(a-c)}{(b+c)^{2}} ; B=\sum \frac{a b+b c+c a}{(b+c)^{2}} ;$$
By the generalized Schur inequality, we deduce that $A \geq 0$. Moreover, $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,620 |
Example 1.10.4. Let $a, b, c$ be non-negative real numbers with sum 4. Prove that
$$5\left(a b^{2}+b c^{2}+c a^{2}\right)+3 a b c \leq 36+3\left(a^{2} b+b^{2} c+c^{2} a\right)$$ | SOLUTION. Because the inequality is cyclic and holds for $a=b=c=\frac{4}{3}$, it suffices to consider it in the case $c=0$ and $a+b=4$. We have to prove that
$$5 a b^{2}-3 a^{2} b \leq 36$$
or
$$a(4-a)(5-2 a) \leq 9$$
Applying the AM-GM inequality, we have the desired result
$$a(4-a)(5-2 a)=\frac{1}{3} \cdot(3 a) \cdo... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,621 |
Example 1.10.5. Let \(a, b, c\) be non-negative real numbers such that \(a+b+c=3\). For each \(k \geq 0\), find the maximum value of
\[a^{2}(k b+c)+b^{2}(k c+a)+c^{2}(k a+b)\] | SOLUTION. Because the expression is cyclic, we can assume first that \( c=0, a+b=3 \) and find the maximum value of
\[ F=k a^{2} b+b^{2} a \]
For \( k=1 \), we have \( F=a b(a+b)=3 a b \leq \frac{27}{4} \) by AM-GM inequality. Otherwise, assume that \( k \neq 1 \). We denote
\[ f(a)=k a^{2}(3-a)+a(3-a)^{2}=(1-k) a^{3}... | \min \left\{3(k+1) ; \frac{2\left(k^{2}-k+1\right)}{(k-1)^{2}}\left(\sqrt{k^{2}-k+1}+k-2\right)+\frac{3(k-2)}{k-1}\right\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,622 |
Example 1.10.6. Let \(a, b, c\) be non-negative real numbers such that \(a+b+c=3\). For each \(k \geq 0\), find the maximum and minimum value of
\[
a^{2}(k b-c)+b^{2}(k c-a)+c^{2}(k a-b) .
\] | SOLUTION. As in the preceding solutions, we will first consider the case $c=0$. Denote
$$k a^{2} b-a b^{2}=k a^{2}(3-a)-a(3-a)^{2}=-(k+1) a^{3}+3(k+2) a^{2}-9 a=f(a)$$
then we get
$$f^{\prime}(a)=-3(k+1) a^{2}+6(k+2) a-9$$
The equation $f^{\prime}(a)=0$ has exactly two positive real roots (in $[0,3]$ )
$$a_{1}=\frac{k... | \min \{3(k-1) ; m\}, \max \{3(k-1) ; M\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,623 |
Example 1.10.7. Let \(a, b, c\) be non-negative real numbers. Prove that
\[a^{2}+b^{2}+c^{2}+2+\frac{4}{3}\left(a^{2} b+b^{2} c+c^{2} a\right) \geq 3(a b+b c+c a)\] | SOLUTION. For $a=b=c=t$, the inequality becomes
$$4 t^{3}-6 t^{2}+2 \geq 0 \Leftrightarrow 2(t-1)^{2}(2 t+1) \geq 0$$
This one is obvious, so we are done. Equality holds for $a=b=c=1$.
For $c=0$, the inequality becomes
$$\frac{4}{3} a^{2} b+a^{2}+b^{2}+2 \geq 3 a b$$
or
$$f(a)=\left(\frac{4}{3} b+1\right) a^{2}-3 b \c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,624 |
Example 1.10.8. Let \(a, b, c\) be non-negative real numbers. Prove that
\[a^{2}+b^{2}+c^{2}+2\left(a^{2} b+b^{2} c+c^{2} a\right)+12 \geq 6(a+b+c)+a b+b c+c a\] | SOLUTION. For $a=b=c=t$, the inequality becomes $6 t^{3}+12 \geq 18 t$, or $6(t-1)^{2}(t+$ $2) \geq 0$, which is obvious. Therefore it suffices to prove the inequality in case $c=0$. In this case, we have to prove that
$$a^{2}+b^{2}+2 a^{2} b+12 \geq 6(a+b)$$
or
$$f(a)=a^{2}(1+2 b)-6 a+\left(b^{2}-6 b+12\right) \geq 0$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,625 |
Example 1.10.9. Let $a, b, c$ be non-negative real numbers. Prove that
$$a^{2}+b^{2}+c^{2}+3+\frac{1}{6}\left(a^{2} b+b^{2} c+c^{2} a+15 a b c\right) \geq a+b+c+2(a b+b c+c a)$$ | SOLUTION. If $a=b=c$, the inequality is obvious. If $c=0$, it becomes
$$a^{2}+b^{2}+3+\frac{a^{2} b}{6} \geq a+b+2 a b$$
or
$$a^{2}\left(1+\frac{b}{6}\right)-(2 b+1) a+\left(b^{2}-b+3\right) \geq 0$$
It is easy to check that
$$\Delta=(2 b+1)^{2}-4\left(1+\frac{b}{6}\right)\left(b^{2}-b+3\right)=-\frac{2}{3} b^{3}+\fra... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,626 |
Example 1.10.10. Let $a, b, c$ be non-negative real numbers. Prove that
$$2\left(a^{3}+b^{3}+c^{3}\right)+a b c+a b+b c+c a+2 \geq 2\left(a^{2} b+b^{2} c+c^{2} a\right)+a^{2}+b^{2}+c^{2}+a+b+c$$ | SOLUTION. If $a=b=c$, the inequality is obvious, with equality for $a=b=c=1$. Therefore we can assume that $c=0$, and the inequality becomes
$$2\left(a^{3}+b^{3}\right)+a b+2 \geq 2 a^{2} b+a^{2}+b^{2}+a+b$$
We may assume that $a \geq b$. Denote
$$f(a)=2\left(a^{3}+b^{3}\right)+a b+2-2 a^{2} b-a^{2}-b^{2}-a-b$$
If $a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,627 |
Example 1.10.11. Let \(a, b, c, d\) be non-negative real numbers. Prove that
\[4\left(a^{3}+b^{3}+c^{3}+d^{3}\right)+15(a b c+b c d+c d a+d a b) \geq(a+b+c+d)^{3}\] | SOLUTION. Because this is a third-degree symmetric inequality of four variables, according to the generalization of the SD3 theorem, it suffices to check this inequality in case \(a=b=c=d=1\) or \(a=0, b=c=d=1\) or \(a=b=0, c=d=1\) or \(a=b=c=0, d=1\). They are all obvious, so we have the desired result. The equality h... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,628 |
Example 1.10.12. Let \(a, b, c, d\) be non-negative real numbers such that \(a+b+c+d=4\).
Prove that
\[
a^{3}+b^{3}+c^{3}+d^{3}+10(ab+bc+cd+da+ac+bd) \leq 64
\] | SOLUTION. The inequality can be rewritten as (homogeneous form)
$$a^{3}+b^{3}+c^{3}+d^{3}+\frac{5}{2}(a+b+c+d)(ab+bc+cd+da+ac+bd) \leq (a+b+c+d)^{3}$$
It is easy to check that the inequality holds for $(a, b, c, d)=(1,1,1,1)$ or $(1,1,1,0)$ or $(1,1,0,0)$ or $(1,0,0,0)$ (and it is an equality for $(1,1,1,1),(1,0,0,0))... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,629 |
Example 1.10.13. Let \(a, b, c, d\) be non-negative real numbers such that \(a+b+c+d=4\).
Prove that
\[
a^{3}+b^{3}+c^{3}+d^{3}+\frac{14}{3}(a b+b c+c d+d a+a c+b d) \geq 32
\] | SOLUTION. The inequality can be rewritten as (homogeneous form)
$$a^{3}+b^{3}+c^{3}+d^{3}+\frac{7}{6}(a+b+c+d)(a b+b c+c d+d a+a c+b d) \geq \frac{1}{2}(a+b+c+d)^{3}$$
By the previous proposition/theorem, it suffices to consider this inequality in the cases $(a, b, c, d) \in\{(1,1,1,1),(1,1,1,0),(1,1,0,0),(1,0,0,0)\}$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,630 |
Example 1.1.6. Let $a, b, c$ be positive real numbers. Prove that
$$\sqrt{\frac{a^{3}+a b c}{(b+c)^{3}}}+\sqrt{\frac{b^{3}+a b c}{(c+a)^{3}}}+\sqrt{\frac{c^{3}+a b c}{(a+b)^{3}}} \geq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$$ | SOLution. Notice that
$$\begin{array}{c}
\sqrt{\frac{a^{3}+a b c}{(b+c)^{3}}}-\frac{a}{b+c}=\frac{\sqrt{a}}{b+c} \cdot\left(\sqrt{\frac{a^{2}+b c}{b+c}}-\sqrt{a}\right) \\
=\frac{\sqrt{a}(a-b)(a-c)}{(b+c) \sqrt{b+c}\left(\sqrt{a^{2}+b c}+\sqrt{a(b+c)}\right)}
\end{array}$$
The inequality can be rewritten as $\sum_{c y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,631 |
Example 1.10.14. Let \(a_{1}, a_{2}, \ldots, a_{n} (n \geq 3)\) be non-negative real numbers. Prove that
\[
\frac{n-1}{2} \sum_{i=1}^{n} a_{i}^{3} + \frac{3}{n-2} \sum_{c y c} a_{1} a_{2} a_{3} \geq \sum_{c y c} a_{1} a_{2} \left(a_{1} + a_{2}\right)
\] | SOLUTION. In this problem, we have $\alpha=\frac{n-1}{2}$ and $\beta=-1$. Because
$$3 \alpha+(n-1) \beta=\frac{3(n-1)}{2}-(n-1)=\frac{n-1}{2}>0,$$
according to the previous theorem (generalization for $n$ variables), we get that it suffices to consider the initial inequality in case some of the variables $a_{1}, a_{2},... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,632 |
Example 1.11.1. Let \(a, b, c\) be non-negative 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. The inequality is cyclic and holds for $a=b=c$, so, according to the previous theorem, we can assume that $b=1, c=0$. In this case, we have to prove that
$$\begin{array}{c}
\frac{a+2}{2}+\frac{1}{a}+\frac{2 a}{1+2 a} \geq 3 \\
\Leftrightarrow \frac{a}{2}+\frac{a+1}{a(1+2 a)} \geq 1
\end{array}$$
Applying AM-... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,633 |
Example 1.11.2. Let $a, b$, c be non-negative real numbers. For each $k \geq 0$, find the minimum of the expression
$$\frac{a+k b}{c+k b}+\frac{b+k c}{a+k c}+\frac{c+k a}{b+k a}$$ | SOLUTION. For $b=1, c=0$, the expression becomes
$$f(a)=\frac{a+k}{k}+\frac{1}{a}+\frac{k a}{1+k a}=2+\frac{a}{k}+\frac{1}{a}-\frac{1}{1+k a}$$
If $k \leq 1$ then, according to AM-GM inequality, we have
$$f(a) \geq 2+\frac{2}{\sqrt{k}}-1 \geq 3$$
We will now consider the main case, when $k \geq 1$. Clearly,
$$f(a)=2+... | 3 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,634 |
Example 1.11.3. Let \(a, b, c\) be non-negative real numbers. Prove that
\[1 \leq \frac{a+b}{a+4b+c} + \frac{b+c}{b+4c+a} + \frac{c+a}{c+4a+b} \leq \frac{4}{3}\] | SOLution. For $b=1, c=0$, the inequality becomes
$$\frac{a+1}{a+4}+\frac{1}{a+1}+\frac{a}{4 a+1} \geq 1(\star)$$
and
$$\frac{a+1}{a+4}+\frac{1}{a+1}+\frac{a}{4 a+1} \leq \frac{4}{3}(\star \star)$$
The inequality $(\star)$ is equivalent to (after expanding)
$$a^{3}-3 a^{2}+7 a+5 \geq 0$$
which is obvious because
$$a^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,635 |
Example 1.11.4. Let \(a, b, c\) be non-negative real numbers. Prove that
\[
\sqrt{\frac{a}{4a + 4b + c}} + \sqrt{\frac{b}{4b + 4c + a}} + \sqrt{\frac{c}{4c + 4a + b}} \leq 1
\] | SOLUTION. It suffices to prove that
$$\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b} \leq \frac{1}{3} \quad(\star)$$
For $c=0$, the inequality becomes
$$\frac{a}{4a+4b}+\frac{b}{4b+a} \leq \frac{1}{3}$$
or (after expanding)
$$(a-2b)^2 \geq 0$$
Therefore we are done according to the proposition. In $(\star)$, t... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,636 |
Example 1.11.5. Let $a, b, c$ be non-negative real numbers. For each $k, l \geq 0$, find the maximal and minimal value of the expression
$$\frac{a}{k a+l b+c}+\frac{b}{k b+l c+a}+\frac{c}{k c+l a+b} \text {. }$$ | SOLUTION. First we will examine the expression in the case $b=1, c=0$. Denote
$$f(a)=\frac{a}{k a+l}+\frac{1}{k+a}$$
then we get
$$f^{\prime}(a)=\frac{l}{(k a+l)^{2}}-\frac{1}{(k+a)^{2}}$$
The equation $f^{\prime}(a)=0$ has exactly one positive real root $a=\sqrt{l}$. So, if $l>k^{2}$ then
$$\begin{aligned}
\min f(a) ... | \min F=\min \left\{\frac{3}{k+l+1} ; \frac{1}{k} ; \frac{2}{k+\sqrt{l}}\right\} \\ \sup F=\max \left\{\frac{3}{k+l+1} ; \frac{1}{k} ; \frac{2}{k+\sqrt{l}}\right\} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,637 |
Example 1.11.6. Let \(a, b, c\) be non-negative real numbers. Prove that
\[
\frac{3a + 2b + c}{a + 2b + 3c} + \frac{3b + 2c + a}{b + 2c + 3a} + \frac{3c + 2a + b}{c + 2a + 3b} \geq 3
\] | SOLUTION. First we have to consider the inequality in case $b=1, c=0$. In this case, the inequality becomes
$$\begin{array}{l}
\frac{3 a+2}{a+2}+\frac{3+a}{3 a+1}+\frac{2 a+1}{2 a+3} \geq 3 \\
\Leftrightarrow \frac{2 a}{a+2}+\frac{3+a}{3 a+1} \geq \frac{2 a+5}{2 a+3}
\end{array}$$
or (after directly expanding)
$$4 a^{3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,638 |
Example 1.11.7. Let \(a, b, c\) be non-negative real numbers with sum 3. Prove that
$$\frac{a+b}{1+b}+\frac{b+c}{1+c}+\frac{c+a}{1+a} \geq 3$$ | SOLUTION. The inequality is equivalent to (homogeneous form)
$$\frac{a+b}{b+\frac{a+b+c}{3}}+\frac{b+c}{c+\frac{a+b+c}{3}}+\frac{c+a}{a+\frac{a+b+c}{3}} \geq 3$$
Since this problem is cyclic, homogeneous, and holds if $a=b=c$, we can assume that $c=0$ and $a+b=3$. In this case, we have to prove that
$$\begin{array}{l}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,639 |
Example 1.12.2. Let $a, b$, $c$ be positive real numbers such that $a+b+c=1$. Prove that
$$(a b+b c+c a)\left(\frac{a}{b^{2}+b}+\frac{b}{c^{2}+c}+\frac{c}{a^{2}+a}\right) \geq \frac{3}{4}$$ | SOLUTION. We will prove first the following result: for all $x \geq 0$
$$\frac{a}{(x+b)^{2}}+\frac{b}{(x+c)^{2}}+\frac{c}{(x+a)^{2}} \geq \frac{1}{(x+a b+b c+c a)^{2}}(\star)$$
Indeed, by Cauchy-Schwarz inequality, we obtain
$$\left(\sum_{c y c} \frac{a}{(x+b)^{2}}\right)\left(\sum_{c y c} a\right) \geq\left(\sum_{c y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,640 |
Example 1.12.3. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{1}{b(1+a b)}+\frac{1}{c(1+b c)}+\frac{1}{a(1+c a)} \geq \frac{1}{\sqrt[3]{a b c}\left(1+\sqrt[3]{a^{2} b^{2} c^{2}}\right)}$$ | SOLUTION. Taking into account example ??, we have
$$\frac{a}{(x+a b)^{2}}+\frac{b}{(x+b c)^{2}}+\frac{c}{(x+c a)^{2}} \geq \frac{3 \sqrt[3]{a b c}}{\left(x+\sqrt[3]{a^{2} b^{2} c^{2}}\right)^{2}}$$
Integrating on $[0,1]$, we get
$$\sum_{c y c} \int_{0}^{1} \frac{a}{(x+a b)^{2}} \geq \int_{0}^{1} \frac{3 \sqrt[3]{a b c... | \frac{1}{b(1+a b)}+\frac{1}{c(1+b c)}+\frac{1}{a(1+c a)} \geq \frac{3}{\sqrt[3]{a b c}\left(1+\sqrt[3]{a^{2} b^{2} c^{2}}\right)} | Inequalities | proof | Yes | Yes | inequalities | false | 737,641 |
Example 1.1.7. Let $a, b, c$ be non-negative real numbers. Prove that
$$\frac{a^{2}}{(2 a+b)(2 a+c)}+\frac{b^{2}}{(2 b+c)(2 b+a)}+\frac{c^{2}}{(2 c+a)(2 c+b)} \leq \frac{1}{3} .$$ | SOLUTION. If $c=0$, the problem is obvious. Suppose that $a, b, c>0$, then we have
$$\begin{aligned}
1-3 \sum_{c y c} \frac{a^{2}}{(2 a+b)(2 a+c)} & =\sum_{c y c}\left(\frac{a}{a+b+c}-\frac{a^{2}}{(2 a+b)(2 a+c)}\right) \\
& =\sum_{c y c} \frac{a(a-b)(a-c)}{(a+b+c)(2 a+b)(2 a+c)} \\
& =\sum_{c y c} \frac{a^{2}\left(\fr... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,642 |
Example 1.13.1. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{1}{4 a}+\frac{1}{4 b}+\frac{1}{4 c}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \geq \frac{3}{3 a+b}+\frac{3}{3 b+c}+\frac{3}{3 c+a}$$ | SOLUTION. There seems to be no purely algebraic solution to this hard inequality, however, the integral method makes up a very impressive one.
According to a well-known inequality (see problem ?? in volume I), we have
$$\left(x^{2}+y^{2}+z^{2}\right)^{2} \geq 3\left(x^{3} y+y^{3} z+z^{3} x\right)$$
Denote $x=t^{a}, y... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,643 |
Example 1.13.2. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{1}{3a}+\frac{1}{3b}+\frac{1}{3c}+\frac{3}{a+b+c} \geq \frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}+\frac{1}{2b+a}+\frac{1}{2c+b}+\frac{1}{2a+c}$$ | SOLUTION. Starting from Schur inequality, we have
$$x^{3}+y^{3}+z^{3}+3 x y z \geq x y(x+y)+y z(y+z)+z x(z+x)$$
Letting now $x=t^{a}, y=t^{b}, z=t^{c}$, the above inequality becomes
$$\frac{1}{t}\left(t^{3 a}+t^{3 b}+t^{3 c}+3 t^{a+b+c}\right) \geq \frac{1}{t}\left(t^{a+b}\left(t^{a}+t^{b}\right)+t^{b+c}\left(t^{b}+t^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,644 |
Example 1.13.4. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{8}{a+b+c} \geq \frac{17}{3}\left(\frac{1}{2 a+b}+\frac{1}{2 b+c}+\frac{1}{2 c+a}\right) .$$ | SOLUTION. Let $k=\frac{8}{9}$, then we need to prove that
$$\frac{1}{3 a}+\frac{1}{3 b}+\frac{1}{3 c}+\frac{3 k}{a+b+c} \geq(k+1)\left(\frac{1}{2 a+b}+\frac{1}{2 b+c}+\frac{1}{2 c+a}\right).$$
According to example 1.10.1, we have
$$x^{3}+y^{3}+z^{3}+3 k x y z \geq(k+1)\left(x^{2} y+y^{2} z+z^{2} x\right)$$
Denote $x=... | \frac{1}{3 a}+\frac{1}{3 b}+\frac{1}{3 c}+\frac{8}{3(a+b+c)} \geq \frac{17}{9}\left(\frac{1}{2 a+b}+\frac{1}{2 b+c}+\frac{1}{2 c+a}\right) | Inequalities | proof | Yes | Yes | inequalities | false | 737,646 |
Example 1.13.8. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{4}{a}+\frac{4}{b}+\frac{4}{c}+\frac{36}{2 a+b}+\frac{36}{2 b+c}+\frac{36}{2 c+a} \geq \frac{45}{a+2 b}+\frac{45}{b+2 c}+\frac{45}{c+2 a}+\frac{9}{a+b+c}$$ | SOLUTION. We use the following familiar result
$$27\left(a b^{2}+b c^{2}+c a^{2}+a b c\right) \leq 4(a+b+c)^{3}$$
to deduce the desired result. Equality holds if and only if $a=b=c$. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,650 |
Example 1.14.4. Let $a, b, c, d$ be non-negative real numbers with sum 1. Prove that
$$a b c + b c d + c d a + d a b \leq \frac{1}{27} + \frac{176}{27} a b c d$$ | SOLUTION. In this problem, we fix \(a+b=m\) and \(c+d=n\). Let \(x=ab\) and \(y=cd\) then
\[ ab c + bc d + cd a + da b - \frac{1}{27} - \frac{176}{27} abcd = m y + n x - \frac{1}{27} - \frac{176 xy}{27} = f(x, y) \]
is a linear (convex) function in both \(x\) and \(y\). It only reaches the maximum at boundary values, n... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,651 |
Example 1.14.5. Let $a, b, c, d$ be non-negative real numbers with sum 4. Prove that
$$a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}+a^{2} c^{2}+b^{2} d^{2}+10 a b c d \leq 16$$ | Solution. We fix $a+b=m$ and $c+d=n$. Let $ab=x$ and $cd=y$ then
$$\sum_{cyc} a^{2} b^{2}+10 abcd=x^{2}+y^{2}+\left(m^{2}-2 x\right)\left(n^{2}-2 y\right)+10 xy$$
are convex functions in each variable $x$ and $y$. Therefore we only need to consider the case
$$x \in\left\{0 ; \frac{m^{2}}{4}\right\} ; y \in\left\{0 ; \f... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,652 |
Example 1.1.8. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{1}{a^{2}+2 b c}+\frac{1}{b^{2}+2 c a}+\frac{1}{c^{2}+2 a b} \leq\left(\frac{a+b+c}{a b+b c+c a}\right)^{2}$$ | Solution. We have
$$\begin{aligned}
\sum_{c y c} \frac{a b+b c+c a}{a^{2}+2 b c}-\frac{(a+b+c)^{2}}{a b+b c+c a} & =\sum_{c y c}\left(\frac{a b+b c+c a}{a^{2}+2 b c}-1\right)+\sum_{c y c} \frac{(c-a)(c-b)}{a b+b c+c a} \\
& =\sum_{c y c}(a-b)(a-c)\left(\frac{1}{a^{2}+2 b c}+\frac{1}{a b+b c+c a}\right)
\end{aligned}$$
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,653 |
Example 1.14.6. Let \(a, b, c, d\) be non-negative real numbers with sum 4. Prove that
$$(1+3a)(1+3b)(1+3c)(1+3d) \leq 125 + 131abcd$$ | SOLUTION. We fix $a+b=m$ and $c+d=n$. Let $x=ab$ and $y=cd$ (we regard $x$ and $y$ as variables). The inequality becomes
$$(1+9y+3n)(1+9x+3m)-125-131xy \geq 0$$
This expression is a linear (and also convex) function in each variable $x$ and $y$, we get that it suffices to consider the case
$$x \in\left\{0 ; \frac{m^2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,654 |
Example 1.14.7. Let \(a, b, c, d\) be positive real numbers such that \(a b c d=1\). Prove that
\[
a^{3}+b^{3}+c^{3}+d^{3}+12 \geq 2(a+b+c+d+a b c+b c d+c d a+d a b)
\] | SOLUTION. We need to prove that $f(x)+f(y)+f(z)+f(t) \geq 4$ where $x, y, z, t$ are $\ln a, \ln b, \ln c, \ln d$ respectively and
$$f(x)=e^{3 x}-2 e^{x}-2 e^{-x} \text {. }$$
Clearly
$$f^{\prime \prime}(x)=9 e^{3 x}-2 e^{x}-2 e^{-x}$$
Denote $t=e^{x}$. The equation $f^{\prime \prime}(x)=0$ is equivalent to $9 t^{4}-2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,655 |
Example 1.14.8. Let \(a, b, c, d\) be positive real numbers with sum 4. Prove that
\[9\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)+56 \geq 15\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\] | SOLUTION. We have to prove that $f(a)+f(b)+f(c)+f(d) \geq 0$ where
$$f(x)=\frac{9}{x}-15 x^{2}$$
Since $f^{\prime \prime}(x)=\frac{18}{x^{3}}-30$ has exactly one positive real root, we infer that $f(x)$ has a single inflection point. Applying SIP theorem, we only need to consider the inequality in case $a=b=c=x \leq \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,656 |
Example 1.14.9. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1} a_{2} \ldots a_{n}=1$. Prove that
$$\frac{1}{n-1+a_{1}}+\frac{1}{n-1+a_{2}}+\ldots+\frac{1}{n-1+a_{n}} \leq 1$$ | SOLUTION. We have to prove that
$$f\left(x_{1}\right)+f\left(x_{2}\right)+\ldots+f\left(x_{n}\right) \leq 1$$
where $x_{i}=\ln a_{i} \forall i \in\{1,2, \ldots, n\}$ and
$$f(x)=\frac{1}{n-1+e^{x}}$$
We have
$$f^{\prime \prime}(x)=\frac{2\left(e^{x}-(n-1)\right)}{\left(n-1+e^{x}\right)^{3}}$$
Since the function $f^{\p... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,657 |
Example 1.14.10. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers with product 1. Prove that
$$a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}-n \geq \frac{2 n}{n-1} \sqrt[n]{n-1}\left(a_{1}+a_{2}+\ldots+a_{n}-n\right)$$ | SOLUTION. For \( k = \frac{2 n}{n-1} \sqrt[n]{n-1} \), we consider the following function
\[ f(x) = e^{2 x} - k e^{x}. \]
We have to prove that \( f(x_1) + f(x_2) + \ldots + f(x_n) \geq (1 - k) n \) where \( x_i = \ln a_i \forall i \in \{1, 2, \ldots, n\} \). Since the function
\[ f''(x) = 4 e^{2 x} - k e^{x} \]
has e... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,658 |
Example 1.14.11. Let \(a, b, c, d, e, f\) be positive real numbers with sum 6. Prove that
\[
\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+d^{2}\right)\left(1+e^{2}\right)\left(1+f^{2}\right) \geq (1+a)(1+b)(1+c)(1+d)(1+e)(1+f)
\] | SOLUTION. Consider the following function in the positive variable $x$
$$f(x)=\ln \left(1+x^{2}\right)-\ln (1+x)$$
We have certainly
$$f^{\prime \prime}(x)=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}+\frac{1}{(1+x)^{2}}$$
The equation $f^{\prime \prime}(x)=0$ is equivalent to
$$g(x)=3 x^{4}+4 x^{3}+2 x^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,659 |
Example 1.14.12. Let \(a, b, c, d\) be positive real numbers with sum 4. Prove that
\[
\left(1+a^{2}\right)\left(1+b^{2}\right)\left(1+c^{2}\right)\left(1+d^{2}\right) \geq \frac{10^{4}}{9^{3}}
\] | SOLUTION. We need to prove that
$$f(a)+f(b)+f(c)+f(d) \geq 4 \ln 10 - 3 \ln 9$$
where \( f(x) = \ln(1 + x^2) \). Since
$$f''(x) = \frac{2(1 - x^2)}{(1 + x^2)^2}$$
has exactly one positive real root \( x = 1 \), we obtain by the SIP theorem that there exists a number \( p \leq 1 \) for which
$$f(a) + f(b) + f(c) + f(d) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,660 |
Example 1.1.9. Let $a, b, c$ be non-negative real numbers such that $a b+b c+c a=1$. Prove that
$$\frac{1+b^{2} c^{2}}{(b+c)^{2}}+\frac{1+c^{2} a^{2}}{(c+a)^{2}}+\frac{1+a^{2} b^{2}}{(a+b)^{2}} \geq \frac{5}{2}$$ | Solution. First we have that
$$\sum_{c y c} \frac{1+b^{2} c^{2}}{(b+c)^{2}}=\sum_{c y c} \frac{(a b+b c+c a)^{2}+b^{2} c^{2}}{(b+c)^{2}}=\sum_{c y c} a^{2}+2 \sum_{c y c} \frac{a b c}{b+c}+\frac{2 b^{2} c^{2}}{(b+c)^{2}}$$
Therefore, our inequality can be rewritten as
$$2\left(\sum_{c y c} a^{2}-\sum_{c y c} a b\right... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,664 |
Example 1.15.1. Let $a, b, c, d$ be non-negative real numbers. Prove that
$$a^{4}+b^{4}+c^{4}+d^{4}+2 a b c d \geq a^{2} b^{2}+b^{2} c^{2}+c^{2} d^{2}+d^{2} a^{2}+a^{2} c^{2}+b^{2} d^{2}$$ | SOLUTION. If we fix $a+b+c=$ const and $a^{2}+b^{2}+c^{2}=$ const, then
$$\text { RHS }- \text { LHS }=\left(a^{2}+b^{2}+c^{2}\right)^{2}-3(a b+b c+c a)^{2}+6 a b c(a+b+c)+2 a b c d-d^{2}\left(a^{2}+b^{2}+c^{2}\right)$$
is certainly an increasing function of $a b c$. By corollary 3, it's enough to prove the inequality ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,665 |
Example 1.15.2. Let $a, b, c, d$ be non-negative real numbers with sum 4. Prove that
$$(1+2 a)(1+2 b)(1+2 c)(1+2 d) \leq 10\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+41 a b c d$$ | SOLUTION. First, notice that if $a \leq b \leq c \leq d$ and $c \leq 1 / 3$ then $a, b \leq 1 / 3$ and $d \geq 3$ and we are done because
$$10\left(a^{2}+b^{2}+c^{2}+d^{2}\right)+41 a b c d \geq 90>(1+2 a)(1+2 b)(1+2 c)(1+2 d)$$
So we may assume that $a \leq b \leq c \leq d, c \geq \frac{1}{3}$. We fix $c=$ const, $a^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,666 |
Example 1.15.3. Let $x_{1}, x_{2}, \ldots, x_{n}$ be non-negative real numbers such that $x_{1}+x_{2}+\ldots+$ $x_{n}=n$. Prove that
$$\left(x_{1} x_{2} \ldots x_{n}\right)^{\frac{1}{\sqrt{n-1}}}\left(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}\right) \leq n$$ | SOLUTION. If we fix $x_{1}+x_{2}+x_{3}$ and $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ then the left-hand expression of the above inequality is clearly a strictly increasing function of $x_{1} x_{2} x_{3}$, so, according to the corollary 2 of $n \mathbf{S M V}$ theorem, we conclude that it suffices to consider the initial inequal... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,667 |
Example 1.15.4. Let \(a, b, c, d\) be positive real numbers such that that
\[2(a+b+c+d)^{2}=5\left(a^{2}+b^{2}+c^{2}+d^{2}\right)\]
Find the minimum value of
\[P=\frac{a^{4}+b^{4}+c^{4}+d^{4}}{a b c d}\] | SOLUTION. First we guess that the equality holds for $b=c=d$ (in this case, we find out $a=2+\sqrt{5}, b=c=d=1$ and permutations) and this is the key to the solution. Indeed, denote $k=(2+\sqrt{5})^{3}+(2+\sqrt{5})^{-1}>78$, we will prove
$$a^{4}+b^{4}+c^{4} \geq 4 a b c d$$
WLOG, assume that $a \geq b \geq d \geq c$.... | k | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,668 |
Example 1.15.5. Suppose that $a_{1}, a_{2}, \ldots, a_{n}$ are positive real numbers satisfying
$$a_{1}+a_{2}+\ldots+a_{n}=a_{1}^{-1}+a_{2}^{-1}+\ldots+a_{n}^{-1}=n+2$$
Find the minimum and maximum value of
$$P=a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}+2 a_{1} a_{2} \ldots a_{n}$$ | SOLUTION. Without loss of generality, we may assume that \(a_{1} \geq a_{2} \geq \ldots \geq a_{n-1} \geq a_{n}\). We fix \(s=a_{1}+a_{2}+a_{n}\) and \(r=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{n}}\). Denote \(x=a_{1} a_{2} a_{n}\) and \(p=a_{3} a_{4} \ldots a_{n-1}\) then the expression \(P\) can be rewritten into... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 737,669 |
Example 1.15.6. Let $a, b, c, d$ be non-negative real numbers with sum 1. Prove that
$$2\left(a^{2}+b^{2}+c^{2}+d^{2}\right) \geq 27 \sqrt[3]{\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)\left(a^{2}+d^{2}\right)\left(b^{2}+c^{2}\right)\left(b^{2}+d^{2}\right)\left(c^{2}+d^{2}\right)}$$ | SOLUTION. We will prove a homogeneous inequality as follows
$$2\left(\sum_{c y c} a\right)^{2}\left(\sum_{c y c} a^{2}\right) \geq 27 \sqrt[3]{\prod_{\mathrm{sym}}\left(a^{2}+b^{2}\right)}$$
WLOG, assume that $a \geq b \geq c \geq d$. If we fix $a+c+d, a^{2}+c^{2}+d^{2}$ and let $a c d=x$ then
$$\left(a^{2}+c^{2}\righ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,670 |
Example 1.15.8. Let $x_{1}, x_{2}, \ldots, x_{n}(n \geq 4)$ be positive real numbers such that $x_{1}+x_{2}+\ldots+x_{n}=n$. Prove that
$$\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}+\frac{\sqrt{n}(\sqrt{n+4}+2 \sqrt{n-1})}{\sqrt{x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}}} \geq n+2 \sqrt{n-1}+\sqrt{n+4}$$ | SOLUTION. Similarly with the previous example, if we fix $x_{1}+x_{2}+x_{3}, x_{1}^{2}+x_{2}^{2}+x_{3}^{2}$ and fix $x_{4}, x_{4}, \ldots, x_{n}$ then the left hand side of the inequality is a decreasing function of $x_{1} x_{2} x_{3}$. So we may assume that $x_{1}=x_{2}=\ldots=x_{n-1}$. For convenience, we may conside... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,672 |
Example 1.17.1. If $f$ is a convex function then
$$f(a)+f(b)+f(c)+f\left(\frac{a+b+c}{3}\right) \geq \frac{4}{3}\left(f\left(\frac{a+b}{2}\right)+f\left(\frac{b+c}{2}\right)+f\left(\frac{c+a}{2}\right)\right)$$ | SOLUTION. WLOG, suppose that $a \geq b \geq c$. Consider the following number sequences
$$(x)=(a, a, a, b, t, t, t, b, b, c, c, c) \quad ; \quad(y)=(\alpha, \alpha, \alpha, \alpha, \beta, \beta, \beta, \beta, \gamma, \gamma, \gamma, \gamma)$$
where
$$t=\frac{a+b+c}{3}, \alpha=\frac{a+b}{2}, \beta=\frac{a+c}{2}, \gamma=... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,673 |
Example 1.17.2 (Jensen Inequality). If \( f \) is a convex function then
\[ f\left(a_{1}\right)+f\left(a_{2}\right)+\ldots+f\left(a_{n}\right) \geq n f\left(\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}\right) \] | SOLUTION. We use property 1 of majorization. Suppose that $a_{1} \geq a_{2} \geq \ldots \geq a_{n}$, then we have $\left(a_{1}, a_{2}, \ldots, a_{n}\right) \gg (a, a, \ldots, a)$ with $a=\frac{1}{n}\left(a_{1}+a_{2}+\ldots+a_{n}\right)$. Our problem is directly deduced from Karamata inequality for these two sequences. | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,674 |
Example 1. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+c}{a+b}+\frac{b+c}{b+a}+\frac{c+a}{c+b}$$ | Solution. Without loss of generality, assume that $c=\min (a, b, c)$. Note that for $x, y, z>0$ we have
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-3=\frac{1}{x y}(x-y)^{2}+\frac{1}{x z}(x-z)(y-z)$$
Therefore the given inequality can be rewritten as
$$\left(\frac{1}{a b}-\frac{1}{(a+c)(b+c)}\right)(a-b)^{2}+\left(\frac{1}{a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,675 |
Example 2. Let $a, b, c$ be positive real numbers such that $a \geq b \geq c$. Prove that
$$a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \geq 0$$ | Solution. We have
$$\begin{aligned}
f(a, b, c)= & a^{2} b(a-b)+b^{2} c(b-c)+c^{2} a(c-a) \\
= & {\left[a^{2} b(a-b)-a b^{2}(a-b)\right]+\left[b^{2} c(b-c)-a b^{2}(b-c)\right]+} \\
& +\left[c^{2} a(c-a)-a b^{2}(c-a)\right] \\
= & a b(a-b)^{2}+\left(a b+a c-b^{2}\right)(a-c)(b-c)
\end{aligned}$$
Clearly the last express... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,676 |
Example 3. Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}+\frac{3(a b+b c+c a)}{(a+b+c)^{2}} \geq 4$$ | Solution. Without loss of generality, assume that $c=\min (a, b, c)$. We have
$$\begin{aligned}
\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}-3 & =\frac{1}{(a+c)(b+c)}(a-b)^{2}+\frac{1}{(a+b)(b+c)}(a-c)(b-c) \\
\frac{3(a b+b c+c a)}{(a+b+c)^{2}}-1 & =-\frac{1}{(a+b+c)^{2}}(a-b)^{2}-\frac{1}{(a+b+c)^{2}}(a-c)(b-c)
\en... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,677 |
I. From SOS: $S=f(a, b, c)=S_{a}(b-c)^{2}+S_{b}(c-a)^{2}+S_{c}(a-b)^{2}$.
Including $S_{a}, S_{b}, S_{c}$ is functions have $a, b, c$ is variables.
1. If $S_{a}, S_{b}, S_{c} \geq 0$ then $S \geq 0$.
2. If $a \geq b \geq c$ and $S_{b}, S_{b}+S_{c}, S_{b}+S_{a} \geq 0$ then $S \geq 0$.
3. If $a \geq b \geq c$ and $S_{b... | 1. Of course that $(a-b)^{2},(b-c)^{2},(c-a)^{2} \geq 0$ but $S_{a}, S_{b}, S_{c} \geq 0$ so $S \geq 0$.
2. Because $a \geq b \geq c$ so $(a-c)^{2}=(a-b)^{2}+(b-c)^{2}+2(a-b)(b-c) \geq (a-b)^{2}+(b-c)^{2}$
So $S_{a}(b-c)^{2}+S_{b}(c-a)^{2}+S_{c}(a-b)^{2}=\left(S_{a}+S_{b}\right)(b-c)^{2}+\left(S_{b}+S_{c}\right)(a-b)^... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,678 |
Problem 1. Let a, b, c, x, y, z be six real (not necessarily nonnegative) numbers. Assume that $a \geq b \geq c$. Also, assume that either $x \geq y \geq z$ or $x \leq y \leq z$. Then,
$$\left(\sum_{\text {cyc }}(a-b)(a-c)\right) \cdot\left(\sum_{\text {cyc }} x^{2}(a-b)(a-c)\right) \geq\left(\sum_{\text {cyc }} x(a-b)... | Solution:
The inequality we have to prove rewrites as
$$\begin{array}{l}
\left(\sum_{\text {cyc }}(a-b)(a-c)\right) \cdot\left(\sum_{\text {cyc }} x^{2}(a-b)(a-c)\right)-\left(\sum_{\text {cyc }} x(a-b)(a-c)\right)^{2} \geq 0 \\
\quad\left(\sum_{\text {cyc }}(a-b)(a-c)\right) \cdot\left(\sum_{\text {cyc }} x^{2}(a-b)(a... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,679 |
(c) Let $a, b, c$ are three positive reals, prove that,
$$\frac{b+c-a}{5 a^{2}+4 b c}+\frac{c+a-b}{5 b^{2}+4 c a}+\frac{a+b-c}{5 c^{2}+4 a b} \geq \frac{1}{a+b+c}$$ | Solution:
(a) We have
$$1-\frac{(b+c-a)(c+a-b)(a+b-c)}{a b c}=\sum \frac{a(a-b)(a-c)}{a b c}=\sum \frac{(a-b)(a-c)}{b c}$$
and
$$\sum \frac{2 a}{b+c}-3=\sum \frac{(a-b)(a-c)}{b+c}\left(\frac{1}{a+b}+\frac{1}{c+a}\right)$$
Thus it suffices to show that $\sum X_{a}(a-b)(a-c) \geq 0$
where
$$X_{a}=\frac{1}{b c}-\frac{1}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,680 |
Problem 6. Let $a, b, c$ be nonnegative real numbers. Find the maximum of $k$ such that the inequality
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+k \cdot \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geq k+\frac{3}{2}$$ | Solution:
We have that equality $\sum \frac{a}{b+c}-\frac{3}{2}=\frac{1}{2} \sum \frac{(a-b)^{2}}{(a+c)(b+c)}$.
That inequality can be rewritten as
$$\sum \frac{(a-b)^{2}}{(a+c)(b+c)} \geq k \sum \frac{(a-b)^{2}}{a^{2}+b^{2}+c^{2}} \Leftrightarrow \sum(a-b)^{2}\left(\frac{a^{2}+b^{2}+c^{2}}{(a+c)(b+c)}-k\right) \geq 0$... | \frac{\sqrt{3}-1}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,681 |
Problem 8a(Nguyen Duy Tung). Let $a, b, c$ be the nonnegative real numbers. Prove that:
$$\frac{a^{3}+b^{3}}{c^{2}+a b}+\frac{b^{3}+c^{3}}{a^{2}+b c}+\frac{c^{3}+a^{3}}{b^{2}+c a} \geq 2\left(\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\right)$$ | The inequality equivalent to
$$\begin{array}{c}
\sum \frac{(a+b)(b+c)(c+a)\left(a^{3}+b^{3}\right)}{c^{2}+a b} \geq 2\left[a^{4}+b^{4}+c^{4}+a b c(a+b+c)+\sum a^{3}(b+c)\right] \\
\Leftrightarrow \sum(a+b)\left(a^{3}+b^{3}\right)\left(1+\frac{c(a+b)}{a^{3}+b^{3}} \geq 2\left[a^{4}+b^{4}+c^{4}+a b c(a+b+c)+\sum a^{3}(b+... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,682 |
Problem $\mathbf{8 b}$ (Nguyen Huy Tung). Let $a, b, c$ be positive real numbers. Prove that
$$a^{3}+b^{3}+c^{3}+3 a b c \cdot \frac{a^{2} b+b^{2} c+c^{2} a}{a b^{2}+b c^{2}+c a^{2}} \geq a b(a+b)+b c(b+c)+c a(c+a) .$$ | Solution:
WLOG, we may assume $b$ is the number between the two numbers $a$ and $c$.
If $a \geq b \geq c$ then $a^{2} b + b^{2} c + c^{2} a \geq a b^{2} + b c^{2} + c a^{2}$. By Schur Inequality we have
$$a^{3} + b^{3} + c^{3} + 3 a b c \cdot \frac{a^{2} b + b^{2} c + c^{2} a}{a b^{2} + b c^{2} + c a^{2}} \geq a^{3} + ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,683 |
Problem 9(Nguyen Duy Tung, Nguyen Trong Tho). Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a^{3}}{2 a^{2}+b^{2}}+\frac{b^{3}}{2 b^{2}+c^{2}}+\frac{c^{3}}{2 c^{2}+a^{2}} \geq \frac{a+b+c}{3} .$$ | Solution:
We have that
$$\begin{array}{c}
\sum \frac{a^{3}-a b^{2}}{2 a^{2}+b^{2}} \geq 0 \Leftrightarrow \sum\left(a^{3}-a b^{2}\right)\left(2 b^{2}+c^{2}\right)\left(2 c^{2}+a^{2}\right) \geq 0 \\
\Leftrightarrow 3 \sum a^{3} b^{2} c^{2}+2 \sum a^{3} c^{4}+2 \sum a^{5} b^{2}+\sum a^{5} c^{2} \geq 4 \sum a b^{4} c^{2}... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,684 |
Problem 10(Nguyen Duy Tung). Let $a, b, c$ be nonnegative real numbers. Prove that
$$\frac{4 a}{a+b}+\frac{4 b}{b+c}+\frac{4 c}{c+a}+\frac{a b^{2}+b c^{2}+c a^{2}+a b c}{a^{2} b+b^{2} c+c^{2} a+a b c} \geq 7$$ | Solution:
We have that
$$\begin{array}{c}
2\left(3-\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\right)+\left(\frac{a b^{2}+b c^{2}+c a^{2}+a b c}{a^{2} b+b^{2} c+c^{2} a+a b c}-1\right) \geq 6 \\
\Leftrightarrow \frac{(a-b)(b-c)(c-a)}{a^{2} b+b^{2} c+c^{2} a+a b c}-\frac{2(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)} \geq 0 \\
\Leftrig... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,685 |
Problem 11(Vasile Cirtoaje). Let $a, b, c$ be nonnegative real numbers. Prove that
$$a^{3}+b^{3}+c^{3}+2\left(a^{2} b+b^{2} c+c^{2} a\right) \geq 3\left(a b^{2}+b c^{2}+c a^{2}\right)$$ | Solution:
WLOG, we may assume $b$ is the number between the two numbers $a$ and $c$.
If $a \geq b \geq c$ then $2\left(a^{2} b+b^{2} c+c^{2} a\right) \geq 2\left(a b^{2}+b c^{2}+c a^{2}\right)$
And $a^{3}+b^{3}+c^{3} \geq a b^{2}+b c^{2}+c a^{2}$ so the inequality is true.
If $c \geq b \geq a$ the inequality is equival... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,686 |
Problem 12(Nguyen Duy Tung) Let $a, b, c$ be nonnegative real numbers. Prove that
$$4(a+b+c)^{3} \geq 27\left(a b^{2}+b c^{2}+c a^{2}+a b c\right)$$ | Solution:
WLOG, we may assume $b$ is the number between the two numbers $a$ and $c$. If $a \geq b \geq c$ then $a b^{2}+b c^{2}+c a^{2}+a b c \geq a b^{2}+b c^{2}+c a^{2}+a b c$ So
$$27\left(a b^{2}+b c^{2}+c a^{2}+a b c\right) \leq \frac{27}{2}\left(a b^{2}+b c^{2}+c a^{2}+a b^{2}+b c^{2}+c a^{2}+a b c\right)$$
So we... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,687 |
Problem 9(Nguyen Duy Tung). Let $a, b, c$ be nonnegative real numbers. Prove that
$$\frac{a^{2}+b c}{b^{2}+c^{2}}+\frac{b^{2}+c a}{c^{2}+a^{2}}+\frac{c^{2}+a b}{a^{2}+b^{2}} \geq \frac{5}{2}+\frac{4 a^{2} b^{2} c^{2}}{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}$$ | Solution:
The inequality equivalent to
$$\begin{array}{c}
2 \sum\left[\left(a^{2}+b c\right)\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)\right] \geq\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)+8 a^{2} b^{2} c^{2} \\
\Leftrightarrow 2 \sum a^{6}+2 \sum b^{3} c^{3}+2 a b c \sum a^{3}+2 a b... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,688 |
Problem 10. Let $a, b$ and $c$ be positive numbers. Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \sqrt{\frac{a^{2}+b^{2}+c^{2}}{a b+a c+b c}}$$ | Solution:
Notice that if $a \geq b \geq c$ then
$$\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$$
so it is enough to consider the case $a \geq b \geq c$, we will prove
$$\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2} \geq 9\l... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,689 |
General: Let $p$ and $q$ be real numbers such that $p q>0$, and let $a, b, c$ be non-negative real numbers. Prove that $S_{0} \cdot S_{p+q} \geq S_{p} \cdot S_{q}$, where $S_{k}=\sum_{c y c} a^{k}(a-b)(a-c)$ Problem 2(Darij Grinberg). If $p$ is an even nonnegative integer, then the inequality $\sum_{c \mathrm{cyc}} a^{... | Solution: Since the inequality in question is symmetric, we can WLOG assume that \(a \geq b \geq c\). Since \(p\) is an even nonnegative integer, we have \(p=2n\) for some nonnegative integer \(n\).
Define a function sign by \(\operatorname{sign} t=\left\{\begin{array}{cc}-1, & \text{if } t < 0 \\ 0, & \text{if } t = ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,690 |
Problem 11(Nguyen Duy Tung). Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a(b+c)}{b^{2}+c^{2}}+\frac{b(c+a)}{c^{2}+a^{2}}+\frac{c(a+b)}{a^{2}+b^{2}} \geq 2+\frac{8 a^{2} b^{2} c^{2}}{\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)}$$ | Solution: $\quad \begin{array}{ll}\text { S }\end{array}$
The inequality equivalent to
$$\begin{aligned}
& \sum\left[a(b+c)\left(a^{2}+b^{2}\right)\left(a^{2}+c^{2}\right)\right] \geq 2\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right)+8 a^{2} b^{2} c^{2} \\
\Leftrightarrow & \sum a^{5}(b+c)+2 \su... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,691 |
Problem 12 (Vo Quoc Ba Can). Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{8}{3} \cdot \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geq \frac{17}{3} .$$ | Solution:
The inequality equivalent
$$\begin{array}{c}
\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}-3\right)+\frac{8}{3} \cdot\left(\frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}}-1\right) \geq 0 \\
\Leftrightarrow M(a-b)^{2}+N(a-b)(a-c) \geq 0
\end{array}$$
With \( M=\frac{1}{a b}-\frac{8}{3\left(a^{2}+b^{2}+c^{2}\right)} \) and ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,692 |
Problem 13(Nguyen Duy Tung, Vo Quoc Ba Can). Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geq \frac{6\left(a^{2}+b^{2}+c^{2}\right)-3(a b+b c+c a)}{a+b+c}$$ | Solution: WLOG, Assume $b$ is number between two numbers $a$ and $c$.
In case $c \geq b \geq a$ then: $\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \geq \frac{a^{2}}{c}+\frac{c^{2}}{b}+\frac{b^{2}}{a}$.
Indeed, it is equivalent to $\sum \frac{a^{2}}{b}-\sum \frac{a^{2}}{c}=\frac{a^{3}-b^{3}}{a b}+\frac{b^{3}-c^{3}}{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,693 |
Problem 14(Nguyen Duy Tung). Let $a, b, c$ be three side-lengths of a triangle. Prove that
$$2\left(\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\right) \geq a+b+c+\frac{b^{2}}{a}+\frac{c^{2}}{b}+\frac{a^{2}}{c} .$$ | Clearly, this one is equivalent to
$$\begin{array}{c}
\frac{(a-b)^{2}}{b}+\frac{(b-c)^{2}}{c}+\frac{(c-a)^{2}}{a} \geq \frac{(a-b)(b-c)(c-a)(a+b+c)}{a b c} \\
\Leftrightarrow \sum_{c y c} a c(a-b)^{2} \geq(a-b)(a-c)(b-c)(a+b+c)
\end{array}$$
The above form shows that we only need to prove it in the case $a \geq b \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,694 |
Problem 21 Let \(a, b, c\) be positive real numbers such that \(a+b+c=3\). Prove that
\[a^{2} b+b^{2} c+c^{2} a \leq 4\] | Solution: The inequality equivalent to
$$\begin{array}{c}
2 \sum_{c y c} a^{2} b \leq 8 \\
\Leftrightarrow\left(\sum_{c y c} a^{2} b+\sum_{c y c} a b^{2}\right)+\left(\sum_{c y c} a^{2} b-\sum_{c y c} a b^{2}\right) \leq 8 \\
\Leftrightarrow \sum_{s y m} a^{2}(b+c)+(a-b)(b-c)(c-a) \leq 8
\end{array}$$
Then we need to ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,695 |
Problem 22 Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Prove that
$$a^{2} b+b^{2} c+c^{2} a+2\left(a b^{2}+b c^{2}+c a^{2}\right) \leq 6 \sqrt{3}$$ | Solution: The inequality equivalent to
$$\begin{array}{c}
\Leftrightarrow 2 \sum_{cyc} a^{2} b+4 \sum_{cyc} a b^{2} \leq 12 \sqrt{3} \\
\Leftrightarrow 3 \sum_{sym} a^{2}(b+c)+\left(\sum_{cyc} a b^{2}-\sum_{cyc} a^{2} b\right) \leq 12 \sqrt{3} \\
\Leftrightarrow 3 \sum_{cyc} a^{2}(b+c)+(a-b)(b-c)(c-a) \leq 12 \sqrt{3}
... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,696 |
Problem 23 Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Prove that
$$k(a+b+c)^{4} \geq\left(a^{3} b+b^{3} c+c^{3} a\right)+\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+a b c(a+b+c)$$ | Solution: Let $a=2, b=1, c=0 \Rightarrow k \geq \frac{4}{27}$.
We'll prove $k \geq \frac{4}{27}$ is the better constant.
The inequality is equivalent to
$$\begin{array}{c}
\frac{4}{27}(a+b+c)^{4} \geq \sum_{c y c} a^{3} b+\sum_{s y m} b^{2} c^{2}+a b c \sum_{s y m} a \\
\Leftrightarrow \frac{8}{27}(a+b+c)^{4} \geq\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,697 |
Problem 24 Let $a, b, c$ be nonnegative real numbers. Find the best constant $k$ such that the inequality always holds
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k \frac{a b+b c+c a}{a^{2}+b^{2}+c^{2}} \geq 3+k$$ | Solution: We have
$$2 \sum_{c y c} \frac{a}{b}=\left(\sum_{c y c} \frac{a}{b}+\sum_{c y c} \frac{b}{a}\right)+\left(\sum_{c y c} \frac{a}{b}-\sum_{c y c} \frac{b}{a}\right)=\frac{\sum_{s y m} a^{2}(b+c)}{a b c}+\frac{(a-b)(b-c)(c-a)}{a b c}$$
Thus, the inequality equivalent to
$$\sum_{c y c} \frac{a}{b}+2 k \frac{a b+... | 3 \sqrt[3]{4}-2 | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,698 |
Problem 25 Let $a, b, c$ be nonnegative real numbers. Find the best constant $k$ such that the inequality always holds:
$$\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}+k(a+b+c) \geq 3(k+1) \frac{a^{2}+b^{2}+c^{2}}{a+b+c}$$ | Solution: We have
$$2 \sum_{c y c} \frac{a^{2}}{b}=\left(\sum_{c y c} \frac{a^{2}}{b}+\sum_{c y c} \frac{b^{2}}{a}\right)+\left(\sum_{c y c} \frac{a^{2}}{b}-\sum_{c y c} \frac{b^{2}}{a}\right)=\frac{\sum_{c y c} a^{3}(b+c)}{a b c}+\frac{(a+b+c)(a-b)(b-c)(c-a)}{a b c}$$
Thus, the inequality is equivalent to
$$\begin{ar... | not found | Inequalities | math-word-problem | Yes | Yes | inequalities | false | 737,699 |
Problem 25(Vo Thanh Van) Let $a, b, c$ be nonnegative real numbers. Prove that
$$\sqrt{\frac{(a+b)^{3}}{8 a b(4 a+4 b+c)}}+\sqrt{\frac{(b+c)^{3}}{8 b c(4 b+4 c+a)}}+\sqrt{\frac{(c+a)^{3}}{8 c a(4 c+4 a+b)}} \geq 1$$ | Solution: Let $P=\sqrt{\frac{(a+b)^{3}}{8 a b(4 a+4 b+c)}}+\sqrt{\frac{(b+c)^{3}}{8 b c(4 b+4 c+a)}}+\sqrt{\frac{(c+a)^{3}}{8 c a(4 c+4 a+b)}}$
$$\begin{aligned}
Q=8 a b(4 a+4 b+c)+ & 8 b c(4 b+4 c+a)+8 c a(4 c+4 a+b)=\sum 32 a b(a+b)+24 a b c \\
& =32(a+b+c)(a b+b c+c a)-72 a b c
\end{aligned}$$
Apply Holder Inequali... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,700 |
Problem 25(APMO 2004) Let \(a, b, c\) be nonnegative real numbers. Prove that
\[
\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right) \geq 9(ab+bc+ca)
\] | Solution: The equivalent to
$$a^{2} b^{2} c^{2}+2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)+4\left(a^{2}+b^{2}+c^{2}\right)+8 \geq 9(a b+b c+c a)$$
We have $a^{2}+b^{2}+c^{2} \geq a b+b c+c a$
$$\begin{array}{c}
\left(a^{2} b^{2}+1\right)+\left(b^{2} c^{2}+1\right)+\left(c^{2} a^{2}+1\right) \geq 2(a b+b c+c a) ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,702 |
Problem 26(Vo Thanh Van) Let $a, b, c$ be nonnegative real numbers. Prove that
$$\frac{a}{b^{3}+c^{3}}+\frac{b}{a^{3}+c^{3}}+\frac{c}{a^{3}+b^{3}} \geq \frac{18}{5\left(a^{2}+b^{2}+c^{2}\right)-ab-ac-bc}$$ | Solution: The equivalent to
$$\begin{array}{c}
\sum \frac{a(a+b+c)}{b^{3}+c^{3}} \geq \frac{18(a+b+c)}{5\left(a^{2}+b^{2}+c^{2}\right)-a b-b c-c a} \\
\Leftrightarrow \sum \frac{a^{2}}{b^{3}+c^{3}}+\frac{a}{b^{2}+c^{2}-b c} \geq \frac{18(a+b+c)}{5\left(a^{2}+b^{2}+c^{2}\right)-a b-b c-c a}
\end{array}$$
Apply Cauchy-S... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,703 |
Problem 27 (Moldova TST 2005) Let $a, b, c$ be nonnegative real numbers such that $a^{4}+b^{4}+c^{4}=3$. Prove that
$$\frac{1}{4-a b}+\frac{1}{4-b c}+\frac{1}{4-c a} \leq 1$$ | Solution: The equivalent to
$$\begin{array}{c}
49-8(a b+b c+c a)+(a+b+c) a b c \leq 64-16(a b+b c+c a)+4(a+b+c) a b c-a^{2} b^{2} c^{2} \\
\Leftrightarrow 16+3(a+b+c) a b c \geq a^{2} b^{2} c^{2}+8(a b+b c+c a)
\end{array}$$
Apply Schur Inequality, we have
$$\begin{aligned}
\left(a^{3}+b^{3}+c^{3}+3 a b c\right)(a+b+c... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,704 |
Problem 28(Vasile Cirtoaje) Let $a, b, c$ be nonnegative real numbers such that $a b+b c+$ $c a=3$. Prove that
$$a^{3}+b^{3}+c^{3}+7 a b c \geq 10$$ | Solution: Apply Schur inequality we have
$$\begin{aligned}
& a^{3}+b^{3}+c^{3}+3 a b c \geq a b(a+b)+b c(b+c)+c a(c+a) \\
\Leftrightarrow & a^{3}+b^{3}+c^{3}+6 a b c \geq(a b+b c+c a)(a+b+c)=p q=3 p
\end{aligned}$$
Ans $r \geq \frac{p\left(4 q-p^{2}\right)}{9}=\frac{p\left(12-p^{2}\right)}{9}$
We need to prove that $3... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,705 |
Problem 29(Nguyen Phi Hung) Let $a, b, c$ be nonnegative real numbers such that $a^{2}+$ $b^{2}+c^{2}=8$. Prove that
$$4(a+b+c-4) \leq a b c$$ | Solution: From the condition we have $p^{2}-2 q=8$
Apply Schur Inequality we have
$$r \geq \frac{\left(4 q-p^{2}\right)\left(p^{2}-q\right)}{6 p}=\frac{\left(p^{2}-16\right)\left(p^{2}+8\right)}{12 p}$$
So we need to prove that
$$\frac{\left(p^{2}-16\right)\left(p^{2}+8\right)}{12 p} \geq 4(p-4) \Leftrightarrow \frac{... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,706 |
Problem 30 Let $a, b, c>0$ and $a+b+c=1$. Prove that
$$\frac{\sqrt{a^{2}+a b c}}{a b+c}+\frac{\sqrt{b^{2}+a b c}}{b c+a}+\frac{\sqrt{c^{2}+a b c}}{c a+b} \leq \frac{1}{2 \sqrt{a b c}}$$ | Solution: Changes $a, b, c$ to $p, q, r$ we have $r \leq \frac{q^{2}(1-q)}{2(2-3 q)}$. Apply Cauchy-Schwarz Inequality we have
$$\left[\sum \frac{\sqrt{a^{2}+a b c}}{(b+c)(b+a)}\right]^{2} \leq\left[\sum \frac{a}{(a+b)(b+c)}\right]\left(\sum \frac{a+c}{b+c}\right)=\frac{\sum a^{2}+\sum a b}{(a+b)(b+c)(c+a)}\left(\sum \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,707 |
Problem 31 Let $a, b, c>0$. Prove that
$$\left(\frac{a}{b+c}\right)^{2}+\left(\frac{b}{c+a}\right)^{2}+\left(\frac{c}{a+b}\right)^{2}+\frac{10 a b c}{(a+b)(b+c)(c+a)} \geq 2$$ | Solution: We have
$$(a+b)(b+c)(c+a) \geq \frac{8}{9}(a b+b c+c a)(a+b+c) \geq \frac{8}{3} \sqrt[3]{a^{2} b^{2} c^{2}}(a+b+c)$$
Letting \( x = \frac{2a}{b+c}, y = \frac{2b}{c+a}, z = \frac{2c}{a+b} \), we have \( xy + yz + zx + xyz = 4 \). Then the inequality is equivalent to
$$x^{2} + y^{2} + z^{2} + 5xyz \geq 8$$
Ta... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,708 |
Problem 4 a) (Cezar Lupu). Let $a, b, c$ be positive real numbers. Prove that
$$\frac{a b}{(a+b)^{2}}+\frac{b c}{(b+c)^{2}}+\frac{c a}{(c+a)^{2}} \leq \frac{1}{4}+\frac{4 a b c}{(a+b)(b+c)(c+a)}$$ | Solution:
WLOG, we may assume that $a \geq b \geq c>0$. We rewrite the original inequality into the following form
$$\sum_{cyc}\left(\frac{a-b}{a+b}\right)^{2} \geq \frac{2 \sum_{cyc} c(a-b)^{2}}{(a+b)(b+c)(c+a)}$$
or equivalently,
$$\sum_{cyc}\left[\frac{1}{(a+b)^{2}}-\frac{2 c}{(a+b)(b+c)(c+a)}\right](a-b)^{2} \geq 0... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,710 |
4d) Let's $a, b, c$ be nonnegative real numbers, no two of which are zero. Prove that:
$$\sum \frac{3(a+b)}{a^{2}+a b+b^{2}} \geq \frac{16(a b+a c+b c)}{(a+b)(a+b)(c+a)}$$ | Solution,
$$\frac{1}{a^{2}+a b+b^{2}} \geq \frac{a b+b c+c a}{(a+b)^{2}(a+b+c)^{2}}$$
Solution:
first, we assume $a+b=2 z, b+c=2 x, c+a=2 y$
this ineq is equivalent to:
$$\sum \frac{3 z}{3 z^{2}+(x-y)^{2}} \geq \frac{2 \sum x y-\sum x^{2}}{x y z}$$
Or
$$\sum(x-y)^{2}\left(1-\frac{2 x y}{3 z^{2}+(x-y)^{2}}\right) \geq... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,711 |
4e) Let $a, b, c$ are three positive reals, prove that,
$$a^{4}+b^{4}+c^{4}+2\left(a b^{3}+b c^{3}+c a^{3}\right) \geq 2\left(a^{3} b+b^{3} c+c^{3} a\right)+a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}$$ | Solution, The inequality
$$\Leftrightarrow \sum a^{2}(a-b)^{2}+2 \sum a b(b-c)^{2} \geq \sum c^{2}(a-b)^{2}$$
Use the identity
$$\sum(a-b)^{2}\left(c^{2}+a b\right)=\sum(a-b)^{2}(a c+b c)$$
It becomes
$$\sum(a-b)^{2}\left(a^{2}+a c+a b-b c\right) \geq 0$$
Or
$$(a-b)^{2}(a+b)^{2}+2(b-a)(b-c)\left(b^{2}+b c+b a-a c\ri... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,712 |
4f) Let $a, b, c$ are three real numbers, prove that,
$$\left(a^{2}+b^{2}+c^{2}\right)^{2} \geq 3\left(a^{3} b+b^{3} c+c^{3} a\right)$$ | Solution,
WLOG, Assume $(b-a)(b-c) \leq 0$. Because RLH is $3\left(a^{3} b+b^{3} c+c^{3} a\right) \leq L H S$ so we only need prove the inequality in case
$$a^{3} b+b^{3} c+c^{3} a \geq a b^{3}+b c^{3}+c a^{3} \Leftrightarrow c \geq b \geq a \text {. }$$
We can easily write
$$\begin{array}{l}
\qquad \sum(a-b)^{2}\left... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,713 |
4g) Let \(a, b, c \in \mathbb{R}\), prove that,
\[4\left(a^{4}+b^{4}+c^{4}\right)+3\left(a^{3} b+b^{3} c+c^{3} a\right) \geq \frac{7}{27}(a+b+c)^{4}\] | Solution, Easy can rewrite
$$\sum(a-b)^{2}\left(87(a+b)^{2}+27 b^{2}+31 c^{2}+106 b c\right)=\sum(a-b)^{2} S_{c} \geq 0$$
Note that
$$\begin{array}{l}
S_{c}+S_{a}=53(a+b+c)^{2}+34(a+c)^{2}+87(b+c)^{2}+31 a^{2}+27 b^{2}+5 c^{2} \\
S_{b}+S_{a}=53(a+b+c)^{2}+34(b+c)^{2}+87(a+c)^{2}+31 b^{2}+27 c^{2}+5 a^{2}
\end{array}$$... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,714 |
Question 1.2 Given $x, y, z \in \mathbf{R}^{+}$, prove:
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \leqslant \frac{y^{2}}{x^{2}}+\frac{z^{2}}{y^{2}}+\frac{x^{2}}{z^{2}} .$$ | $$\begin{aligned}
& \frac{y^{2}}{x^{2}}+\frac{z^{2}}{y^{2}}+\frac{x^{2}}{z^{2}} \\
= & \frac{1}{2}\left(\frac{y^{2}}{x^{2}}+\frac{z^{2}}{y^{2}}\right)+\frac{1}{2}\left(\frac{z^{2}}{y^{2}}+\frac{x^{2}}{z^{2}}\right)+\frac{1}{2}\left(\frac{x^{2}}{z^{2}}+\frac{y^{2}}{x^{2}}\right) \\
\geqslant & \sqrt{\frac{y^{2}}{x^{2}} ... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,715 |
Question 1.3 Given $x, y, z \in \mathbf{R}^{+}$, prove:
$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x} \leqslant \frac{y^{3}}{x^{3}}+\frac{z^{3}}{y^{3}}+\frac{x^{3}}{z^{3}} .$$ | Prove that by the three-variable mean inequality, we get
$$\frac{y^{3}}{x^{3}}+\frac{z^{3}}{y^{3}}+1 \geqslant 3 \sqrt[3]{\frac{y^{3}}{x^{3}} \cdot \frac{z^{3}}{y^{3}} \cdot 1}$$
which means $\frac{y^{3}}{x^{3}}+\frac{z^{3}}{y^{3}} \geqslant 3 \cdot \frac{z}{x}-1$,
Similarly, $\frac{z^{3}}{y^{3}}+\frac{x^{3}}{z^{3}} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,716 |
Question 2.3 Given that $a, b, c$ are positive real numbers, prove:
$$\frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1} \leqslant 1 \text {. }$$ | Prove that by using the method of difference comparison, and gradually finding a common denominator, we get
$$\begin{aligned}
& \frac{a}{a b+a+1}+\frac{b}{b c+b+1}+\frac{c}{c a+c+1}-1 \\
= & \frac{a}{a b+a+1}+\frac{b}{b c+b+1}-\frac{c a+1}{c a+c+1} \\
= & \left(\frac{a}{a b+a+1}-\frac{c a}{c a+c+1}\right) \\
& +\left(\... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,717 |
Example 1 Given $x, y, z \in \mathbf{R}^{+}$, and $x y z=1$, prove: $\frac{1}{1+x+y}+\frac{1}{1+y+z}+\frac{1}{1+z+x} \leqslant 1$ | Proof: Without loss of generality, let $a b c=1$ (if $a b c=k \neq 1$, we can use $\frac{a}{\sqrt[3]{k}}, \frac{b}{\sqrt[3]{k}}, \frac{c}{\sqrt[3]{k}}$ to replace $\left.a, b, c\right)$.
First, we prove $\frac{a b}{a^{2}+a b+b c} \leqslant \frac{b}{b c+b+1}$, which is equivalent to
$$a b(b c+b+1) \leqslant b\left(a^{2... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,718 |
Example 2 Let $a, b, c$ be positive numbers, and satisfy $a b c=1$, prove:
$$\begin{array}{l}
\frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1}+ \\
\frac{2}{(c+1)^{2}+a^{2}+1} \leqslant 1
\end{array}$$ | $$\begin{array}{l}
\frac{2}{(a+1)^{2}+b^{2}+1}+\frac{2}{(b+1)^{2}+c^{2}+1} \\
+\frac{2}{(c+1)^{2}+a^{2}+1} \\
=\frac{2}{a^{2}+b^{2}+2 a+2}+\frac{2}{b^{2}+c^{2}+2 b+2} \\
+\frac{2}{c^{2}+a^{2}+2 c+2} \\
\leqslant \frac{2}{2 a b+2 a+2}+\frac{2}{2 b c+2 b+2}+\frac{2}{2 c a+2 c+2} \\
=\frac{1}{a b+a+1}+\frac{1}{b c+b+1}+\f... | 1 | Inequalities | proof | Yes | Yes | inequalities | false | 737,719 |
$\begin{array}{c}\text { Example } 2 \text { Given } x, y, z \in \mathbf{R}_{+} \text {. Prove: } \\ \frac{x y z}{(1+5 x)(4 x+3 y)(5 y+6 z)(z+18)} \leqslant \frac{1}{5120} .\end{array}$ | Solve: For the function
$$f(t)=\frac{t}{(a t+b)(c t+d)}, t \in \mathbf{R}_{+}, a, b, c, d>0$$
Differentiating gives
$$\begin{array}{l}
f^{\prime}(t)=p[(a t+b)(c t+d)-t(2 a c t+b c+a d)] \\
=p\left(b d-a c t^{2}\right) . \\
\text { where } p=(a t+b)^{-2}(c t+d)^{-2} \text { represents a positive }
\end{array}$$
quantit... | \frac{1}{5120} | Inequalities | proof | Yes | Yes | inequalities | false | 737,720 |
Example 3 Given that $a, b, c$ are non-negative real numbers, and $a + b + c = 1$. Prove:
$$\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2} \geqslant 2$$ | Proof: Note that, when $a=1, b=c=0$, the inequality to be proved holds. Without loss of generality, assume $a \geqslant b \geqslant c$. Fix $b$, then $c=1-b-a$ is a function of $a$.
Consider the function of $a$
$$f(a)=\left(1-a^{2}\right)^{2}+\left(1-b^{2}\right)^{2}+\left(1-c^{2}\right)^{2},$$
$f(a)$ is decreasing $... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 737,721 |
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