Ovchinnikov Denis wrote: For positive $a,b,c$, such that $abc=1$ prove the inequality: $4(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$. try $a=1,b=4,c=\frac{1}{4}$.

kuing wrote: Ovchinnikov Denis wrote: For positive $a,b,c$, such that $abc=1$ prove the inequality: $4(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$. try $a=1,b=4,c=\frac{1}{4}$. Sorry, in left side cubic squares Corrected

Ovchinnikov Denis wrote: For positive $a,b,c$, such that $abc=1$ prove the inequality: $4(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$. let $a = \frac{y}{x},b = \frac{z}{y},c = \frac{x}{z},x,y,z > 0$, then $\Leftrightarrow 4\left( {\sqrt[3]{{\frac{{y^2 }}{{zx}}}} + \sqrt[3]{{\frac{{z^2 }}{{xy}}}} + \sqrt[3]{{\frac{{x^2 }}{{yz}}}}} \right) \le 3\left( {2 + \frac{x}{y} + \frac{y}{z} + \frac{z}{x} + \frac{y}{x} + \frac{z}{y} + \frac{x}{z}} \right)^{\frac{2}{3}} $ $ \Leftrightarrow 4\left( {\frac{{x + y + z}}{{\sqrt[3]{{xyz}}}}} \right) \le 3\left( {\frac{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}}{{xyz}}} \right)^{\frac{2}{3}} $ $ \Leftrightarrow 64xyz\left( {x + y + z} \right)^3 \le 27\left( {x + y} \right)^2 \left( {y + z} \right)^2 \left( {z + x} \right)^2 $ $ \Leftrightarrow xyz\left( {\frac{{x + y + z}}{3}} \right)^3 \le \left( {\frac{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}}{8}} \right)^2, $ by $\frac{{\left( {x + y + z} \right)\left( {xy + yz + zx} \right)}}{9} \le \frac{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}}{8}$, so we just need to prove $xyz\left( {\frac{{x + y + z}}{3}} \right)^3 \le \left( {\frac{{\left( {x + y + z} \right)\left( {xy + yz + zx} \right)}}{9}} \right)^2 $ $ \Leftrightarrow 3xyz\left( {x + y + z} \right) \le \left( {xy + yz + zx} \right)^2, $ obvious true, done.

2006丝绸之路不等式

Attachments:
2006年丝绸之路.pdf (54kb)

Ovchinnikov Denis wrote: For positive $a,b,c$, such that $abc=1$ prove the inequality: $4(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$. See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=594965#p594965

See also 5th Silk Road Mathematical Competition SRMC2006