SRIDEV wrote: a,b,c are positive & abc=1, then show that a^(b+c) .b^(c+a) .c^(a+b) ≤ 1 By Chebyshov we obtain: $$\prod_{cyc}a^{b+c}=e^{\sum\limits_{cyc}(b+c)\ln{a}}\leq e^{\frac{1}{3}\sum\limits_{cyc}(b+c)\sum\limits_{cyc}\ln{a}}=1.$$

Let $a,b,c$ be positive real numbers with $abc=1$. Prove the inequality $$a^{b+c}b^{c+a}c^{a+b}\le1.$$

$a^{b+c}b^{c+a}c^{a+b} \leq 1$ $1=(abc)^{a+b+c} \leq a^a b^b c^c$ $a^ab^bc^c =(\sqrt[a+b+c]{a^ab^bc^c})^{a+b+c} \geq (\frac{a+b+c}{a*\frac{1}{a}+b*\frac{1}{b}+c*\frac{1}{c}})^{a+b+c} =(\frac{a+b+c}{3})^{a+b+c} \geq 1$

Cool inequality problem.