Inequality between real numbers $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$ let $a,b$ and $c$ positive reals. Shows that $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$
How  prove this inequality. Thanks
a suggestion please to prove this inequality

The inequality $a^ab^bc^c\le(abc)^{\frac{a+b+c}{3}}$ is incorrect. Let $a=b=1$, $c=4$
Then 
$$a^ab^bc^c=4^4$$ 
while
$$(abc)^{\frac{a+b+c}{3}}=4^2$$. 
However, the opposite inequality $a^ab^bc^c\ge(abc)^{\frac{a+b+c}{3}}$ is correct as demonstrated in Angel's answer.
