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

Let's transform $a^ab^bc^c\geq(abc)^{(a+b+c)/3}$ by taking $\log$:
$$
a^ab^bc^c\geq(abc)^{(a+b+c)/3}\iff a\log a+b\log b+c\log c\geq\frac{a+b+c}{3}\log(abc)\\
\iff\frac{a}{a+b+c}\log a+\frac{b}{a+b+c}\log b+\frac{c}{a+b+c}\log c\geq\log(\sqrt[3]{abc}).
$$
Let's prove the last inequality above. With $f(x)=-\log x=\log(1/x)$ being a convex function on $(0,\infty)$, we can use Jensen's inequality to infer that
$$
\frac{a}{a+b+c}f(1/a)+\frac{b}{a+b+c}\log f(1/b)+\frac{c}{a+b+c}\log f(1/c)\\
\geq f\left(\frac{3}{a+b+c}\right)=\log\left(\frac{a+b+c}{3}\right)\geq \log(\sqrt[3]{abc}).
$$
The last inequality above follows from the AM-GM inequality and the monotonicity of $\log$.
