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
 A: 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 $\log$.
A: Without loss of generality suppose $a<b<c$. Then $\text{log} (a )<\text{log}( b) <\text{log} (c)$ and inequality of Chebyshev. 
$\frac{a+b+c}{3}\frac{\text{log}(a)+\text{log}(b)+\text{log}(c)}{3}<\frac{a\text{log}(a)+b\text{log}(b)+c\text{log}(c)}{3}$.
Where you have to $a\text{log}(a)+b \text{log}(b)+c\text{log}(c)>\frac{a+b+c}{3}(\text{log}(a)+ \text{log}(b)+\text{log}(c)).$
And therefore $a^ab^bc^c<(abc)^{\frac{a+b+c}{3}}$.
A: 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.
A: An elementary proof. No logs etc.
Notice that the inequality stays the same if we replace $a,b,c$ with $pa, pb, pc$: 
$$ (pa)^{pa}(pb)^{pb}(pc)^{pc} = p^{pa+pb+pc} (a^ab^bc^c)^p$$
and
$$ (pa pb pc)^{(pa+pb+pc)/3}  = p^{pa+pb+pc} ((abc)^{(a+b+c)/3})^p$$
Thus we can assume that $abc = 1$ (choose $p = \frac{1}{\sqrt[3]{abc}}$).
Now replace $c = 1/ab$ to try and prove
$$ \frac{a^a b^b}{(ab)^{\frac{1}{ab}}} \ge 1$$
Now there are two cases:
Case I)
$a \ge b \ge 1 \ge c$
In that case we easily have have $x = a- 1/ab \ge 0$ and so $a^a/a^{1/ab} = a^x \ge 1$ (and similarly with $b$).
Case II)
$a \le b \le 1 \le c$
In this case, we have that $x = a - 1/ab \le 0$, thus $a^x \ge 1$ (remember $a \le 1$). Similarly with $b$.
We are done!
