Proving $a^ab^bc^c\ge(abc)^{(a+b+c)/3}$ for positive real numbers. Prove that 
$$a^ab^bc^c\ge(abc)^{(a+b+c)/3}$$
where $a,b,c\in\mathbb{R^+}$
I tried using powered AM-GM but didn't get anything. please give me a hint to solve it.
 A: Since $\log(x)$ is monotonically increasing, we have
$$
(a-b)(\log(a)-\log(b))\ge0\\
(c-a)(\log(c)-\log(a))\ge0\\
(b-c)(\log(b)-\log(c))\ge0\\
$$
Add these and
$$
(a\log(a)+b\log(b)+c\log(c))-(a\log(a)+b\log(b)+c\log(c))\ge0
$$
to get
$$
3(a\log(a)+b\log(b)+c\log(c))-(a+b+c)(\log(a)+\log(b)+\log(c))\ge0
$$
Divide by $3$ and exponentiate to get
$$
a^ab^bc^c\ge(abc)^{\frac{a+b+c}3}
$$
A: $$a^ab^bc^c\ge{(abc)^\frac{a+b+c}{3}} \stackrel{\log (\cdot)}{\iff} a\log a+b\log b+c\log c \geq (\log a+\log b+\log c)\cdot \frac{a+b+c}{3}$$
You can use Chebyshev's sum inequality to finish it.
A: Take the $\log$ of both sides. This is equivalent to:
$$
a\log a + b\log b + c\log c \ge \frac{a+b+c}3 (\log a + \log b + \log c)
$$
now use the rearrangement inequality twice to get:
$$
a\log a + b\log b + c\log c \ge b\log a + c\log b + a\log c ;\\
a\log a + b\log b + c\log c \ge c\log a + a\log b + b\log c;\\
a\log a + b\log b + c\log c = a\log a + b\log b + c\log c.
$$
The sum of these inequalities is the right result.

Alternative:
$$
\frac{a+b+c}3 \frac{\log a + \log b + \log c}3
\le \frac{a+b+c}3 \log   \frac{a+b+c}3
$$using the concavity of $\log$;
$$
\le \frac{a\log a + b\log b + c\log c }3
$$using convexity of $x\to x\log x$.
A: First Prove that $$a^ab^b\ge a^bb^a$$
Then w have
$$a^ab^b\ge a^bb^a$$
 $$a^ac^c\ge a^cc^a$$
 $$c^cb^b\ge b^cc^b$$
$$a^ab^bc^c\ge a^a b^bc^c$$
Multiplying together
$$a^{3a}b^{3b}c^{3c}\ge (abc)^{a+b+c}$$
$$a^{a}b^{b}c^{c}\ge (abc)^{(a+b+c)/3}$$
