<s>[INST] 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.
 [/INST] 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 $x\to x\log x$.
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