Prove $a^ab^bc^c\ge (abc)^{\frac{a+b+c}3}$ for positive numbers. 
Prove that the following inequality holds
  $$a^a b^b c^c\ge (abc)^{\frac{a+b+c}{3}}$$
  if $a,b,c$ are positive.

I'm not sure how to handle these kinds of powers. Are there any "famous" but not so advanced inequalities that involve this kinds of expressions?
I was trying to do something like ordering them without lost of generality because it's symmetric, but I get nowhere and I just make the thing even more complicated.
I'm asking for solution rather than hints since this is my first time encountering these kinds of problems.
 A: Take logarithms. The function $x\log x$ is convex.  The result now follows from Jensen's Inequality  (use "weights" $1, 1, 1$).
A: Take the logs to get rid of the powers ;)
$$a \ln(a) + b \ln (b) + c \ln (c) \geq \frac{a+b+c}{3} [ \ln (a)+ \ln(b) + \ln(c)] $$
Then a relatively simple but less known inequality, Chebyshev inequality: 

If $ a_1 \geq a_2 \geq \cdots \geq a_n$, $b_1 \geq b_2 \geq \cdots \geq b_n$, then  $${1\over n} \sum_{k=1}^n a_k \cdot b_k \geq \left({1\over n}\sum_{k=1}^n a_k\right)\left({1\over n}\sum_{k=1}^n b_k\right)$$

solves the problem.

If you are not familiar to Chebyshev, try proving directly that 
$$a \ln(a) + b \ln (b) \geq a \ln(b) + b \ln(a) $$
Do the same for $(a,c), (b,c)$ and add them togeter....

P.S. The last observation leads to the following "elementary" but more complicated solution.
We prove first that 
$$a^ab^b \geq a^bb^a \,.$$
Indeed since the equation is symmetric, we can assume WLOG that $a \geq b$ and then
$$a^{a-b} \geq b^{a-b} \,.$$
Which is exactly the desired inequality.
Then we have
$$a^ab^b \geq a^bb^a \,.$$
$$a^ac^c \geq a^cc^a \,.$$
$$b^bc^c \geq b^cc^b \,.$$
$$a^ab^bc^c \geq a^ab^bc^c \,.$$
Multiplying togeter, you get
$$a^{3a}b^{3b}c^{3c} \geq (abc)^{a+b+c} \,.$$
What I did here was to reprove the Chebyshev inequality in this particular case, without writing the logs....
A: This might be a bit similar to N.S.'s answer but here goes anyway:
Take logarithms. Without loss of generality, take $a\leq b\leq c$. Define:
$x_1 = x_2 = x_3 = \frac a3, \\x_4 = x_5 = x_6 = \frac b3, \\x_7 = x_8 = x_9 = \frac c3, \\y_1 = y_2 = y_3 = \log a, \\y_4 = y_5 = y_6 = \log b, \\y_7 = y_8 = y_9 = \log c.$
Clearly, $x_1 \leq x_2 \leq \cdots \leq x_9$ and $y_1 \leq y_2 \leq \cdots \leq y_9$.
Now just use the rearrangement inequality on this and you're done.
A: First apply weighted GM-HM inequality on $a,b,c$ with weights $a,b,c$ respectively to get $$(a^ab^bc^c)^{\frac{1}{a+b+c}} \ge \frac{a+b+c}{3}$$ Now apply usual AM-GM inequality on $a,b,c$, each with weight $1$ to get $$\frac{a+b+c}{3}\ge (abc)^{\frac{1}{3}}$$ Combining both the results and raising both sides to the power of $a+b+c$, we get what was desired, id est $$a^a b^b c^c\ge (abc)^{\frac{a+b+c}{3}}$$
