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....