Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$ Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$
I tried it to do using $AM \ge GM$ but don't know how to proceed.
Please help.

There is a nice solution of this using convexity. Jensen's inequality states that if $f$ is a concave function and $p_1, ... ,p_n$ is a distribution, then for any numbers $x_1, ... ,x_n$ we have:
$$
f\left(\sum_{i=1}^n{p_i x_i}\right) \geq \sum_{i=1}^n{p_i f(x_i)} .
$$
We shall use this and the fact that $\log$ is a concave function. Let us take the log of the left hand side and see where that takes us. We get:
$$
(a+b+c) \log\left(\frac{ab+bc+ac}{a+b+c}\right) = 
$$
$$
(a+b+c) \log\left(\left(\frac{a}{a+b+c}\right) b  +
\left(\frac{b}{a+b+c}\right) c + \left(\frac{c}{a+b+c}\right) a \right) \geq
$$
Now we use Jensen.
$$
(a+b+c) \left[ \frac{a}{a+b+c} \log(b)  + \frac{c}{a+b+c} \log(a) + 
\frac{b}{a+b+c} \log(c)\right] = 
$$
$$
a\log(b)+b\log(c) + c\log(a) = \log\left(b^a c^b a^c\right).
$$
This gives us the inequality 
$$
\left(\frac{ab+bc+ac}{a+b+c}\right)^{a+b+c} \geq b^a c^b a^c .
$$
In exactly the same way we can show that 
$$
\left(\frac{ab+bc+ac}{a+b+c}\right)^{a+b+c} \geq a^b b^c c^a ,
$$
And so the inequality holds for the geometric mean of these two quantities.
$$
\left(\frac{ab+bc+ac}{a+b+c}\right)^{a+b+c} \geq \sqrt{b^a c^b a^c a^b b^c c^a} = 
\sqrt{(ab)^c(ac)^b(bc)^a} .
$$
