Prove that $(\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a(ac)^b(ab)^c}$ 
Prove that $$\left( \frac{bc+ac+ab}{a+b+c} \right)^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}$$
  where $a,b,c>0$

My attempt:

I couldn't proceed after that.  Please help me in this regard, thanks!
 A: We need to prove that 
$$(a+b+c)\ln\frac{ab+ac+bc}{a+b+c}\geq\frac{1}{2}\sum\limits_{cyc}a\ln{bc}$$ or
$$\ln\left(\frac{ab+ac+bc}{a+b+c}\right)^2\geq\sum\limits_{cyc}\frac{a}{a+b+c}\ln{bc}.$$
But by Jensen $$\sum\limits_{cyc}\frac{a}{a+b+c}\ln{bc}\leq\ln\sum\limits_{cyc}\frac{abc}{a+b+c}$$
Thus, it remains to prove that $$\frac{3abc}{a+b+c}\leq\left(\frac{ab+ac+bc}{a+b+c}\right)^2,$$
which is 
$$\sum\limits_{cyc}c^2(a-b)^2\geq0$$
Done!
A: $a^bb^cc^a\ge\sqrt{(ab)^c(bc)^a(ca)^b}$
$\implies a^{2b}b^{2c}c^{2a}\ge a^{b+c}b^{c+a}c^{a+b}$
$\implies a^bb^cc^a\ge a^cb^ac^b$
It doesn't seems true always because of simmatry..
A: If you want to use the weighted AM-GM inequality, considering
\begin{align*}
  \frac{bc+ca+ab}{a+b+c} &=
  \frac{2(bc+ca+ab)}{2(a+b+c)} \\
  &= \frac{(b+c)\color{red}{\boldsymbol{a}}+
           (c+a)\color{green}{\boldsymbol{b}}+
           (a+b)\color{blue}{\boldsymbol{c}}}{(b+c)+(c+a)+(a+b)} \\
  \left( \frac{bc+ca+ab}{a+b+c} \right)^{2(a+b+c)}
  &\ge
  \color{red}{\boldsymbol{a}}^{b+c}
  \color{green}{\boldsymbol{b}}^{c+a}
  \color{blue}{\boldsymbol{c}}^{a+b} \\
  \left( \frac{bc+ca+ab}{a+b+c} \right)^{a+b+c}
  &\ge \sqrt{a^{b+c}b^{c+a}c^{a+b}} \\
  &= \sqrt{(bc)^{a}(ca)^{b}(ab)^{c}}
\end{align*}
