<s>[INST] 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!
 [/INST] 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*}
</s>