Find the minimum of : $P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})$ $a;b;c\in \mathbb{R}^+$ such that $a+b+c=6$. 

Find the minimum of : $P=(1+\frac{1}{a^3})(1+\frac{1}{b^3})(1+\frac{1}{c^3})$

Thanks :)
I have no ideas about this problem ! :(

You can use Lagrange multipliers as @ Martín-Blas Pérez Pinilla said. This is an alternative solution.
Using AM-GM, we have $6=a+b+c\geq 3\sqrt[3]{abc}\iff \frac1{\sqrt[3]{abc}}\geq \frac12$.
On the other hands, we have 
\begin{equation}
1+\frac1{a^3}=\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{2^3}+\frac1{a^3}\geq 9\frac1{\sqrt[3]{2^8a}}.
\end{equation}
I am sure, from this point you can get continue it by yourself to get minimum $P={\frac{9^3}{8^3}}$
