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 ! :(
 A: Do you want to find the minimum value? Because $P\to\infty$. For example we can take $a=5,b=1-\frac1n$, and $c=\frac1n$. In this case you have
\begin{equation}
P>1+\frac1{c^3}>n^3\to\infty
\end{equation}
A: 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 $P={\frac{9^3}{8^3}}$
A: Use Holder inequality and we have:
$$\left(1 + \frac 1{a^3}\right)\left(1 + \frac 1{b^3}\right)\left(1 + \frac 1{c^3}\right) \ge \left(1 + \frac 1{abc}\right)^3$$
From $AM-GM$ we have:
$$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$$
$$\frac 63 \ge \sqrt[3]{abc}$$
$$2^3 \ge {abc}$$
$$8 \ge abc$$
So using this we minimize the right hand side:
$$\left(1 + \frac 1{a^3}\right)\left(1 + \frac 1{b^3}\right)\left(1 + \frac 1{c^3}\right) \ge \left(1 + \frac 1{abc}\right)^3 \ge \left(1 + \frac 1{8}\right)^3 = \left(\frac 9{8}\right)^3 = \frac{729}{512}$$
Now to find when the equality happens. Note that we have equality in the first inequality when $\frac{1}{a^3} = k\cdot 1$, $\frac{1}{b^3} = k\cdot 1$ and $\frac{1}{c^3} = k\cdot 1$.
From this we conclude that equality holds when $a=b=c$. For the second equality we have inequality when $8=abc \implies 8=a^3 \implies a=b=c=2$
So the minimum value occurs at $a=b=c=2$ and it's $\frac{729}{512}$
