Proving inequality using Lagrange multipliers. I started learing about Lagrange Multipliers and I got the following question:
Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}≥\frac{3}{2}$, for each $a,b,c>0$.
I'm not sure how to use Lagrange multipliers for it... any ideas?

Let $f(x,y,z) = \frac{x}{y+z} +\frac{y}{x+z} +\frac{z}{x+y} $ 
$$\begin{eqnarray*}
 \vec \nabla f \cdot \hat x &=& +\frac{1}{(y+z)} - \frac{y}{(x+z)^2}- \frac{z}{(x+y)^2}
\\ \vec \nabla f \cdot \hat y &=& -\frac{x}{(y+z)^2} + \frac{y}{(x+z)}- \frac{z}{(x+y)^2}
\\ \vec \nabla f \cdot \hat z &=& -\frac{x}{(y+z)^2} - \frac{y}{(x+z)^2}+ \frac{z}{(x+y)}
\end{eqnarray*}$$
You should be able to demonstrate that $$\vec\nabla f=0 \implies x=y=z$$
and since $f(x,x,x)=\frac 32$ you are pretty well done. 
