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?
 A: Since our inequality is homogeneous, we can assume $a+b+c=3$. 
Also let $f(a,b,c,\lambda)=\sum\limits_{cyc}\frac{a}{3-a}+\lambda(a+b+c-3)-\frac{3}{2}$
$$\frac{\partial f}{\partial a}=\frac{3}{(3-a)^2}+\lambda$$.
Let in $(a,b,c)$ our $f$ gets a minimal value 
($f$ is a continuous function and this thing happens on the compact).
Thus, in this point $\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0$, which gives
$$\frac{3}{(3-a)^2}=\frac{3}{(3-b)^2}$$
and since $a+b<3$, we obtain: $a=b$, which says that $c=3-2a$ and it remains to prove that 
$$\frac{2a}{3-a}+\frac{3-2a}{2a}\geq\frac{3}{2},$$
where $0<a<\frac{3}{2}$, which gives $(a-1)^2\geq0$.
Also for $a\rightarrow0^+$... our inequality is obviously true.
Done!
A: 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. 
A: This is Nesbitt's inequality. There are a number of proofs for that, e.g.
here. I don't think that Lagrange Multipliers apply here.
A: Let the objective function be 
$$f = a(c+a)(a+b) + b(b+c)(a+b) + c(b+c)(c+a)$$
and the constraint be 
$$g = (a+b)(b+c)(c+a) - k.$$
Note that if $g=0$, then 
$$\frac{f}{k} = \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}.$$
Solving $\nabla f-\lambda g = 0$ and $g=0$ we find $f$ is minimized for 
$a=b=c=k^{1/3}/2$, with value $3k/2$. 
Thus, 
$f/k \ge 3/2$, 
which proves the inequality. 
