Solving Inequalities similar to Nesbitt's Is it possible to use inequalities like Cauchy-Schwarz or QM-AM-GM-HM to find the minimum value of $\frac{a}{b+c}+\frac{b}{a+c}+\frac{2c}{a+b}$ for $a,b,c\gt0$? 
From just trying different values, the minimum seems to be $11\over6$,
but how would one prove this? I tried setting 
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{2c}{a+b}=S$, giving $S+6=(a+b+c)(\frac{1}{a+b}+\frac{1}{a+c}+\frac{2}{a+b})$, but I'm not sure how to proceed from here, or if this is even the right first step.

Enforcing the substitution $A=b+c, B=a+c, C=a+b$ the problem boils down to finding the minumum of
$$\begin{eqnarray*}&& \frac{B+C-A}{2A}+\frac{A+C-B}{2B}+\frac{A+B-C}{C}\\&=&-2+\left(\frac{B}{2A}+\frac{A}{2B}\right)+\left(\frac{C}{2A}+\frac{A}{C}\right)+\left(\frac{B}{C}+\frac{C}{2B}\right)\end{eqnarray*}$$
By setting $\frac{A}{B}=x$ and $\frac{B}{C}=y$, that boils down to studying
$$ f(x,y)= -2+\frac{1}{2x}+\frac{x}{2}+\frac{1}{2xy}+xy+y+\frac{1}{2y} $$
over $\mathbb{R}^+\times\mathbb{R}^+$. Such function has a unique stationary point at $(x,y)=\left(1,\frac{1}{\sqrt{2}}\right)$, hence the minimum of our expression is achieved at $(A,B,C)=(1,1,\sqrt{2})$ and it equals $\color{red}{2\sqrt{2}-1}$.
