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.
On
Yes of course! We can use C-S.
For $a=b=1+\sqrt2$ and $c=1$ we get a value $2\sqrt2-1$.
By C-S $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{2c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{ab+bc}+\frac{2kc^2}{kac+bkc}\geq$$ $$\geq\frac{(a+b+\sqrt{2k}c)^2}{ab+ac+ab+bc+kac+kbc}$$ The equality occurs for $\frac{1}{b+c}=\frac{1}{a+c}=\frac{1}{\sqrt{\frac{k}{2}}(a+b)}$, which gives $k=1$.
Id est, it remains to prove that $(a+b+\sqrt2c)^2\geq2(2\sqrt2-1)(ab+ac+bc)$, which is $2c^2-2(\sqrt2-1)(a+b)c+a^2+b^2-4(\sqrt2-1)ab\geq0$,
for which it's enough to prove that $$(\sqrt2-1)^2(a+b)^2-2\left(a^2+b^2-4(\sqrt2-1)ab\right)\leq0$$ which is $(a-b)^2\geq0$
Done!
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 $\color{red}{2\sqrt{2}-1}$.