<s>[INST] 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.
 [/INST] 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!
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