Nesbitt's inequality states that $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \;\ge\;\frac{3}{2}\tag{Nes}$$ holds for all $\,x,y,z\in\mathbb{R}^{>0}$. Applying Cauchy-Schwarz to it yields $$\left(x^2+y^2+z^2\right) \left[\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right] \;\ge\;\frac{9}{4}\tag{1}$$ and my question: Does this remain true upon replacing the first factor by $xy+yz+zx\,$?
Please prove or reject $$(xy+yz+zx)\left[\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right] \;\ge\;\frac{9}{4}\tag{2} $$ with $\,x,y,z>0$ as before.
There's the estimate $\,xy+yz+zx\le\left(x^2+y^2+z^2\right)\,$ obtained from $\operatorname{(AM\ge GM)}$. It can get very coarse, e.g. choose $x=1$ and both $y,z$ close to zero, but it appears as if the other factor in brackets can compensate such that $(2)$ still holds true.
This is a pro, but not a proof.When trying to prove $(2)$ along the same lines as above I am stuck: Applying Cauchy-Schwarz to $$\sum_{\text{cyc}}\frac{\sqrt{xy}}{x+y}$$ gives the needed LHS of $(2)$, but by $\operatorname{(AM\ge GM)}$ the cyclic sum is bounded above by $\frac{3}{2}$.
So this is a con, but not a counterexample.
