<s>[INST] prove thatt $\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$ 
prove that $$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge 9/4$$ where $x,y,z$ are positives such that $xy+yz+xz=1$

By Holder;$$\left(\sum_{cyc} \frac{1}{{(x+y)}^2} \right){\left(\sum yz+zx \right)}^2\ge {\sum \left(z^{2/3} \right)}^{3}$$.
Hence it suffices to prove $$\sum {\left(z^{2/3} \right)}^{3}\ge 3^{2/3}$$ which is falsse.
I was able to get a weaker bound.since $xy+yz+zx=1$ we have $x^2+y^2\le 2$. By C-S and $x^2+y^2<2$
$$\sum_{cyc} \frac{1}{{(x+y)}^2}\ge \frac{1}{2}\sum \frac{1}{x^2+y^2}>3/4$$
I am interested on a solution using standard inequalities (C-S AM-GM,chebyshov etc) rather than fully expanding and using uvw/pqr/schur.
 [/INST] This is the solution by Vo Quoc Ba Can.
Need to prove:
$$(xy+yz+zx)\left(\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2}\right)\geqq \frac{9}{4}$$
Wlog, assume $x\geq y \geq z,$ we have the following lemma:
$$\sum\limits_{cyc} \frac{1}{(x+y)^2}\geqq \frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\quad \quad(1)$$
$$\Leftrightarrow  \left(\frac{1}{x+z}-\frac{1}{y+z}\right)^2\geqq \frac{(x-y)^2}{4xy(x+y)^2}$$
$$\Leftrightarrow 4xy(x+y)^2 \geqq (x+z)^2(y+z)^2$$
Since $x\geqq y \geqq z \therefore 4xy\geqq 4y^2\geqq (y+z)^2;(x+y)^2 \geqq (x+z)^2.$
So $(1)$ is true, need to prove:
$$(xy+yz+zx)\left(\frac{1}{4xy}+\frac{2}{(x+z)(y+z)}\right)\geqq\frac{9}{4}$$
But it's $$(x+y)(y+z)(z+x)\geq 8xyz,$$
which is true by AM-GM.
</s>