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.

We need to prove that:
$$\sum_{cyc}\frac{1}{(x+y)^2}\geq\frac{9}{4(xy+xz+yz)}$$ or
$$\sum_{cyc}\left(\frac{1}{(x+y)^2}-\frac{3}{4(xy+xz+yz)}\right)\geq0$$ or
$$\sum_{cyc}\frac{4xz+4yz-2xy-3x^2-3y^2}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}\frac{(z-x)(3x+y)-(y-z)(3y+x)}{(x+y)^2}\geq0$$ or
$$\sum_{cyc}(x-y)\left(\frac{3y+z}{(y+z)^2}-\frac{3x+z}{(x+z)^2}\right)\geq0$$ or
$$\sum_{cyc}(x-y)^2(3xy+xz+yz-z^2)(x+y)^2\geq0.$$
Now, let $x\geq y\geq z$.
Thus, $$\sum_{cyc}(x-y)^2(3xy+xz+yz-z^2)(x+y)^2\geq$$
$$\geq\sum_{cyc}(x-y)^2(xz+yz-z^2)(x+y)^2\geq$$
$$\geq(x-z)^2y(x+z-y)(x+z)^2+(y-z)^2x(y+z-x)(y+z)^2\geq$$
$$\geq(y-z)^2y(x-y)(x+z)^2+(y-z)^2x(y-x)(y+z)^2=$$
$$=(x-y)^2(y-z)^2(xy-z^2)\geq0.$$
