How prove this inequality $\frac{3(x^2+y^2+z^2)}{(x+y+z)^2+2(yz+xz+xy)}\ge\sum\limits_{cyc}\frac{x^2}{x^2+(y+z)^2}$

Question

Question:

let $x,y,z\ge 0$.prove or disprove

$$\dfrac{3(x^2+y^2+z^2)}{(x+y+z)^2+2(yz+xz+xy)}\ge\sum\limits_{cyc}\dfrac{x^2}{x^2+(y+z)^2}$$

My idea: let $x+y+z=1$, then we can only $$\sum_{cyc}\dfrac{x^2}{x^2+(y+z)^2}=\sum_{cyc}\dfrac{x^2}{2x^2-2x+1}\le\dfrac{3(x^2+y^2+z^2)}{1+2(xy+yz+xz)}$$

then I can't.Thank you

Answer 1

It's wrong! Try $x=4$ and $y=z=9$.