show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$ let $x,y,z\ge 0$ and such $x^2+y^2+z^2=3$ show that
$$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$
try:
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$
and
$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{5}{3}z^2+\frac{2}{3}x^2+\frac{2}{3}y^2+(x-y)^2}\le\sum\dfrac{3(x+y)^2}{2(x^2+y^2)+5z^2}$$
following I want use C-S,But I don't Success

Also, uvw helps.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition gives $$3u^2-2v^2=1,$$ which does not depend on $w^3$.
In another hand, we need to prove that $f(w^3)\geq0$, where $f$ is a concave function:
$f(w^3)\geq0$ is a quadratic inequality of $w^3$ with a coefficient before $w^6$ is equal to $-8$. 
But the concave function gets a minimal value for an extreme value of $w^3$, 
which happens in the following cases. 


*

*$w^3=0$.


Let $z=0$.
Thus, after homogenization we can assume that $y=1$ and we need to prove that
$$\sum_{cyc}\frac{1}{5(x^2+y^2+z^2)-6xy}\leq\frac{1}{x^2+y^2+z^2}$$ or 
$$\frac{1}{5x^2-6x+5}+\frac{2}{5x^2+5}\leq\frac{1}{x^2+1}$$ or
$$\frac{1}{5x^2-6x+5}\leq\frac{3}{5(x^2+1)}$$ or
$$5x^2-9x+5\geq0,$$ which is obvious;


*Two variables are equal.


Thus, after homogenization we can assume $y=z=1$ (for $y=z=0$ the inequality is obviously true),
which gives
$$\frac{2}{5x^2-6x+10}+\frac{1}{5x^2+4}\leq\frac{1}{x^2+2}$$ or
$$(x-1)^2(5x^2-2x+2)\geq0$$ and we are done!
Now we see that the starting inequality is true for any reals $x$, $y$ and $z$ such that $x^2+y^2+z^2=3$.
