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
 A: SOS helps.
For $a^2+b^2+c^2=1$ after $x=\sqrt3a$, $y=\sqrt3b$ and $z=\sqrt3c$ we need to prove that:
$$\frac 1{5-6ab}+\frac 1{5-6bc}+\frac 1{5-6ca}\leq 1$$ or$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{5-6ab}\right)\geq0$$ or
$$\sum_{cyc}\frac{2-6ab}{5-6ab}\geq0$$ or $$\sum_{cyc}\frac{3(a-b)^{2}+2c^{2}-a^{2}-b^{2}}{5-6ab}\geq0$$ or 
$$\sum_{cyc}\left(\frac{3(a-b)^{2}}{5-6ab}+\frac{(c-a)(a+c)}{5-6ab}-\frac{(b-c)(b+c)}{5-6ab}\right)\geq0$$ or
$$\sum_{cyc}\frac{3(a-b)^{2}}{5-6ab}+\sum_{cyc}(a-b)\left(\frac{a+b}{5-6bc}-\frac{a+b}{5-6ac}\right)\geq0$$ or
$$\sum_{cyc}(a-b)^{2}\left(\frac{1}{5-6ab}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}\right)\geq0.$$
Now, let $a\geq b\geq c.$ 
Thus, $$S_{c}=\frac{1}{5-6ab}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}\geq$$
$$\geq\frac{1}{5-6ac}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}=\frac{5-8bc-2ac}{(5-6ac)(5-6bc)}=$$
$$=\frac{4(b-c)^{2}+(a-c)^{2}+4a^{2}+b^{2}}{(5-6ac)(5-6bc)}\geq0;$$
$$S_{b}=\frac{1}{5-6ac}-\frac{2(a+c)b}{(5-6ab)(5-6bc)}\geq$$
$$\geq\frac{1}{5-6bc}-\frac{2(a+c)b}{(5-6ab)(5-6bc)}=\frac{5-8ab-2bc}{(5-6ab)(5-6bc)}=$$
$$=\frac{4(a-b)^{2}+(b-c)^{2}+4c^{2}+a^{2}}{(5-6ab)(5-6bc)}\geq0$$
and $$S_{a}+S_{b}=\frac{1}{5-6bc}-\frac{2(b+c)a}{(5-6ab)(5-6ac)}+\frac{1}{5-6ac}-\frac{2(a+c)b}{(5-6ab)(5-6bc)}=$$
$$=\frac{1}{5-6ac}-\frac{2(b+c)a}{(5-6ab)(5-6ac)}+\frac{1}{5-6bc}-\frac{2(a+c)b}{(5-6ab)(5-6bc)}=$$
$$=\frac{5-8ab-2ac}{(5-6ab)(5-6ac)}+\frac{5-8ab-2bc}{(5-6ab)(5-6bc)}\geq0.$$
Id est, $$\sum_{cyc}(a-b)^{2}\left(\frac{1}{5-6ab}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}\right)=\sum_{cyc}(a-b)^2S_c\geq$$
$$\geq S_b(a-c)^2+S_a(b-c)^2\geq S_b(b-c)^2+S_a(b-c)^2=(b-c)^2(S_b+S_a)\geq0.$$
A: 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 $x^2+y^2+z^2=3$.
A: We need to prove
$$ \sum \dfrac{xy}{5-2xy} \leqslant 1,$$
or
$$ \sum \dfrac{3xy}{5(x^2+y^2+z^2)-6xy} \leqslant 1,$$
Indeed, because
$$[13(x^2+y^2)+10z^2-18z(x+y)+36xy][5(x^2+y^2+z^2)-6xy]-108xy (x^2+y^2+z^2)$$
$$=\left[63(x^2+y^2)+\frac{109z^2}{2}+112xy-72z(x+y)\right](x-y)^2+\frac{(2x+2y-5z)^2(x+y-2z)^2}{2}$$
$$\geqslant 0.$$
Therefore
$$ \sum \dfrac{3xy}{5(x^2+y^2+z^2)-6xy} \leqslant \sum \frac{13(x^2+y^2)+10z^2-18z(x+y)+36xy}{36(x^2+y^2+z^2)}=1.$$
Done.
