show this inequality $\sum_{cyc}\frac{1}{5-2xy}\le 1$

Question

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

Answer 1

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, $$\sum_{cyc}(a-b)^{2}\left(\frac{1}{5-6ab}-\frac{2(a+b)c}{(5-6ac)(5-6bc)}\right)=\sum_{cyc}(a-b)^2S_c\geq$$ $$\geq S_b(a-c)^2+S_a(b-c)^2\geq S_b(b-c)^2+S_a(b-c)^2=(b-c)^2(S_b+S_a)\geq0.$$

Answer 2

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.

1. $w^3=0$.

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;

2. 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 $x^2+y^2+z^2=3$.

Answer 3

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.