Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied

Question

Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied: $(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \geq t$.

If there were such an inequality: $ t_{1} \leq (xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \leq t_{2}$, it is obvious that $t_{1}=0$ is suitable. And if $x=y=z$, then we have: $(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2=3x^2 \cdot \dfrac{9}{4x^2}=\dfrac{27}{4}$. Next question is this value is the maximum or minimum? If $x=1, y=1, z=0$, then we have $F(1,1,0)=(\dfrac{1}{2}+2)^2=\dfrac{25}{4}< \dfrac{27}{4}$, but if $x=1, y=1, z=n$, then we have $F(1,1,n)=(1+2n)(\dfrac{1}{2}+\dfrac{2}{1+n})^2>(1+2n)(\dfrac{1}{4}) \rightarrow +\infty$, which means, if I'm not mistaken, that this function has no extremum?

What to do with the case when $x,y,z$ are different I have not yet figured out.

Maybe in my case one of the following inequalities could be applied, but I'm stumped: $1) \ xyz \geq(x+y-z)(x+z-y)(y+z-x)$

$2) \ x^3+y^3+z^3 \geq x^2y+x^2z+y^2x+y^2z+z^2x+z^2y-3xyz$

$3) \ x^3+y^3+z^3 \geq 3xyz$.

Maybe someone can give an idea where to start with my case? Thanks!!

Answer 1

For $z\rightarrow0^+$ and $x=y=1$ we obtain: $t\leq\frac{25}{4}$.

We'll prove that $\frac{25}{4}$ is valid.

Indeed, we need to prove that $$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2\geq\frac{25}{4}.$$ Now, let $x+y+z=2u$ and since our inequality is homogeneous, we can assume $xy+xy+xz=1$

and we need to prove that: $$\sum_{cyc}\frac{1}{x+y}\geq\frac{5}{2}.$$

1. Let $u\geq1$.

Thus, by AM-GM we obtain: $$\sum_{cyc}\frac{1}{x+y}=\sum_{cyc}\frac{xy+xz+yz}{x+y}=2u+\sum_{cyc}\frac{xy}{x+y}\geq2u+\sum_{cyc}\frac{xy}{x+y+z}=$$ $$=2u+\frac{1}{2u}=\frac{3}{2}u+\frac{u}{2}+\frac{1}{2u}\geq\frac{3}{2}+2=\frac{5}{2}.$$ 2.Let $u\leq1$.

Here, by Schur, we obtain: $$xyz\geq\frac{1}{9}(4(x+y+z)(xy+xz+yz)-(x+y+z)^3)=\frac{8}{9}(u-u^3),$$ which gives $$\sum_{cyc}\frac{1}{x+y}=\frac{\sum\limits_{cyc}(x+y)(x+z)}{\prod\limits_{cyc}(x+y)}=\frac{\sum\limits_{cyc}(x^2+3xy)}{(x+y+z)(xy+xz+yz)-xyz}=$$ $$=\frac{x^2+y^2+z^2+3}{2u-xyz}=\frac{4u^2+1}{2u-xyz}\geq\frac{4u^2+1}{2u-\frac{8}{9}(u-u^3)}=$$ $$=\frac{9(4u^2+1)}{2(4u^3+5u)}-\frac{5}{2}+\frac{5}{2}=\frac{(1-u)(20u^2-16u+9)}{2(4u^3+5u)}+\frac{5}{2}\geq\frac{5}{2}$$ and we are done.

Answer 2

Another way.

By my previous post we need to prove that: $$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2\geq\frac{25}{4},$$ which is true by Muirhead: $$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2-\frac{25}{4}=$$ $$=\frac{\sum\limits_{sym}\left(4x^5y-x^4y^2-3x^3y^3+x^4yz+14x^3y^2z+\frac{19}{3}x^2y^2z^2\right)}{4\prod\limits_{cyc}(x+y)^2}\geq0.$$