$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-3\geq k\left ( \frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}-1 \right )$

Question

Let $x,y,z>0$. Find the maximum value of $k$ such that the inequality $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-3\geq k\left ( \frac{x^{2}+y^{2}+z^{2}}{xy+yz+zx}-1 \right )$ is true for all $x,y,z>0$.

Here is my incorrect way to solve: Since the inequality is true for all $x,y,z>0$, let $b=c$ then we have $\frac{x}{y}+\frac{y}{x}-2\geq k(\frac{x^2+2y^2}{2xy+y^2}-1)\Leftrightarrow \frac{(x-y)^2}{xy}\geq k\frac{(x-y)^2}{2xy+y^2}\Rightarrow k\leq \frac{2xy+y^2}{xy}=2+\frac{y}{x}$.

If we fix $y$, let $x\rightarrow +\infty $ then $k\leq 2$. Howerver, the problem is that I can not prove the case $k=2$ is true and maybe the case $k=2$ is never true. And in my opinion, we must prove $k\leq1$ in any way. And the case $k=1$ is easy to solve.

Can anyone help me to show that $k\leq1$ please, thank all!

Answer 1

For $k=1$ our inequality is true.

Thus, by your work $1\leq k_{max}\leq2.$

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Thus, we need to prove that $$\frac{\sum\limits_{cyc}(x^2z-xyz)}{xyz}\geq\frac{k\sum\limits_{cyc}(x^2-xy)}{xy+xz+yz}$$ or $$v^2\sum_{cyc}(2x^2z-2xyz)\geq6k(u^2-v^2)w^3$$ or $$v^2\sum_{cyc}(x^2y+x^2z-2xyz)-6k(u^2-v^2)w^3\geq v^2\sum_{cyc}x^2y-x^2z)$$ or $$v^2(9uv^2-9w^3)-6k(u^2-v^2)w^3\geq v^2(x-y)(x-z)(y-z).$$ Now, for all $k\leq2$ we obtain: $$v^2(9uv^2-9w^3)-6k(u^2-v^2)w^3\geq v^2(9uv^2-9w^3)-12(u^2-v^2)w^3=$$ $$=3(3uv^4-4u^2w^3+v^2w^3)\geq0$$ by $uvw$.

Indeed, the last inequality is a linear inequality of $w^3$, which says it's enough to prove it for an extreme value of $w^3,$ which happens in the following cases.

1. $w^3\rightarrow0^+$.

In this case $3uv^4-4u^2w^3+v^2w^3\geq0$ is obviously true;

2. Two variables are equal.

Since this inequality is homogeneous, it's enough to assume that $y=z=1$ and we need to prove that $$\frac{(x+2)(2x+1)^2}{9}-\frac{4(x+2)^2x}{9}+\frac{(2x+1)x}{3}\geq0$$ or $$(x-1)^2\geq0.$$ Id est, we proved that $$v^2(9uv^2-9w^3)-6k(u^2-v^2)w^3\geq0$$ and it's enough to prove that $$\left(v^2(9uv^2-9w^3)-6k(u^2-v^2)w^3\right)^2\geq v^4\prod_{cyc}(x-y)^2$$ or $$\left(3v^2(uv^2-w^3)-2k(u^2-v^2)w^3\right)^2\geq3v^4(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $$(k^2(u^2-v^2)^2+3kv^2(u^2-v^2)+3v^6)w^6+3uv^4(u^2-3v^2-k(u^2-v^2))w^3+3v^{10}\geq0.$$ Now, let $u^2=tv^2$.

Thus, $t\geq1$ and if we want that our inequality will be true we need $$9t(t-3-k(t-1))^2-4\cdot3\cdot(k^2(t-1)^2+3k(t-1)+3)\leq0$$ or $$(t-1)^2((3t-4)k^2-6(t-2)k+3(t-4))\leq0.$$

Let $t=\frac{4}{3}.$

Thus,$$(3t-4)k^2-6(t-2)k+3(t-4)=4k-8\leq0;$$ Let $1\leq t<\frac{4}{3}.$

Thus, we need to prove that $f(k)\geq0$ for all $1\leq k\leq 2$, where
$$f(k)=(4-3t)k^2-6(2-t)k+3(4-t),$$ which is true because $$f(1)=4>0;$$ $$f(2)=4-3t>0$$ and $$\frac{3(2-t)}{4-3t}>1;$$ Let $t>\frac{4}{3}.$

Thus, we need $$(3t-4)k^2-6(t-2)k+3(t-4)\leq0,$$ for which we need $$ k\leq\frac{3(t-2)+\sqrt{9(t-2)^2-3(3t-4)(t-4)}}{3t-4}$$ or $$k\leq\frac{3(t-2)+2\sqrt{3(t-1)}}{3t-4},$$ which gives that we need $$ k\leq\inf_{t>\frac{4}{3}}\frac{3(t-2)+2\sqrt{3(t-1)}}{3t-4}=1$$ and we are done!