Findind the best constant $k$ satisfying the following inequality$$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+3k\ge (k+1)\cdot\frac{x+y+z}{\sqrt[3]{xyz}},$$holds for all $x,y,z>0.$
It was in AOPS. See here.
Firstly, I hope you explain me the difference between "greatest constant" and "best constant".
As @arqady's comment, the desired $k$ is not nice and I got stuck to full understand how $uvw$ works here.
I truly wish @Michael Rozenberg makes it more clearly here (idAOPS-arqady is Michael Rozenberg).
I can prove it when $0$, which is actually not a best $k$.
Indeed, we'll prove $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge \frac{x+y+z}{\sqrt[3]{xyz}}. \tag{1}$$ By using AM-GM $$\frac{x}{y}+\frac{x}{y}+\frac{y}{z}\ge 3\sqrt[3]{\frac{x}{y}\cdot\frac{x}{y}\cdot\frac{y}{z}}=3\sqrt[3]{\frac{x^2}{yz}}=3\cdot\frac{x}{\sqrt[3]{xyz}}.$$Similarly, $$\frac{y}{z}+\frac{y}{z}+\frac{z}{x}\ge3\cdot\frac{y}{\sqrt[3]{xyz}}.$$$$\frac{z}{x}+\frac{z}{x}+\frac{x}{y}\ge3\cdot\frac{z}{\sqrt[3]{xyz}}.$$Sum up these inequalities, we prove $(1).$
All ideas and comments about the starting inequality are welcome. Thank you for your help.
Proof.
Due to homopgenization, we may assume that $xyz=1.$
Let $x=\dfrac{b}{a};y=\dfrac{c}{b};z=\dfrac{a}{c}$ then $a,b,c>0$ and the OP becomes $$\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}+3k\ge (k+1)\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)$$ $$\iff a^3+b^3+c^3-3abc\ge (k+1)(a^2b+b^2c+c^2a-3abc).\tag{1}$$ In case $c\rightarrow 0,$ it's $$a^3+b^3\ge (k+1)a^2b.$$Notice that by AM-GM $$a^3+b^3=b^3+\frac{a^3}{2}+\frac{a^3}{2}\ge \dfrac{3}{\sqrt[3]{4}}a^2b\ge (k+1)a^2b.$$Thus $k\le \dfrac{3}{\sqrt[3]{4}}-1.$
Now, it's enough to prove $(2)$ is true when $k=\dfrac{3}{\sqrt[3]{4}}-1$ $$\iff a^3+b^3+c^3-3abc\ge \dfrac{3}{\sqrt[3]{4}}(a^2b+b^2c+c^2a-3abc).\tag{2}$$WLOG, assuming that $c=\min\{a,b,c\}.$ Denote $b=c+u; a=c+v$ where $u,v\ge 0.$
$(2)$ turns out $$(u+c)^3+(v+c)^3+c^3-3c(u+c)(v+c)\ge\dfrac{3}{\sqrt[3]{4}} [(c+u)(c+v)^2+(c+u)^2c+c^2(c+v)-3c(u+c)(v+c)]$$ or $$\left(3-\dfrac{3}{\sqrt[3]{4}}\right)(u^2-uv+v^2)c+u^3+v^3-\dfrac{3}{\sqrt[3]{4}}uv^2\ge 0,$$which is true.
Hence, $k_{\max}=\dfrac{3}{\sqrt[3]{4}}-1$ is the desired constant.