Inequality $\frac{xy+z}{x+yz}+\frac{yz+x}{y+zx}+\frac{zx+y}{z+xy}-\frac{x+y+z}{3}\leq 1$

Question

For $x,y,z \in [2,\infty)$, prove that

$\frac{xy+z}{x+yz}+\frac{yz+x}{y+zx}+\frac{zx+y}{z+xy}-\frac{x+y+z}{3}\leq 1$

I tried to group the terms and prove that $\frac{xy+z}{x+yz} - \frac{y}{3}\leq \frac{1}{3}$

Answer 1

using the lemma:if $a,b\ge 2$ then $ab\ge a+b$

$$\sum_{cyc} \frac{xy+z}{x+\color{red}{yz}}-\sum \frac{x}{3}$$ $$\le \sum_{cyc} \frac{xy+z}{x+\color{red}{y+z}}-\sum\frac{x}{3}=1-\dfrac{1}{6}\Big( \dfrac{{(x-y)}^2+{(y-z)}^2+{(z-x)}^2}{x+y+z}\Big)\le 1$$