Let $\left(\frac{x}{d},\frac{y}{d},\frac{z}{d}\right)=(a,b,c)$. Then $\frac{x+1}{y}+\frac{y+1}{z}+\frac{z+1}{x}=$ $\frac{x^2z+xz+y^2x+yx+z^2y+zy}{xyz}=\frac{d^3(a^2c+b^2a+c^2b)+xy+yz+zx}{d^3abc}$ It follows that $d^3\mid xy+yz+zx\implies d^3\le xy+yz+zx\implies d\le\sqrt[3]{xy+yz+zx}$.

I think we can generalise this.This is a bit general form of:If $\frac {a+1} b+\frac {b+1} a$ is an integer and $gcd(a,b)=d$ then $d^2\le a+b$ .And the proof is same.