# Prove that $d\le{(xy+yz+xz)^{\frac{1}{3}}}$.

Let $x,y,z$ be positive integers such that $\frac{x+1}{y}+\frac{y+1}{z}+\frac{z+1}{x}$ is an integer. Let $d$ be the greatest common divisor of $x,y,z$. Prove that

$$d\le{(xy+yz+xz)^{\frac{1}{3}}}$$.

I have no idea how to proceed in this question.

For simplicity, $$\sum _{ cyc }^{ }{ f(a,b,c) }=f(a,b,c)+f(b,c,a)+f(c,a,b)$$
Notice that the equation abve becomes $$\frac{\sum _{ cyc }^{ }{ x^{ 2 }y } +\sum _{ cyc }^{ }{ xy }}{xyz}$$.
However, because $d^3$ divides both $\sum _{ cyc }^{ }{ x^{ 2 }y}$ and $xyz$, $d^3$ must divide $${\sum _{ cyc }^{ }{ xy }}$$ for the equation to be a interger.
This implies that $d^3 \le \sum _{ cyc }^{ }{ xy }$