Prove $x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2$

Question

Let $x>1$, $y>1$ and $z>1$ be such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. Prove that: $$x^{y+1}z+y^{z+1}x+z^{x+1}y\geq x^2y^2z^2.$$

When $x=y=z=3$, both sides are equal to $3^6$. The difficulty is how to deal with the variables appearing in the exponents, and how to use the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.

Answer 1

It's just AM-GM: $$\sum_{cyc}x^{y+1}z=xyz\sum_{cyc}\frac{1}{y}x^{y}\geq xyz\prod_{cyc}\left(x^y\right)^{\frac{1}{y}}=x^2y^2z^2.$$