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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 $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.

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

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  • $\begingroup$ Please clarify the am-gm step. $\endgroup$
    – vadim123
    Commented Aug 12, 2017 at 5:49
  • $\begingroup$ @vadim123 AM-GM it's the following thing. For $x_i>0$ and $\alpha_i>0$ such that $\sum\limits_{i=1}^n\alpha_i=1$ prove that $\sum\limits_{i=1}^n\alpha_ix_i\geq\prod\limits_{i=1}^nx_i^{\alpha_i}$. The proof follows from Jensen for $\ln$. $\endgroup$
    – Michael Rozenberg
    Commented Aug 12, 2017 at 5:53

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