Inequality using Hölder $x=\frac{y}{y-1}$ $y>1$

Question

Prove that : $$x^{x}+xy^{y-1}>(xy)^{\frac{1}{n}}$$ where $$x=\frac{y}{y-1}$$and $\min(y,n)>1$.

I know that : Young inequality $xy<\frac{x^p}{p}+\frac{y^q}{q}$

$q=\frac{p}{p-1}$

but how can I apply it here ?

Answer 1

Since $$\frac{1}{x}+\frac{1}{y}=1,$$ by Holder we obtain: $$x^{x-1}+y^{y-1}=\frac{x^x}{x}+\frac{y^y}{y}\geq xy,$$ which gives $$x^x+xy^{y-1}\geq x^2y$$ and it's enough to prove that $$(x^2y)^n\geq xy,$$ which is true becase $$(x^2y)^n\geq(xy)^n\geq xy.$$