Minimum without using of differential calculus Find minimum of $$x + y^5$$ where $x>0,y>0 $ $xy=1$ without using of differential calculus.

$x+y^5=\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+y^5$. By inequality between arithmetic mean and geometric mean we get $$\frac{x+y^5}{6}\geq \sqrt[6]{\frac{x^5}{5^5}y^5}=5^{-5/6}\Rightarrow x+y^5\geq 6\cdot 5^{-5/6}$$ and equality holds iff $\frac{x}{5}=y^5$, so corresponding values of arguments are $x=\sqrt[6]{5},y=\sqrt[6]{\frac{1}{5}}$
