Find minimum of $$x + y^5$$ where $x>0,y>0 $ $xy=1$ without using of differential calculus.
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No calculus? Why? – hmakholm left over Monica Nov 13 '14 at 15:27
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Because I want to know if this is possible. – Nov 13 '14 at 15:29
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1Find the $y>0$ where $$\lim_{h\to0}\frac{(y+h)^{-1}+(y+h)^5 - y^{-1}-y^5}{h} = 0$$ – peterwhy Nov 13 '14 at 15:48
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@peterwhy Fine comment !!!. +1. – Felix Marin Nov 13 '14 at 15:50
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$x+y^5=\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+\frac{x}{5}+y^5$. 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 $x=\sqrt[6]{5},y=\sqrt[6]{\frac{1}{5}}$