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Question Number 189858 by Shrinava last updated on 23/Mar/23

Answered by mr W last updated on 23/Mar/23

$$f\left(x\right)=x^{\frac{1}{x}}isstrictlyincreasingfor0<x<e$$$$since1<\sqrt{e}<\sqrt{\pi }<e,$$$${\left(\sqrt{e}\right)}^{\frac{1}{\sqrt{e}}}<{\left(\sqrt{\pi }\right)}^{\frac{1}{\sqrt{\pi }}}$$$$\Rightarrow {\left(\sqrt{e}\right)}^{\frac{\sqrt{\pi }}{\sqrt{e}}}<\left(\sqrt{\pi }\right)$$$$\Rightarrow {\left(\sqrt{e}\right)}^{\sqrt{\pi }}<{\left(\sqrt{\pi }\right)}^{\sqrt{e}}✓$$

Commented by Shrinava last updated on 24/Mar/23

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

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