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Question Number 108000 by ZiYangLee last updated on 13/Aug/20

$$\mathrm{If}\mathrm{x},\mathrm{y}>0$$ $$\log \mathrm{x}+\log \mathrm{y}=2,$$ $$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}$$

Answered by mr W last updated on 13/Aug/20

$$\ln xy=2$$ $$\Rightarrow xy=e^2$$ $$AM⩾GM$$ $$\frac{1}{x}+\frac{1}{y}⩾2\sqrt{\frac{1}{x}\times \frac{1}{y}}=\frac{2}{\sqrt{xy}}=\frac{2}{e}$$ $$i.e.{(\frac{1}{x}+\frac{1}{y})}_{min}=\frac{2}{e}$$

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