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

If x,y>0   log x+log y=2,  What is the minimum value of (1/x)+(1/y)

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

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

ln xy=2  ⇒xy=e^2   AM≥GM  (1/x)+(1/y)≥2(√((1/x)×(1/y)))=(2/( (√(xy))))=(2/e)  i.e. ((1/x)+(1/y))_(min) =(2/e)

$$\mathrm{ln}\:{xy}=\mathrm{2} \\ $$ $$\Rightarrow{xy}={e}^{\mathrm{2}} \\ $$ $${AM}\geqslant{GM} \\ $$ $$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\geqslant\mathrm{2}\sqrt{\frac{\mathrm{1}}{{x}}×\frac{\mathrm{1}}{{y}}}=\frac{\mathrm{2}}{\:\sqrt{{xy}}}=\frac{\mathrm{2}}{{e}} \\ $$ $${i}.{e}.\:\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right)_{{min}} =\frac{\mathrm{2}}{{e}} \\ $$

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