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Prove that if \(x+y=1\) then \((1+\frac{1}{x})(1+\frac{1}{y})\geq9\).

Note by Joshua Ong 3 years, 10 months ago

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Cauchy Schwarz gives \[\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\ge \left(1+\dfrac{1}{\sqrt{xy}}\right)^2\]

However, \(\dfrac{1}{\sqrt{xy}}\ge \dfrac{2}{x+y}=2\) by AM-GM, so \[\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\ge \left(1+\dfrac{1}{\sqrt{xy}}\right)^2\ge (1+2)^2=9\] and we are done.

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TopNewestCauchy Schwarz gives \[\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\ge \left(1+\dfrac{1}{\sqrt{xy}}\right)^2\]

However, \(\dfrac{1}{\sqrt{xy}}\ge \dfrac{2}{x+y}=2\) by AM-GM, so \[\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\ge \left(1+\dfrac{1}{\sqrt{xy}}\right)^2\ge (1+2)^2=9\] and we are done.

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