AM-GM Practice 2

$(1+\frac{1}{x})(1+\frac{1}{y})=\frac{1+x+y+xy}{xy}=\frac{9+xy}{xy}$ From AM-GM $\frac{x+y}{2}\ge\sqrt{xy}\Rightarrow4\ge\sqrt{xy}\Rightarrow16\ge xy$ $\Rightarrow \frac{1}{16}\le\frac{1}{xy}$ hence $\frac{9}{xy}+1\le\frac{25}{16}$

therefore minimum vakue of $(1+\frac{1}{x})(1+\frac{1}{y})=\boxed{1.5625}$

(1+1/x)(1+1/y)= 9/xy +1, so for the L.H.S to be minimum, xy has to be maximum, which is 16. So the answer is 25/16= 1.562. It's not hard to see why 16 is the maximum. xy= 16, when x=4, y=4, now if I increase x by an amount 'a', then y has to be also decreased by an amount 'a', so that x+y remains 8. Now xy= (4+a)(4-a)= 16-a^2, so whatever non-zero value of 'a' you choose, xy will always be less than 16. So taking a=0 gives the maximum value, 16.