problem in inequalities

Question

If $x,y$ are positive real numbers such that $x+y=1$ then

a minimum value of $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ is??

Where am I wrong??

$(1+1/x)(1+1/y)=1+(1/x)+(1/y)+(1/xy). $

By AM-GM it is more than equal to$ (4/√(xy))$

By AM-GM $x+y\geq2\sqrt{xy} $ or $ \frac{1}{\sqrt{xy}}\geq2$.

Then $(1+1/x)(1+1/y)\geq\frac{4}{\sqrt{xy}}\geq8$

Minimum value is 8.

But answer key says it is 9.

Where am I wrong???

Answer 1

By C-S and AM-GM we obtain: $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\geq\left(1+\frac{1}{\sqrt{xy}}\right)^2\geq\left(1+\frac{1}{\frac{x+y}{2}}\right)^2=9$$ You can not use AM-GM by your way because the minimal value occurs for $x=y=\frac{1}{2}$,

but with your way the equality "occurs" for $1=\frac{1}{x}=\frac{1}{y}=\frac{1}{xy}$, which is impossible.

Answer 2

Hint:

$$(1+1/x)(1+1/y)=1+\dfrac2{xy}$$

Now $4xy-(x+y)= -(x-y)^2\le0\implies4xy\le(x+y)^2=?$

Or WLOG $x=\sin^2t$

Answer 3

The function and the constraint are symmetrical in $x,y$. So a local max/min shall be at $x=y=1/2, f(x,y)=9$, and you can easily demonstrate that it is a minimum.

Answer 4

Your way is quite close: $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=$$ $$1+\frac{x+y}{x}+\frac{x+y}{y}+\frac{1}{xy}=$$ $$3+\underbrace{\frac{y}{x}+\frac{x}{y}}_{\ge 2}+\underbrace{\frac{1}{xy}}_{\ge 4}\ge 9.$$