# 1971 Canadian MO Problems/Problem 2

## Problem

Let $x$ and $y$ be positive real numbers such that $x+y=1$. Show that $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\ge 9$.

## Solution

### Solution 1 $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \ge 9$$ $$1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9$$ $$\frac{xy+x+y+1}{xy}\ge 9$$ $$\frac{xy+2}{xy}\ge 9$$ $$1 + \frac{2}{xy}\ge 9$$ $$\frac{2}{xy} \ge 8$$ $$1\ge 4xy$$ which is true since by AM-GM, we get: $$x^2 + y^2 \ge 2xy$$ $$x^2 + 2xy + y^2 \ge 4xy$$ $$(x+y)^2 \ge 4xy$$ and we are given that $x + y = 1$, so $$(x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy$$

### Solution 2

Let $f(x)=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{1-x}\right)$. Since we want to find the minimum of the function over the interval $(0,1)$, we can take the derivative. Using the product rule, we get $\frac{2(2x-1)}{x^2(1-x)^2}$. Since we want this value to be zero, the numerator must be zero. But the only root of the equation $2(2x-1)$ is $1/2$, and so plugging the answer back in we have $$(1+\frac{1}{\frac{1}{2}})(1+\frac{1}{\frac{1}{2}})$$ $$(1+2)(1+2)$$ $$3\cdot 3$$ $$9$$ Thus the minimum is $\boxed{9}$ and we are done.