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

$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \ge 9$ $1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \ge 9$ $\frac{xy+x+y+1}{xy}\ge 9$ $\frac{xy+2}{xy}\ge 9$ $1 + \frac{2}{xy}\ge 9$ $\frac{2}{xy} \ge 8$ $1\ge 4xy$ 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 $(x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy$

1971 Canadian MO (Problems)
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 •	Followed by Problem 3

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1971 Canadian MO Problems/Problem 2

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