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 \[(x+y)^2 \ge 4xy \Rightarrow 1 \ge 4xy\]

See Also

1971 Canadian MO (Problems)
Preceded by
Problem 1
1 2 3 4 5 6 7 8 Followed by
Problem 3
Invalid username
Login to AoPS