Let x and y be positive real numbers such that x + y = 1. Show that

( 1 + 1/x ) ( 1 + 1/y ) \ge 9.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Applying the AM-GM inequality, we have

(1)    x + y \ge 2\sqrt{xy},

Now successively,

(x + y)^{2} \ge 4xy,
1 \ge 4xy,
xy \ge 4x^{2}y^{2},
\sqrt{xy} \ge 2xy,
\sqrt{xy} \ge 8xy,

Since x + y = 1, (1) can also be written as 1 \ge 2\sqrt{xy}

2 \ge \sqrt{xy} \ge 8xy,
x + y + 1 \ge 8xy,
(x + y + 1) / (xy) \ge 8,
1/x + 1/y + 1/ (xy) \ge 8,
1 + 1/x + 1/y + 1/ (xy) \ge 9,
( 1 + 1/x ) ( 1 + 1/y ) \ge 9.

Solution #2

We just write down a sequence of equivalent inequalities

( 1 + 1/x ) ( 1 + 1/y ) \ge 9,
( x + 1 ) ( y + 1 ) \ge 9xy,
1 + x + y \ge 8xy,
2 \ge 8xy,
1 \ge 4xy,
(x + y)^{2} \ge 4xy,
(x - y)^{2} \ge 0.

Since the last inequality is correct, so is the first one. This is of course possible to proceed backwards.

Solution by Mike Wray

Lett x = (1 + t)/2, y = (1 t)/2, t \in (-1,1). Then

LHS = (3+t)(3-t)/((1+t)(1-t)) = (9 - t^{2})/(1- t^{2}) = (8 + 1 - t^{2} )/(1- t^{2}) \ge 8 + 1 = 9.