Let a, b, c, d be positive real numbers such that abcd = 1. Show that (1+a)(1+b)(1+c)(1+d) \ge 16

Discussion:

Concept: Inequality (see this link for some background information).

Using A.M. – G.M. inequality we see that

\frac{1+a}{2} \ge \sqrt {1 \times a}
\frac{1+b}{2} \ge \sqrt {1 \times b}
\frac{1+c}{2} \ge \sqrt {1 \times c}
\frac{1+d}{2} \ge \sqrt {1 \times d}

Hence \displaystyle {\frac{1+a}{2} \times \frac{1+b}{2} \times \frac{1+c}{2} \times \frac{1+d}{2}\ge \sqrt {abcd} = 1 }

Therefore (1+a)(1+b)(1+c)(1+d) \ge 16