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