• No products in the cart.

# Inequality on four positive real numbers (TOMATO Subjective 82)

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

April 21, 2015