Question

If $a, b, c, d$ are four positive real numbers such that $abcd = 1$, what is the minimum value of $(1+a)(1+b)(1+c)(1+d)$?

• A. 4
• B. 1
• C. 16
• D. 18

Hint:

When minimizing symmetric expressions with constant product, AM-GM is often the quickest approach.

Answer 1

The Correct Option is C

By AM-GM inequality: $(1+a) \ge 2\sqrt{a}$, similarly for other factors. Product $(1+a)(1+b)(1+c)(1+d) \ge 2^4 \sqrt{abcd} = 16 \times \sqrt{1} = 16$. Equality holds when $a=b=c=d=1$.