Question

State True or False.

If a, b, c, d are four positive real numbers such that $abcd=1$, then
$(1+a)(1+b)(1+c)(1+d)\ge 16$

• A. True
• B. False

Answer 1

The correct option is A True

Since $AM\ge GM$, therefore
$\frac{1+a}{2}\ge \sqrt{\left(1.a\right)},\frac{1+b}{2}\ge \sqrt{\left(1.b\right)}$,$\frac{1+c}{2}\ge \sqrt{\left(1.c\right)},\frac{1+d}{2}\ge \sqrt{\left(1.d\right)}$
Multiplying corresponding sides of theabove inequalities, we have
$(1+a)(1+b)(1+c)(1+d)\ge 16\sqrt{\left(abcd\right)}=16$
since $abcd=1$
Hence $(1+a)(1+b)(1+c)(1+d)\ge 16$