Question

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

Answer 1

a, b, c, d are positive and real.

To minimize $(1+a)(1+b)(1+c)(1+d)$ when $abcd=1$

Let product to be minimized be P by $AM-GM$,

$\frac{1+a+1+b+1+c+1+d}{4}\ge (1+a)(1+b)(1+c)(1+d)$

$1+\frac{a+b+c+d}{4}\ge P^{1/4}$

For P to be minimum, $\frac{a+b+c+d}{4}$ is minimum

By $AM-GM$

$\frac{a+b+c+d}{4}\ge (abcd{)}^{1/4}$

$a+b+c+d\ge 4$

This is true when $a=b=c=d=1$

$P_{min}^{1/4}=1+1=2$

$P=16$.