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

Finding the minimum value:

Given: $a,b,c,d$ are four positive real numbers such that $abcd=1$,

Let $x=(1+a)(1+b)(1+c)(1+d)$

$\Rightarrow$ $x=1+a+b+c+d+(ab+bc+cd+ad+bd+ac)+(abc+bcd+cda+dab)+abcd$

$\Rightarrow$ $x=1+a+b+c+d+ab+bc+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}+ac+\frac{1`}{d}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1\left[abcd=1\right]$

$\Rightarrow$ $x=2+\left(a+\frac{1}{a}\right)+\left(b+\frac{1}{b}\right)+\left(c+\frac{1}{c}\right)+\left(d+\frac{1}{d}\right)+\left(ab+\frac{1}{ab}\right)+\left(bc+\frac{1}{bc}\right)+\left(ac+\frac{1}{ac}\right)$

Since according to rule of arithmetic mean any number $z$ will always satisfy, $z+\frac{1}{z}\ge 2$.

$\Rightarrow$$x_{min}=2+2+2+2+2+2+2+2$

$\Rightarrow$$x_{min}=16$

Therefore, the minimum value of is $16$.