Question

If $a,b,c,d$ be positive numbers such that $a+b+c+d=2$, then the minimum value of $\frac{1}{abc}+\frac{1}{bcd}+\frac{1}{cda}+\frac{1}{dab}$ is

• A. 4
• B. 64
• C. 16
• D. 32

Answer 1

The correct option is D 32

$a+b+c+d>=4\ast (abcd{)}^{1/4}$
$2>=4\ast (abcd{)}^{1/4}$
$1/2>=(abcd{)}^{1/4}$
$abcd<=1/16$
Dividing the equation $a+b+c+d=2$ by abcd
We get $1/bcd+1/acd+1/abd+1/abc=2/abcd$
After substituting the maximum value of $abcd$ in the above equation, we obtain the minimum value of left hand side.
So the minimum value of the expression is 32.
Option D is correct.