Question

If $x,y,z$ are positive numbers, then the minimum value of $(x+y)(y+z)(z+x)(\frac{1}{x}+\frac{1}{y})(\frac{1}{y}+\frac{1}{z})(\frac{1}{z}+\frac{1}{x})$ is

• A. $32$
• B. $64$
• C. $128$
• D. $356$

Answer 1

The correct option is B $64$

Using the relation, $A.M.\ge H.M.$, we get
$\frac{x+y}{2}\ge \frac{2}{\frac{1}{x}+\frac{1}{y}}\Rightarrow (x+y)(\frac{1}{x}+\frac{1}{y})\ge 4\dots \dots \left(1\right)$
Similarly,
$(y+z)(\frac{1}{y}+\frac{1}{z})\ge 4\dots \dots \left(2\right)$
$(z+x)(\frac{1}{z}+\frac{1}{x})\ge 4\dots \dots \left(3\right)$

From equation $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$(x+y)(y+z)(z+x)(\frac{1}{x}+\frac{1}{y})(\frac{1}{y}+\frac{1}{z})(\frac{1}{z}+\frac{1}{x})\ge 64$

Hence, the minimum value of the given expression is $64$.