Question

If three positive real numbers x,y,z are in A.P such that xyz = 4, then what will be the minimum value of y?

Answer 1

Since, x,y and z are positive real numbers,
$\frac{x+y+z}{3}\ge (xyz{)}^{\frac{1}{3}}$
x,y and z are in AP; $\Rightarrow$ y - d, y, y+d
$\Rightarrow \frac{y-d+y+y+d}{3}\ge (xyz{)}^{\frac{1}{3}}$
$\Rightarrow \frac{3y}{3}\ge (xyz{)}^{\frac{1}{3}}$ (Since, xyz=4)
$\Rightarrow y\ge (4{)}^{\frac{1}{3}}$
$\Rightarrow$ Minimum value of y is $(4{)}^{\frac{1}{3}}$.