If `xyz = 1` and `x, y, z gt 0` then the minimum value of the expression `(x+2y)(y+2z)(z+2x)` is

Options

A. `81`
B. `9`
C. `18`
D. `27`

Answer

D

Explanation

If `xyz = 1` and ……….
apply `AM ge GM`
`(x+y+y)/(3)ge(xy^(2))^((1)/(3))` ….(i)
`(y+z+z)/(3)ge (yz^(2))^((1)/(3))` ….(ii)
`(z+x+x)/(3)ge (zx^(2))^((1)/(3))` ….(iii)
From equation (i), (ii), and (iii)
`((x+2y)/(3))((y+2z)/(3))((z+2x)/(3))ge xyz`
`implies (x+2y) (y+2z) (z+2x) ge 27`