Let x,y,z be positive real numbers such that xyz=8. Find the minimum value of x+2y+4z.
Solution — click to reveal
By AM-GM, x+2y+4z≥33(x)(2y)(4z)=338xyz=338⋅8=12.Equality occurs when x=2y=4z and xyz=8. We can solve to get x=4,y=2, and z=1, so the minimum value is 12.