Let a,b, and c be positive real numbers such that a>b and a+b+c=4. Find the minimum value of 4a+3b+(a−b)bc3.
Solution — click to reveal
By AM-GM, (a−b)+b+(a−b)bc3≥33(a−b)⋅b⋅(a−b)bc3=3c.Hence, 4a+3b+(a−b)bc3=3a+3b+[(a−b)+b+(a−b)bc3]≥3a+3b+3c=12.Equality occurs when a=2 and b=c=1, so the minimum value is 12.