Let 0≤a,b,c≤1. Find the maximum value of abc+(1−a)(1−b)(1−c).
Solution — click to reveal
Since 0≤c≤1,c≤1 and 1−c≤1, so abc+(1−a)(1−b)(1−c)≤ab+(1−a)(1−b).Then by AM-GM, ab≤2a+band (1−a)(1−b)≤2(1−a)+(1−b)=22−a−b,so ab+(1−a)(1−b)≤2a+b+22−a−b=1.Equality occurs when a=b=c=0, so the maximum value is 1.