Question

设x1.x2.x3大于0,且x1+x2+x3=1,则x1*x2^2*x3+x1*x2*x3^2的最大值是多少?

Answer 1

x1*x2^2*x3+x1*x2*x3^2

=x1*x2*x3*(x2+x3)

=(3-3x2-3x3)*(2x2)*(2x3)*(x2+x3)/12

≤{[(3-3x2-3x3)*(2x2)*(2x3)*(x2+x3)]/4}^4/12

=(3/4)^4/12

=27/1024.

当3-3x2-3x3=2x2=2x3=x2+x3时,即x1=1/4,x2=x3=3/8时,等号成立,

所以,x1*x2^2*x3+x1*x2*x3^2的最大值是27/1024.