Question

已知x1,x2,x3均为正数,且x1+x2+x3=1,求证x1^2\(x1+x2)+x2^2\(x2+x3)+x3^2\

Answer 1

柯西不等式

[x1^2(x1+x2)+x2^2(x2+x3)+x3^2(x1+x3)]

*[(x1+x2)+(x2+x3)+(x1+x3)]

>=[根号(x1^2(x1+x2)*(x1+x2))+根号(x2^2(x2+x3)*(x2+x3))+根号(x3^2(x3+x1)*(x3+x1))]^2

=[x1+x2+x3]^2=1

而(x1+x2)+(x2+x3)+(x1+x3)=2(x1+x2+x3)=2

所以

x1^2(x1+x2)+x2^2(x2+x3)+x3^2(x1+x3)≥12

等号成立时,

x1^2(x1+x2)(x1+x2)=x2^2(x2+x3)(x2+x3)=x3^2(x3+x1)(x3+x1)

可得x1=x2=x3=1/3