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\
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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\(x1+x3)≥1\2
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)≥1\2
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
[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)≥1\2
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
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