解关于x得不等式(a^2)x+b^2(1-x)>=[ax+b(1-x)]^2,其中a不等于b

a^2-b^2)x+b^2>=[(a-b)x+b]^2
(a-b)^2x^2+[2(a-b)b-(a^2-b^2)]x<=0
(a-b)^2x^2+(a-b)(2b-a-b)x<=0
(a-b)^2*(x^2-x)<=0
(a-b)^2*(x-1)x<=0
因为a不等于b,所以(a-b)^2>0,(x-1)x<=0
解得0<=x<=1
原关于x的不等式的解集为[0,1]