Question

设b²-2ab>0,关于x的不等式a²x² b²(1-a)≥[ax+b(1-x)

Answer 1

a^2x^2-b^2x+b^2≥[(a-b)x+b]^2

a^2x^2-b^2x+b^2≥(a-b)^2x^2+2b(a-b)x+b^2

[(a-b)^2-a^2]x+[2b(a-b)+b^2]x≤0

(b^2-2ab)x^2+(2ab-b^2)x≤0

(b^2-2ab)x^2-(b^2-2ab)x≤0

因为b^-2ab>0,所以

x^2-x≤0

故得 0≤x≤1.