The first point: when 0≤x≤ 1 and x 3 ≤ x 2, that is, x 3/2 ≤ x 2/2, so (a+2) x-x 2-x 3/2 ≥ (a+2) x-x.

Second place: Actually, we need to find an X to make -3/2x[x-2/3(a+2)] greater than 0.

Because 0