设x＞0,0＜α＜1,利用最值证明不等式:x^α≦αx+(1-α)

Question

设x＞0,0＜α＜1,利用最值证明不等式:x^α≦αx+(1-α)

Answer 1

令f(x) = x^α - αx - (1 - α) = x^α - αx + (α-1)
f'(x) = αx(α-1)-α = α[x^(α-1)-1]
∵0＜α＜1
∴-1＜α-1＜0
0＜x＜1时,x^(α-1)＞1,f'(x)=x^(α-1)-1＞0,f(x)单调增
x＞1时,x^(α-1)＜1,f'(x)=x^(α-1)-1＜0,f(x)单调减
当x=1时有极大值f(1) = x^1 - α*1 + α-1 = 0
即f(x) = (x^α - αx) - (1 - α) ≤ 0
∴(x^α - αx) ≤ (1 - α)