Question

证明:当x＞0,0＜α＜1时,不等式x∧α－αx≤1－α成立

Answer 1

【 当x＞0,0＜α＜1时,不等式x^α - αx ≤ 1 - α成立 】

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)