一道嗖嘎的证明题

f ( x ) f(x) f(x)在区间[0,1]上连续可微,满足 0 < f ′ ( x ) ≤ 1 00<f(x)1,任给 x ∈ [ 0 , 1 ] x\in[0,1] x[0,1] f ( 0 ) = 0 f(0)=0 f(0)=0,试证:
∫ 0 1 ( f ( x ) ) 3 d x ≤ ( ∫ 0 1 f ( x ) d x ) 2 \int_0^1\left(f\left(x\right)\right)^3\mathrm{d}x\leq\left(\int_0^1f(x)\mathrm{d}x\right)^2 01(f(x))3dx(01f(x)dx)2
证明:

0 < f ′ ( x ) ≤ 1 00<f(x)1知:
0 < f ( x ) − f ( 0 ) x − 0 = f ′ ( ξ 1 ) ≤ 1 0<\frac{f(x)-f(0)}{x-0}=f'(\xi_1)\leq 1 0<x0f(x)f(0)=f(ξ1)1
因此:
0 < f ( x ) ≤ x 00<f(x)x
那么
2 f ( t ) ⋅ f ′ ( t ) ≤ 2 f ( t ) 2f(t)\cdot f'(t)\leq2f(t) 2f(t)f(t)2f(t)

f 2 ( x ) − 2 ∫ 0 x f ( t ) d t x = 2 f ( x ) f ′ ( ξ 2 ) − 2 f ( ξ 2 ) ≤ 0 \frac{f^2(x)-2\int ^x_0f(t)\mathrm{d}t}{x}=2f(x)f'(\xi_2)-2f(\xi_2)\leq 0 xf2(x)20xf(t)dt=2f(x)f(ξ2)2f(ξ2)0
于是有
f 2 ( x ) ≤ 2 ∫ 0 x f ( t ) d t f^2(x)\leq 2\int^x_0f(t)\mathrm{d}t f2(x)20xf(t)dt
易得
f 3 ( x ) ≤ 2 f ( x ) ∫ 0 x f ( t ) d t f^3(x)\leq2f(x)\int^x_0f(t)\mathrm{d}t f3(x)2f(x)0xf(t)dt
由于
∫ 0 x f 3 ( t ) d t − [ ∫ 0 x f ( t ) d t ] 2 x = f 3 f ( ξ 3 ) − 2 f ( ξ 3 ) ∫ 0 x f ( t ) d t ≤ 0 \frac{\int^x_0f^3(t)\mathrm{d}t-\left[\int ^x_0f(t)\mathrm{d}t\right]^2}{x}=f^3f(\xi_3)-2f(\xi_3)\int^x_0 f(t)\mathrm{d}t\leq0 x0xf3(t)dt[0xf(t)dt]2=f3f(ξ3)2f(ξ3)0xf(t)dt0

∫ 0 x f 3 ( t ) d t ≤ [ ∫ 0 x f ( t ) d t ] 2 , ∀ x ∈ [ 0 , 1 ] \int^x_0f^3(t)\mathrm{d}t\leq \left[\int^x_0f(t)\mathrm{d}t\right]^2,\quad \forall x\in[0,1] 0xf3(t)dt[0xf(t)dt]2,x[0,1]
当我们取 x = 1 x=1 x=1时,题目得证。

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