设
f
(
x
)
f(x)
f(x)在区间[0,1]上连续可微，满足
0
<
f
′
(
x
)
≤
1
0<f'(x)\leq1
0<f′(x)≤1,任给
x
∈
[
0
,
1
]
x\in[0,1]
x∈[0,1]且
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≤(∫01?f(x)dx)2
证明：
由
0
<
f
′
(
x
)
≤
1
0<f'(x)\leq 1
0<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<x?0f(x)?f(0)?=f′(ξ1?)≤1
因此：
0
<
f
(
x
)
≤
x
0<f(x)\leq x
0<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)
又
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)?2∫0x?f(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)≤2∫0x?f(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)∫0x?f(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
x∫0x?f3(t)dt?[∫0x?f(t)dt]2?=f3f(ξ3?)?2f(ξ3?)∫0x?f(t)dt≤0
故
∫
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]
∫0x?f3(t)dt≤[∫0x?f(t)dt]2,?x∈[0,1]
当我们取
x
=
1
x=1
x=1时，题目得证。