思路分析：
看到要证明积分不等式的问题，首先想到构造函数，那构造什么样的函数呢？题目要证明什么，我们就构造什么,本题我们就可以构造 F ( x ) = ( x ? a ) f ( a + x 2 ) ? ∫ a x f ( t ) d t F(x)=(x-a)f(\frac{a+x}{2})-\int_{a}^{x}f(t)dt F(x)=(x?a)f(2a+x?)?∫ax?f(t)dt
证明：
必要性： 令
F
(
x
)
=
(
x
?
a
)
f
(
a
+
x
2
)
?
∫
a
x
f
(
t
)
d
t
F(x)=(x-a)f(\frac{a+x}{2})-\int_{a}^{x}f(t)dt
F(x)=(x?a)f(2a+x?)?∫ax?f(t)dt ，很明显F(a)=0,
F
′
(
x
)
=
f
(
a
+
x
2
)
+
1
2
(
x
?
a
)
f
′
(
a
+
x
2
)
?
f
(
x
)
F^{\prime}(x)=f(\frac{a+x}{2})+\frac{1}{2}(x-a)f^{\prime}(\frac{a+x}{2})-f(x)
F′(x)=f(2a+x?)+21?(x?a)f′(2a+x?)?f(x)
=
1
2
(
x
?
a
)
f
ˊ
(
a
+
x
2
)
?
[
f
(
x
)
?
f
(
a
+
x
2
)
]
=\frac{1}{2}(x-a)\acute{f}(\frac{a+x}{2})-[f(x)-f(\frac{a+x}{2})]
=21?(x?a)fˊ?(2a+x?)?[f(x)?f(2a+x?)]
=
1
2
(
x
?
a
)
f
′
(
a
+
x
2
)
?
1
2
(
x
?
a
)
f
(
ξ
)
ˊ
=\frac{1}{2}(x-a)f^{\prime}(\frac{a+x}{2})-\frac{1}{2}(x-a)\acute{f(\xi)}
=21?(x?a)f′(2a+x?)?21?(x?a)f(ξ)ˊ?,
ξ
∈
(
a
+
x
2
,
x
)
\xi\in(\frac{a+x}{2},x)
ξ∈(2a+x?,x)
=
1
2
(
x
?
a
)
[
f
ˊ
(
a
+
x
2
)
?
f
′
(
ξ
)
)
]
=\frac{1}{2}(x-a)[\acute{f}(\frac{a+x}{2})-f^{\prime}(\xi))]
=21?(x?a)[fˊ?(2a+x?)?f′(ξ))]
因为
f
′
′
(
x
)
>
0
,
f
′
(
a
+
x
2
)
?
f
′
(
ξ
)
<
0
f^{\prime{\prime}}(x)>0,f^{\prime}(\frac{a+x}{2})-f^{\prime}(\xi)<0
f′′(x)>0,f′(2a+x?)?f′(ξ)<0,故而
F
′
(
x
)
<
0
F^{\prime}(x)<0
F′(x)<0, 所以
F
(
b
)
≤
F
(
a
)
=
0
F(b) \leq F(a)=0
F(b)≤F(a)=0,所以
F
(
b
)
=
(
b
?
a
)
f
(
a
+
b
2
)
?
∫
a
b
f
(
t
)
d
t
≤
0
,
即
f
(
a
+
b
2
)
<
1
b
?
a
∫
a
b
f
(
t
)
d
t
F(b)=(b-a)f(\frac{a+b}{2})-\int_{a}^{b}f(t)dt \leq 0,即f(\frac{a+b}{2})<\frac{1}{b-a}\int_{a}^{b}f(t)dt
F(b)=(b?a)f(2a+b?)?∫ab?f(t)dt≤0,即f(2a+b?)<b?a1?∫ab?f(t)dt
充分性：
?
x
0
∈
(
?
∞
,
+
∞
)
\forall x_0\in(-\infty,+\infty)
?x0?∈(?∞,+∞),取
a
=
x
0
?
h
,
b
=
x
0
+
h
a=x_0-h,b=x_0+h
a=x0??h,b=x0?+h,其中
h
>
0
h>0
h>0,那我们的式子
f
(
a
+
b
2
)
≤
1
b
?
a
∫
a
b
f
(
t
)
d
t
f(\frac{a+b}{2})\leq\frac{1}{b-a}\int_{a}^{b}f(t)dt
f(2a+b?)≤b?a1?∫ab?f(t)dt就可以转化成
f
(
x
0
)
≤
1
2
h
∫
x
0
?
h
x
0
+
h
f
(
x
)
d
x
?
∫
x
0
?
h
x
0
+
h
f
(
x
)
d
x
?
2
h
f
(
x
0
)
2
h
≥
0
f(x_0)\leq\frac{1}{2h} \int_{x_0-h}^{x_0+h}f(x)dx \Leftrightarrow \frac{\int_{x_0-h}^{x_0+h}f(x)dx-2hf(x_0)}{2h}\geq0
f(x0?)≤2h1?∫x0??hx0?+h?f(x)dx?2h∫x0??hx0?+h?f(x)dx?2hf(x0?)?≥0,
为了证明题目中的
f
′
′
(
x
)
≥
0
,
f^{\prime\prime}(x)\geq 0,
f′′(x)≥0,我们只能想到2种办法
①从导数的定义式上入手，想办法构造导数的定义式
②利用泰勒公式(因为泰勒公式里面还有一阶导和二阶导)
③洛必达法则（不停地分子分母洛下去，必定能出现高阶导数）
对于本题而言，我们已经引入了邻域
(
x
0
?
h
,
x
0
+
h
)
(x_0-h,x_0+h)
(x0??h,x0?+h),因为要证明
f
′
′
(
x
)
f^{\prime\prime}(x)
f′′(x),所以我们不妨假设a和b隔得非常非常近，即
lim
?
h
→
0
\lim_{h\to 0}
limh→0?，对于这种求极限的情况，我们首先想到洛必达法则，故而对于式子
∫
x
0
?
h
x
0
+
h
f
(
x
)
d
x
?
2
h
f
(
x
0
)
2
h
≥
0
\frac{\int_{x_0-h}^{x_0+h}f(x)dx-2hf(x_0)}{2h}\geq0
2h∫x0??hx0?+h?f(x)dx?2hf(x0?)?≥0,我们可以想办法用洛必达法则解决，对于分子而言，必须要进行3次求导才能出现
f
′
′
(
x
)
f^{\prime\prime}(x)
f′′(x)二阶导，而分母却经不起3次求导，那怎么办呢，我们把分母的次幂加大到3阶
∫
x
0
?
h
x
0
+
h
f
(
x
)
d
x
?
2
h
f
(
x
0
)
2
h
3
≥
0
\frac{\int_{x_0-h}^{x_0+h}f(x)dx-2hf(x_0)}{2h^3}\geq0
2h3∫x0??hx0?+h?f(x)dx?2hf(x0?)?≥0,然后上下求导
lim
?
h
→
0
∫
x
0
?
h
x
0
+
h
f
(
x
)
d
x
?
2
h
f
(
x
0
)
2
h
3
\lim_{h\to0}\frac{\int_{x_0-h}^{x_0+h}f(x)dx-2hf(x_0)}{2h^3}
limh→0?2h3∫x0??hx0?+h?f(x)dx?2hf(x0?)?
=
lim
?
h
→
0
f
(
x
0
+
h
)
?
f
(
x
0
?
h
)
?
2
f
(
x
0
)
6
h
2
=\lim_{h\to0}\frac{f(x_0+h)-f(x_0-h)-2f(x_0)}{6h^2}
=h→0lim?6h2f(x0?+h)?f(x0??h)?2f(x0?)?
=
lim
?
h
→
0
f
′
(
x
0
+
h
)
+
f
′
(
x
0
?
h
)
12
h
=\lim_{h\to0} \frac{f^{\prime}(x_0+h)+f^{\prime}(x_0-h)}{12h}
=h→0lim?12hf′(x0?+h)+f′(x0??h)?
=
lim
?
h
→
0
f
′
′
(
x
0
+
h
)
?
f
′
′
(
x
0
?
h
)
12
=\lim_{h\to0}\frac{f^{\prime\prime}(x_0+h)-f^{\prime\prime}(x_0-h)}{12}
=h→0lim?12f′′(x0?+h)?f′′(x0??h)?
当
h
→
0
h\to0
h→0时，分子就是
f
′
′
(
x
0
)
+
f
′
′
(
x
0
)
=
2
f
′
′
(
x
0
)
f^{\prime\prime}(x_0)+f^{\prime\prime}(x_0)=2f^{\prime\prime}(x_0)
f′′(x0?)+f′′(x0?)=2f′′(x0?)
所以
=
lim
?
h
→
0
f
′
′
(
x
0
+
h
)
+
f
′
′
(
x
0
?
h
)
12
=
f
′
′
(
x
0
)
6
=\lim_{h\to0} \frac{f^{\prime\prime}(x_0+h)+f^{\prime\prime}(x_0-h)}{12}=\frac{f^{\prime\prime}(x_0)}{6}
=h→0lim?12f′′(x0?+h)+f′′(x0??h)?=6f′′(x0?)?
所以
f
′
′
(
x
0
)
6
≥
0
\frac{f^{\prime\prime}(x_0)}{6}\geq0
6f′′(x0?)?≥0，由极限保号性，
f
′
′
(
x
0
)
≥
0
f^{\prime\prime}(x_0)\geq0
f′′(x0?)≥0. 本题得证
