2022 number two real problem

Thought analysis ：

See the problem of proving integral inequalities , First think of constructors , So what kind of function to construct ？ What is the title to prove , We just build something , We can construct $F(x)=(x-a)f(\frac{a+x}{2})-{\int }_a^{x}f(t)dt$

prove ：

The need for ： Make $F(x)=(x-a)f(\frac{a+x}{2})-{\int }_a^{x}f(t)dt$ , Obviously F(a)=0,
$F^{\prime }(x)=f(\frac{a+x}{2})+\frac{1}{2}(x-a)f^{\prime }(\frac{a+x}{2})-f(x)$
$$=\frac{1}{2}(x-a)\acute{f}(\frac{a+x}{2})-[f(x)-f(\frac{a+x}{2})]$$$$=\frac{1}{2}(x-a)f^{\prime }(\frac{a+x}{2})-\frac{1}{2}(x-a)\acute{f(\xi )}$$,$\xi \in (\frac{a+x}{2},x)$$$=\frac{1}{2}(x-a)[\acute{f}(\frac{a+x}{2})-f^{\prime }(\xi ))]$$
because $f^{\prime \prime }(x)>0,f^{\prime }(\frac{a+x}{2})-f^{\prime }(\xi )<0$, so $F^{\prime }(x)<0$, therefore $F(b)\le F(a)=0$, therefore $F(b)=(b-a)f(\frac{a+b}{2})-{\int }_a^{b}f(t)dt\le 0,namelyf(\frac{a+b}{2})<\frac{1}{b-a}{\int }_a^{b}f(t)dt$

adequacy ：
$\forall x_0\in (-\infty ,+\infty )$, take $a=x_0-h,b=x_0+h$, among $h>0$, Then our formula $f(\frac{a+b}{2})\le \frac{1}{b-a}{\int }_a^{b}f(t)dt$ It can be transformed into $f(x_0)\le \frac{1}{2h}{\int }_{x_0-h}^{x_0+h}f(x)dx\iff \frac{{\int }_{x_0-h}^{x_0+h}f(x)dx-2hf(x_0)}{2h}\ge 0$,

To prove... In the title $f^{\prime \prime }(x)\ge 0,$ We can only think of 2 Kind of way
① Start with the definition of derivative , Find a way to construct the definition of derivative
② Using Taylor's formula ( Because there are first-order derivatives and second-order derivatives in Taylor's formula )
③ The law of lophida （ Keep the numerator and denominator rolling down , There must be higher derivatives ）
For this question , We have introduced neighborhood $(x_0-h,x_0+h)$, Because to prove $f^{\prime \prime }(x)$, So let's assume a and b Very, very close , namely ${\lim }_{h\to 0}$, For this case of finding the limit , We first think of the law of lobida , Therefore, for the formula $\frac{{\int }_{x_0-h}^{x_0+h}f(x)dx-2hf(x_0)}{2h}\ge 0$, We can find a way to solve the problem with lobida's law , For molecules , It has to be done 3 The second derivative can appear $f^{\prime \prime }(x)$ The second derivative , And the denominator can't stand 3 Subderivative , Then what shall I do? , We increase the power of the denominator to 3 rank $\frac{{\int }_{x_0-h}^{x_0+h}f(x)dx-2hf(x_0)}{2h^3}\ge 0$, Then take the derivative up and down
${\lim }_{h\to 0}\frac{{\int }_{x_0-h}^{x_0+h}f(x)dx-2hf(x_0)}{2h^3}$
$$=\underset{h\to 0}{\lim }\frac{f(x_0+h)-f(x_0-h)-2f(x_0)}{6h^2}$$$$=\underset{h\to 0}{\lim }\frac{f^{\prime }(x_0+h)+f^{\prime }(x_0-h)}{12h}$$$$=\underset{h\to 0}{\lim }\frac{f^{\prime \prime }(x_0+h)-f^{\prime \prime }(x_0-h)}{12}$$
When $h\to 0$ when , Molecules are $f^{\prime \prime }(x_0)+f^{\prime \prime }(x_0)=2f^{\prime \prime }(x_0)$
therefore $$=\underset{h\to 0}{\lim }\frac{f^{\prime \prime }(x_0+h)+f^{\prime \prime }(x_0-h)}{12}=\frac{f^{\prime \prime }(x_0)}{6}$$
therefore $\frac{f^{\prime \prime }(x_0)}{6}\ge 0$, By limit number preservation ,$f^{\prime \prime }(x_0)\ge 0$. This question is proved