### 证明

“$\Rightarrow$” 令$F(x)=(x - a)f\left(\frac{a + x}{2}\right)-\int_{a}^{x}f(t)dt$，则$F(a)=0$.

$F'(x)=f\left(\frac{a + x}{2}\right)+\frac{1}{2}(x - a)f'\left(\frac{a + x}{2}\right)-f(x)$

\[

\begin{aligned}

&= \frac{1}{2}(x - a)f'\left(\frac{a + x}{2}\right) + f\left(\frac{a + x}{2}\right) - f(x) \\

&= \frac{1}{2}(x - a)f'\left(\frac{a + x}{2}\right) - f'(\xi)\frac{1}{2}(x - a) \\

&= \frac{1}{2}(x - a)\left[f'\left(\frac{a + x}{2}\right) - f'(\xi)\right]

\end{aligned}

\]

由于$f''(x)\geq0$，所以$f'(x)$单增，从而$f'\left(\frac{a + x}{2}\right)<f'(\xi)$，故$F'(x)<0$，$F(x)$单 调递减.

$x > a$，$F(x)<0$，则$F(b)<0$，及$f\left(\frac{a + b}{2}\right)\leq\frac{1}{b - a}\int_{a}^{b}f(x)dx$.

“$\Leftarrow$” $\forall x_{0}\in(-\infty,+\infty)$，取$a = x_{0}-h$，$b = x_{0}+h$，其中$h > 0$，则

\[ \begin{aligned} f\left(\frac{a + b}{2}\right) \leq \frac{1}{b - a}\int_{a}^{b}f(x)dx &\Leftrightarrow \frac{\int_{x_{0}-h}^{x_{0}+h}f(x)dx - 2f(x_{0})h}{2h} \geq 0, \end{aligned} \]

从而 ### 证明

“$\Rightarrow$” 令$F(x)=(x - a)f\left(\frac{a + x}{2}\right)-\int_{a}^{x}f(t)dt$，则$F(a)=0$.

$F'(x)=f\left(\frac{a + x}{2}\right)+\frac{1}{2}(x - a)f'\left(\frac{a + x}{2}\right)-f(x)$

\[

\begin{aligned}

&= \frac{1}{2}(x - a)f'\left(\frac{a + x}{2}\right) + f\left(\frac{a + x}{2}\right) - f(x) \\

&= \frac{1}{2}(x - a)f'\left(\frac{a + x}{2}\right) - f'(\xi)\frac{1}{2}(x - a) \\

&= \frac{1}{2}(x - a)\left[f'\left(\frac{a + x}{2}\right) - f'(\xi)\right]

\end{aligned}

\]

由于$f''(x)\geq0$，所以$f'(x)$单增，从而$f'\left(\frac{a + x}{2}\right)<f'(\xi)$，故$F'(x)<0$，$F(x)$单 调递减.

$x > a$，$F(x)<0$，则$F(b)<0$，及$f\left(\frac{a + b}{2}\right)\leq\frac{1}{b - a}\int_{a}^{b}f(x)dx$.

“$\Leftarrow$” $\forall x_{0}\in(-\infty,+\infty)$，取$a = x_{0}-h$，$b = x_{0}+h$，其中$h > 0$，则

\[ \begin{aligned} f\left(\frac{a + b}{2}\right) \leq \frac{1}{b - a}\int_{a}^{b}f(x)dx &\Leftrightarrow \frac{\int_{x_{0}-h}^{x_{0}+h}f(x)dx - 2f(x_{0})h}{2h} \geq 0, \end{aligned} \]

从而

 \[ \frac{\int_{x_{0}-h}^{x_{0}+h}f(x)dx - 2f(x_{0})h}{2h^{3}}\geq0. \] 由 \[ \begin{aligned} \lim_{h \to 0}\frac{\int_{x_{0}-h}^{x_{0}+h}f(x)dx - 2f(x_{0})h}{2h^{3}} &= \lim_{h \to 0}\frac{f(x_{0}+h)+f(x_{0}-h)-2f(x_{0})}{6h^{2}} \\ &= \lim_{h \to 0}\frac{f'(x_{0}+h)-f'(x_{0}-h)}{12h} \\ &= \lim_{h \to 0}\frac{f''(x_{0}+h)+f''(x_{0}-h)}{12} \\ &= \frac{1}{6}f''(x_{0}), \end{aligned} \] 同时由极限的保号性知$f''(x_{0})\geq0$.