I'm sure that this has been here before, but it didn't come up when I entered the title.
This was inspired by Prove that all the tangents of a function do not touch a given point
Here is my proof. I'm hoping for a simpler one.
Suppose $f''(x) > 0$. Show that the tangent to $f$ is below the curve.
Formally.
The tangent at $(a, f(a))$ is $y = f(a)+f'(a)(x-a) $.
Putting in $x=b, y=f(b)$, we want to show that when $b \ne a$ we have $f(b) \gt f(a)+f'(a)(b-a) $ or $f(b)- f(a) \gt f'(a)(b-a) $.
Suppose $b > a$. Then, if $a < x < b$, $f'(x) =f'(a)+\int_a^x f''(t) dt $ so that
$\begin{array}\\ f(b)-f(a) &=\int_a^b f'(x) dx\\ &=\int_a^b (f'(a)+\int_a^x f''(t) dt) dx\\ &=\int_a^b f'(a)dx+\int_a^b \int_a^x f''(t) dt dx\\ &=(b-a) f'(a)+\int_a^b \int_a^x f''(t) dt dx\\ &\gt(b-a) f'(a) \qquad\text{since } \int_a^b \int_a^x f''(t) dt dx > 0 \\ \end{array} $
Similarly, suppose $b < a$.
(This took a little work to get right.)
If $b < x < a$ then
$\begin{array}\\ f'(x) &=f'(b)+\int_b^x f''(t) dt\\ &=f'(b)+\int_b^a f''(t) dt-\int_x^a f''(t) dt\\ &=f'(b)+(f'(a)-f'(b)-\int_x^a f''(t) dt\\ &=f'(a)-\int_x^a f''(t) dt\\ \end{array} $
so that
$\begin{array}\\ f(a)-f(b) &=\int_b^a f'(x) dx\\ &=\int_b^a (f'(a)-\int_x^a f''(t) dt) dx\\ &=\int_b^a f'(a)dx-\int_b^a \int_x^a f''(t) dt dx\\ &=(a-b)f'(a)-\int_b^a \int_x^a f''(t) dt dx\\ &<(a-b)f'(a) \qquad\text{since }\int_b^a \int_x^a f''(t) dt dx > 0\\ \text{or}\\ f(b)-f(a) &>(b-a)f'(a) \end{array} $
Note that this same proof shows that if $f''(x) < 0$ then the tangent is above the curve or $f(b)-f(a) \lt (b-a)f'(a) $.
Let $\varphi(x)=f(x)-f(a)-(x-a)f'(a)$, then $\varphi'(x)=f'(x)-f'(a)$ and $\varphi''(x)=f''(x)>0$, thus $\varphi'$ is a strictly non decreasing function. But $\varphi'(a)=0$, thus $\varphi'(x)>0$ for $x>a$ and $\varphi'(x)<0$ for $x<a$. By the same argument, $\varphi(x)>\varphi(a)=0$ for all $x\neq a$.