Generalization of convexity This is not a homework, this is just something that came to my mind recently. Assume $f$ is a sufficiently nice function.
We know that
$$\frac{df}{dx} \geq 0 \iff f(x_2) \geq f(x_1) \text{ for } x_2 \geq x_1$$
$$\frac{d^2f}{dx^2} \geq 0 \iff \frac{f(x_1) + f(x_2)}{2} \geq f(\frac{x_1 + x_2}{2})$$
Is there any nice way to generalize this for for higher derivatives? A generalization of nonnegativity of higher order derivatives being equivalent to some short nice condition not involving any derivatives at all? If not generalization, is there at least a nice extension to $\frac{d^3f}{dx^3}$?

I think we can play with the Schwarzian derivative (we work on $[0;\infty[$):
We have :
$$(Sf)(x)=\frac{f'''(x)}{f'(x)}-1.5(\frac{f''(x)}{f'(x)})^2$$
Here we assume that the Schwarzian is positive so :
$$\frac{f'''(x)}{f'(x)}\geq1.5(\frac{f''(x)}{f'(x)})^2$$
We can rewrite the inequaliy like this :
$$\frac{f'''(x)}{f''(x)}\geq1.5(\frac{f''(x)}{f'(x)})$$
We integrate between $0$ and $x$ we have (always) :
$$ln(|f''(x)|)\leq 1.5ln(|f'(x)|)$$
We take the exponential  we get :
$$|f''(x)|\geq (|f'(x)|)^{1.5}$$
We can rewrite the inequality like this :
$$\frac{|f''(x)|}{\sqrt{|f'(x)|}}\geq(|f'(x)|)$$
We integrate it give :
$$2\sqrt{|f'(x)|}\geq |f(x)|$$
Or 
$$4|f'(x)|\geq f(x)^2$$
Or :
$$\frac{4|f'(x)|}{f(x)^2}\geq 1$$
Now we assume that the derivative is always positive we get :
$$\frac{4 f'(x)}{f(x)^2}\geq 1$$
Or :
$$ 4f'(x)\geq f(x)^2$$
And it implies convexity of the function $f(x)$ if you solve the inequality (see the lemma of Gronwall)
