The inequality you wrote down is a special case of a general principle about higher-order convexity: any "simple enough" linear inequality of the type you wrote down will be true as long as it is true for every polynomial function which satisfies your convexity condition. Furthermore, these simple inequalities can always be proved directly by a divided difference computation.
To make this claim precise, I have to define higher-order convexity, and clarify what I mean by "simple enough". I'll start by reviewing the elementary theory of divided differences. Divided differences are defined inductively by the rules
- $[a;f] = f(a)$,
- $[a,b;f] = \frac{f(a) - f(b)}{a-b}$,
- $[a_0, ..., a_n, a_{n+1}; f] = \frac{[a_0, ..., a_{n-1}, a_n; f] - [a_0, ..., a_{n-1}, a_{n+1}; f]}{a_n - a_{n+1}}$ if $a_n \ne a_{n+1}$.
The divided difference $[a_0, ..., a_n; f]$$[a_0, \dotsc, a_n; f]$ turns out to be a symmetric function of the points $a_0, ..., a_n$ -$a_0, \dotsc, a_n$ — in fact, we have the explicit formula
$[a_0, ..., a_n; f] = \sum_{i=0}^n \frac{f(a_i)}{\prod_{j\ne i}(a_i - a_j)}$,
which $$ [a_0, \dotsc, a_n; f] = \sum_{i=0}^n \frac{f(a_i)}{\prod_{j\ne i}(a_i - a_j)}, $$ which can easily be proved by induction. If $p(x)$ is a polynomial of degree $n$ with leading coefficient $cx^n$, then the divided difference $[a_0, ..., a_n; p]$$[a_0, \dotsc, a_n; p]$ will be equal to the constant $c$.
A function $f$ is called $(k-1)$th-order convex if every divided difference $[a_0, ..., a_k; f]$$[a_0, \dotsc, a_k; f]$ is positive, so ordinary convexity corresponds to first-order convexity. For $k \ge 3$, something nice happens:
Proposition: If $k \ge 3$, then $f$ is $(k-1)$th-order convex if and only if $f$ is differentiable and $f'$ is $(k-2)$th-order convex.
This fact can be proved directly -— it's a fun exercise. As a consequence, a function $f$ is $(k-1)$th-order convex for $k \ge 2$ if and only if $f \in C^{k-2}$ and $f^{(k-2)}$ is convex. So the functions that showed up in this question in are exactly the third-order convex functions on $[0,1]$.
Now for the sorts of inequalities we'd like to consider: suppose that $a_i, b_i, c_i$$a_i$, $b_i$, $c_i$ are constants with $b_i \in \mathbb{N}$, and consider the functional inequality
$\sum_i c_if^{(b_i)}(a_i) \stackrel{?}{\ge} 0.\tag{$*$}$
Define $$ \sum_i c_if^{(b_i)}(a_i) \stackrel{?}{\ge} 0.\tag{$*$}\label{star} $$ Define the complexity of the inequality ($*$)\eqref{star} to be
$\mathcal{C}(*) = \sum_{a \in \{a_i\}} \max\{b_i+1 \mid a_i = a\}$.
The $$ \mathcal{C}(*) = \sum_{a \in \{a_i\}} \max\{b_i+1 \mid a_i = a\}. $$ The general fact is:
Theorem: If $\mathcal{C}(*) \le k+1$, then the following are equivalent:
- the inequality ($*$)\eqref{star} is true for all $(k-1)$th-order convex functions $f$,
- the inequality ($*$) has\eqref{star} is an equality when we plug in $f(x) = x^m$ for $m < k$, and ($*$)\eqref{star} is true when we plug in $f(x) = x^k$,
- there is a constant $C \ge 0$ such that $\sum_i c_if^{(b_i)}(a_i) = C\cdot[\underbrace{a, ..., a,}_{\max\{b_i+1 \mid a_i = a\}\text{ times}} ...; f]$$\sum_i c_if^{(b_i)}(a_i) = C\cdot[\underbrace{a, \dotsc, a,}_{\text{$\max\{b_i+1 \mid a_i = a\}$ times}} \dotsc; f]$, where the divided difference is interpreted as a limit of divided differences at infinitesimally shifted points if any $b_i > 0$.
The proof of this general fact ends up being a straightforward exercise in linear algebra.
Things get a bit more complex once we jump to $\mathcal{C}(*) \le k+2$ -— I used to be very interested in these sorts of inequalities a long time ago, and I have an article articleInequalities and higher order convexity on the arxivarXiv with lots of tricks for verifying these sorts of low complexity functional inequalities.
