Does this strong convexity estimate hold?

If suffices to require that $F$ is strictly convex and differentiable on an interval $I \subset \Bbb R$. (Even the differentiability requirement can be dropped, see the remarks at the end of the answer.)

For $a, b \in I$ with $a < b$ and $c = \lambda a + (1 - \lambda) b$ with $0 \le \lambda \le 1$ we can write $$ D(a, b, c) = \lambda F(a)+(1-\lambda)F(b)-F(c) \\ = \lambda \bigl \{ F(a) - F(c) - (a-c)F'(c) \bigr\} + (1- \lambda) \bigl \{F(b) - F(c) - (b-c)F'(c)\bigr\} \, . $$

This suggests to introduce $$ H(u, v) = F(u) - F(v) - (u-v) F'(v) $$ for $u, v \in I$. $H$ has the following properties:

1. $H(u, v) > 0$ if $u \ne v$.
2. $H(u_1, v) > H(u_2, v)$ if $u_1 < u_2 \le v$, i.e. $H(u,v)$ is decreasing in $u$ as long as $u \le v$.
3. $H(u, v_1) < H(u, v_2)$ if $u \le v_1 < v_2$, i.e. $H(u, v)$ is increasing in $v$ as long as $u \le v$.

Property (1) is a direct consequence of the strict convexity: $F(u)$ is larger than the corresponding value of the tangent line at $x=v$.

For property (2) we assume $u_1 < u_2 \le v$ and compute $$ H(u_1, v) - H(u_2, v) = F(u_1) - F(u_2) - (u_1 - u_2) F'(v) \\ \ge F(u_1) - F(u_2) - (u_1 - u_2) F'(u_2) = H(u_1, u_2) > 0 \, . $$ Here we used that $F'$ is increasing.

For property (3) we assume $u \le v_1 < v_2$ and compute $$ H(u,v_1) - H(u, v_2) = -F(v_1) - (u-v_1)F'(v_1) + F(v_2) + (u-v_2) F'(v_2) \\ \le -F(v_1) - (u-v_1)F'(v_2) + F(v_2) + (u-v_2) F'(v_2) \\ = -H(v_1, v_2) < 0 \, . $$

With these tools, estimating $D(a, b, c)$ from below becomes easy. If $a \le r_0 < r_1 \le c < b$ then $$ D(a, b, c) = \lambda H(a, c) + (1-\lambda)H(b,c) \\ \ge \lambda H(a, c) \ge \lambda H(r_0, r_1) \ge \lambda(1- \lambda) H(r_0, r_1) \\ = m \lambda(1-\lambda) (r_1-r_0)^2 $$ with $m$ defined as $$ m = \frac{H(r_0, r_1)}{(r_1-r_0)^2} = \frac{F(r_0) - F(r_1) - (r_0 - r_1) F'(r_1)}{(r_1-r_0)^2} > 0 \, . $$

Remarks:

• The assumption that $F$ is only defined on $[0, \infty)$ with values in $[0, \infty)$ was not used in the proof.
• The differentiability requirement can also be dropped. A convex function has one-sided derivatives in every inner point of the interval. The above proof still works if we replace $F'$ by the right (or left) derivative.