Does this bound hold beyond the domain where the function is convex? Let  $F:(0,\infty) \to [0,\infty)$ be a continuous  function satisfying $F(1)=0$, which is strictly increasing on $[1,\infty)$, and strictly decreasing on $(0,1]$.
Suppose also that $F|_{(1-\epsilon,1+\epsilon)}$ is convex and smooth for some $\epsilon>0$.
Choose some $\delta \in (0,1)$, such that $F$ is convex at every point $y \in (\delta,1)$, where by convexity at a point $y$, I mean that that for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =y$, we have
$$
F(y)=F\left(\alpha x_1 + (1- \alpha)x_2 \right) \leq \alpha F(x_1) + (1-\alpha)F(x_2). \tag{1}
$$
Such a $\delta$ always exists.

Question: Let $X$ be a probability space and let $g:X \to (0,\infty)$ be measurable. Suppose that $\int_X g < \delta$. Is it true that $\int_X F \circ g \ge F(\delta)$?

If $F$ were convex at the point $\int_X g$, then by Jensen inequality, we would have
$$
\int_X F \circ g \ge F(\int_X g) \ge F(\delta),
$$
where in the last step, we have used the fact that
$$
0<\int_X g \le \delta<1
$$
together with the fact that $f$ is decreasing on $(0,1]$.
Since $F$ does not need to be convex at $\int_X g$, I suspect that the answer can be negative in general.
 A: If $F$ is convex at a point $y$, then $F$ is bounded below by some tangent line $T_{y}$, so
$$\int_X F \circ g \ge \int_X T_{\delta} \circ g =T_{\delta}(\int_X g) > T_{\delta}(\delta) = F(\delta)$$
Strictness comes from the fact that $T_{\delta}' < 0$ in your example since $F(\delta) > F(1) = 0$.


Here's how to construct such a tangent line $T_{y}$.
If $F$ is convex at a point $y$, then for any $h_{1}, h_{2}>0$:
$$\frac{F(y)-F(y-h_{1})}{h_{1}}\leq \frac{F(y+h_{2})-F(y)}{h_{2}}$$
$$\Rightarrow b_{y} \equiv \sup_{h_{1}>0} \frac{F(y)-F(y-h_{1})}{h_{1}}\leq \inf_{h_{2}>0} \frac{F(y+h_{2})-F(y)}{h_{2}} \equiv c_{y}$$
Choose $T_{y}$ to be any line equal to $F$ at point $y$ with a slope between $b_{y}$ and $c_{y}$.

We can show this construction bounds $F$ from below.
Take any $x>y$, then:
$$F(x) \geq F(y) + c_{y}(x-y) \geq T_{y}(x)$$
Take any $x<y$, then:
$$F(x) \geq F(y) + b_{y}(x-y) \geq T_{y}(x)$$
(Since $x-y$ is negative in the last line and the slope of $T_{y}$ is greater than $b_{y}$.)


You only said $F$ is convex at points in $(\delta,1)$, but I argue this implies $F$ is convex at the point $\delta$.
Take any $x_1>x_2>0, \alpha \in [0,1]$, satisfying $\alpha x_1 + (1- \alpha)x_2 = \delta$. Then, for any $\hat{\alpha} > \alpha$ sufficiently close, $\hat{\alpha} x_1 + (1- \hat{\alpha})x_2 \in (\delta,1)$, so:
$$\hat{\alpha} F(x_1) + (1- \hat{\alpha})F(x_2) \geq F(\hat{\alpha} x_1 + (1- \hat{\alpha})x_2)$$
By the continuity of $F$, taking the limit as $\hat{\alpha} \rightarrow \alpha$, this implies:
$$\alpha F(x_1) + (1- \alpha)F(x_2) \geq F(\alpha x_1 + (1- \alpha)x_2)$$
