It is known that, given constants $a, b$ with $a \geq 1$ and $-1 < b < 0$, one can construct on the interval $[0, 1]$ a cubic spline
$$ g(x) = c_{0} + c_{1}(x-1) + c_{2}(x-1)^{2} + c_{3}(x-1)^{3} $$
satisfying the boundary conditions
$$ g(0) = 0, \qquad g'(0) = 0, \qquad g(1) = a+b, \qquad g'(1) = b.$$
Essentially, this spline "smoothly connects" the rays
$$ \{ (x, 0) \; : \; x \leq 0 \} \mbox{ and } \{(x, ax + b) \; : \; x \geq 1 \} $$
in the $xy$-plane. Since the rays don't intersect, the spline $g(x)$ (being the configuration of an elastic beam clamped at the ends $(0,0), (1, a+b)$) should be convex. And, solving for the coefficients $c_{k}$ we can verify, brute force, the convexity in the form $g''(x) \geq 0$ on $]0, 1[$. But, is there a mathematically rigorous way that is more elegant? It is ok to "broaden" the problem and allow $g(x)$ to be some other smooth interpolant with the said boundary conditions, just as long as it is convex.