Let $f$ be a differentiable real function on $[1, +\infty)$ such that $f^\prime$ is convex.
Now, what can be said about $f$ ?
I know that, if $f$ is differentiable on $(a, b)$, then $f$ is convex if and only if $f^\prime$ is increasing.
Also, if $f^{\prime\prime}$ exists on $(a, b)$, then $f$ is convex if and only if $f^{\prime\prime}\geq 0$.
If $f'$ is convex on $\left]1,+\infty\right[$, then the running average $$g\colon \left]1,+\infty \right[ \to \mathbb{R} : t \mapsto \frac{1}{t-1}\int_{1}^{t}f'(x)dx$$ is convex. (Hint: you may use the technique in Running average of a convex function is convex to prove this). Since, for every $t >1$, we have $g(t) = (f(t)-f(1))/(t-1)$, one may conclude that $$\frac{f(t)-f(1)}{t-1} \text{ is convex on } \left]1,+\infty\right[.$$ Moreover, if $f(1)>0$, then $f(1)/(t-1)$ is convex (second-order test), and thus, you get $f(t)/(t-1)$ is convex.