Suppose $f(x)$ has continuous second-order derivative over $(a,+\infty)$, and $f(x)>0$,$f''(x)\leq 0$. Prove $f'(x)\geq 0$.
Proof
Consider proving by contradiction. If the conclusion does not hold, then $$ \exists x_0 \in (a,+\infty):f'(x_0)<0.$$ Since $f''(x)\leq 0$, $f'(x)$ is nonincreasing. Therefore $$\forall x \in [x_0,+\infty):f'(x)\leq f'(x_0)<0.$$ One can claim that $f'(x)$ has a limit as $x \to +\infty$, which is either a finite negtive number or the negative infinity.Thus, by L'Hôpital's rule, $$\lim_{x \to +\infty}\frac{f(x)}{x}=\lim_{x \to +\infty}f'(x)<0,$$ which contradicts, just noticing that $f(x)/x$ is positive with a sufficiently large $x$, according to the assumption condition.
Please correct me if I'm wrong!
Your proof is OK, but you could also argue without using L'Hôpital's rule: For all $x > x_0$ $$ \tag{*} f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac 12 f''(c) (x-x_0)^2 $$ with some $c \in (x_0, x)$, according to Taylor's theorem. If $f'(x_0) < 0$ then it follows that $$ 0 < f(x) \le f(x_0) + \underbrace{f'(x_0)}_{< 0}\cdot (x-x_0) $$ This is a contradiction since the right-hand side is negative for sufficiently large $x > x_0$.
In other words: The graph of a concave function lies below the tangent at the graph at any point. If there is a tangent with negative slope then $f(x)$ is negative sufficiently large $x$.
It is sufficient to require the second derivative exists and is negative. The continuity of $f''$ is not needed for the proof.
Or as a direct proof, without contradiction: From $(*)$ it follows that for all $x > x_0$ $$ 0 < f(x) \le f(x_0) + f'(x_0) (x-x_0) \\ \implies f'(x_0) \ge -\frac{f(x_0)}{x-x_0} \\ \implies f'(x_0) \ge \lim_{x \to \infty} -\frac{f(x_0)}{x-x_0} = 0 \, . $$