I come to you with yet another qualifying problem we can't seem to solve...
Let $f:$ $(0,\infty) \to \Bbb R$ be convex, and let $\lim_{x \to 0}f(x)=0$. Show that $g(x)$ = $f(x) \over x$ is increasing on $(0,\infty)$.
Tried using the standard definition of convex using $\lambda$=$x\over y$ for $x<y$, to no avail. Tried an assortment of other lambdas, contradiction methods, and got nothing. Any ideas?
Since the function lies above its tangents, we have $$f(x)\ge f(a)+f'(a)(x-a)$$ As $x\to 0$ we get $$0\ge f(a)+-af'(a)\implies af'(a)-f(a)\ge 0$$ And $$\frac{d}{dx}\left({f(x)\over x}\right)=\frac{f'(x)x-f(x)}{x^2}\ge0$$ which establishes that $g$ is increasing (I'm not sure if you should perhaps use strict inequalities instead).