A proof for deducing Lypaunov's inequality seems to be centered on showing that if $f$ is convex on $0< a \le x \le b$, then $f(x)/x$ must be non-decreasing on the interval. I am not aware that $f$ needs to be differentiable.
Starting from a definition of convexity, I can write $$ f(x) \le f(a) + \frac{(x-a)[f(b) - f(a)]}{b-a} \implies \frac{f(x)}{x} \le \frac{f(a)}{a} + \frac{(x-a)[f(b) - f(a)]}{x(b-a)}. $$ We can define $g(x) = f(x)/x$ to obtain $$ \frac{g(x) - g(a)}{x-a} \le \frac{f(b)-f(a)}{x(b-a)}. $$ None of this seems to be heading in the right direction, what am I doing wrong? Is there another requirement on the function $f$?
Without extra assumption, it is false. Let $f:(1,3)\rightarrow \mathbb{R}$ be defined as $f(x) = |x-2|$. Clearly, $f$ is convex. However, $x\mapsto f(x)/x$ is not non-decreasing. Denote $g(x) = f(x)/x$. Obviously $g(1) = 1$ but $g(2) =0$.