If I know that for every $y>x>0$, $f(y)-f(x)\geq (y-x)x$ - does that mean the $f$ is not uniformy continuous in $(0,\infty)$? If it's true, how can I prove that?
I know that the definition of uniform continuity is that for every $\epsilon>0$ there exists $\delta>0$ for which every $x,y$, if $|y-x|<\delta \implies |f(y)-f(x)|<\epsilon$.
Though I'm having difficulties finding the connection to the definition and how to use it in this case.
Assume $f$ is uniformly continuous and that $\delta$ "works" for some $\epsilon$. Then for all $x$, we have $$\epsilon > |f(x+\tfrac\delta2)-f(x)|\ge f(x+\tfrac\delta2)-f(x)\ge \tfrac\delta2\cdot x$$ This clearly leads to a contradiction for $x=\frac{2\epsilon}\delta$.