Consider on $(0, 1]$ the function $f$ defined by $$f(w)=\inf_{\overset{uv=w\leq1}{u>0,~0<v\leq\frac{1}{2}}}(g(u)+h(v)),$$ where $$g(u)= \left\{ \begin{array}{lc} (\log u)^{\delta},& u\geq 2, \\cu^{-1},& u<2\end{array}\right.$$ where $\delta<0$ and $c$ is a constant chosen so that $g$ is continuous, and $$h(v)=v^{p-1},\quad 0\leq v\leq \frac{1}{2},$$ where $p\in (0, 1)$. Note that both $g$ and $h$ are monotone decreasing.
Is there a simple way to get a lower bound for $f$ (up to constant)?
I am looking for a sharper lower bound than just a constant, that means that $f(w)\to \infty$ as $w\to 0$.
Is it true that the infimum is attained when $f$ and $g$ are comparable?
Thanks in advance!
The problem reduces to finding
$$\inf_{\substack{0<w\leq 1\\0<v\leq \frac{1}{2}}}(\frac{c v}{w}+v^{p-1})= \inf_{0<v\leq \frac{1}{2}}(c v+v^{p-1})$$