This question is related to this recent other question, where two intervals $[a,b] $ and $[c,d]$ were considered. Here I ask about a simpler version with just one interval $[a,b]$.
Consider the following optimization problem : $f$ is a positive continuous function $[a,b] \to ]0,+\infty[$, satisfying the Lipschitz condition $|f(x)-f(y)| \leq L|x-y|$ for any $x,y \in [a,b]$. Given the constraint $\int_{a}^{b}\frac{dt}{f(t)}=\alpha$ (where $\alpha$ is a positive constant), the problem is then to find the maximum (or supremum) value $M$ of $\int_{a}^{b}f(t)dt$ under this constraint, and find the functions for which this maximum is attained, if any.
Although my evidence for this is still incomplete, I believe the maximum is attained when $f$ is a decreasing affine function with coefficient $-L$. In this case,
$ f(t)=L\bigg(b-t+\frac{b-a}{e^{L\alpha}-1}\bigg), \ \int_a^{b} f(t)dt= L(b-a)^2\bigg(\frac{1}{2}+\frac{1}{e^{L\alpha}-1}\bigg) $
By the Stone-Weierstrass theorem, when looking for the maximum we may assume that $f$ is differentiable (we may even assume that $f$ is a polynomial). In this case, one may apply the methods explained in the abovementioned post : $|f'| \leq L$, so $|(f^2)'| =|2ff'| \leq 2L|f|$ and hence $f^2(x) \leq f^2(a)+2LF(x)$, where we put $F(x)=\int_a^x f(t)dt$. So $f(x)=\frac{f^2(x)}{f(x)} \leq \frac{ f^2(a)+2LF(x)}{f(x)}$. Putting $G(x)=f^2(a)+2LF(x)$, we deduce $\frac{G'(x)}{G(x)} \leq \frac{2L}{f(x)}$. integrating between $a$ and $b$, we obtain ${\sf ln}\big(\frac{G(b)}{G(a)}\big) \leq 2L\alpha$. Now $G(a)=f^2(a)$ and $G(b)=f^2(a)+2L\int_a^b f(t)dt$, so that one finally obtains
$ \int_a^b f(t)dt \leq \frac{e^{2L\alpha}-1}{2L}f^2(a) $
Any feedback appreciated on the following questions : what is the maximum/supremum, which functions attain equality.