Let $\mathcal{F}$ be the set of continuous and strictly increasing functions from $[0,1]$ to $[0,1]$ with $f(0)=0$. Is there a closed form for $$\inf_{f\in\mathcal{F}}\sup_{0\le x\le 1}\frac{(1-x)f(x)}{\int_0^1f(t)\,\mathrm{d}t}?$$
That is, the ratio of the following dark area to the integral area.
Note: this question is similar to this one, but this time I want to minimize the ratio when $f$ varies.
The desired infimum is zero. For $\epsilon\in (0,1/2)$, consider the strictly increasing continuous function $$f(x)=\begin{cases} \frac{\epsilon x}{1-x}& \text{if $x\in [0,1-\epsilon]$},\\ x& \text{if $x\in [1-\epsilon,1]$.} \end{cases}$$ Then $$\int_0^1f(t)\,dt =\frac{\epsilon^2}{2}-\epsilon\log(\epsilon)$$ and $\sup_{x\in [0,1]}(1-x)f(x)=\epsilon(1-\epsilon)$. Hence, as $\epsilon \to 0^+$ $$\frac{\sup_{x\in [0,1]}(1-x)f(x)}{\int_0^1f(t)\,dt }=\frac{\epsilon(1-\epsilon)}{\frac{\epsilon^2}{2}-\epsilon\log(\epsilon)}=\frac{1-\epsilon}{\frac{\epsilon}{2}+\log(1/\epsilon)}\to 0^+.$$