Let $f\_s,f\_l:(0,\infty) \mapsto {\mathbb R}$ be monotonic increasing functions with $f\_s(x) \leq f\_l(x)$ and let $x\_s,x\_l$ be the smallest positive roots of the equation $1-f\_{s,l}(x)=0$ respectively (assuming they exist, are real and no double roots). Then I believe the following is true
$$ \int\_0^{x\_s} \frac{{\rm d}x}{\sqrt{1-f\_s(x)}} \geq \int\_0^{x\_l} \frac{{\rm d}x}{\sqrt{1-f\_l(x)}}$$
and if it is true I would like to proof it.
I just don‘t know where to start.

Heuristically I think it is because the increase in integration interval $x\_s \geq x\_l$ outweights the decrease of the integrand values $$\frac{{\rm d}x}{\sqrt{1-f\_s(x)}} \leq \frac{{\rm d}x}{\sqrt{1-f\_l(x)}} \, .$$

Observe for small deviations $f\_l = f\_s + \delta f$ we have to first order for the RHS
$$ 
\int\_0^{x\_l} \frac{{\rm d}x}{\sqrt{1-f\_s(x)}} \left( 1 + \frac{\delta f(x)}{2(1-f\_s(x))} \right)
$$
and so this requires
$$
\int\_{x\_l}^{x\_s} \frac{{\rm d}x}{\sqrt{1-f\_s(x)}} \geq \int\_0^{x\_l} \frac{{\rm d}x}{\sqrt{1-f\_s(x)}} \left( \frac{\delta f(x)}{2(1-f\_s(x))} \right) \, .
$$
Any suggestions?

**edit:** You are right! I was a bit optimistic and also found a counterexample: Choosing $f\_l(x)$ to become very close to $1$ very quickly and then only increasing slowly for a large interval makes the $f\_l$ integral very large. On the other hand one can choose $f\_s(x)$ negative and large enough, so that the integral for $f\_s$ is very small. $f\_s$ then increases just before $x\_l$ to unity at $x\_s=x\_l$. This contribution should be small in comparison when the interval is large enough.