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Let $f$ be a continuous nonincreasing function on $[0,1]$ with $f(1)=0$ and $\int_0^1 f(x)dx=1$. Does there exist a constant $k$ for which we can always draw a rectangle with area at least $k$, with sides parallel to the axes, in the area bounded by the two axes and the curve $f$?

If we choose the bottom-left corner to be at the origin and the bottom-right corner at $x$, the height of the rectangle is $f(x)$. So we want to maximize $xf(x)$, which amounts to choosing $x=-f(x)/f'(x)$. But it is hard to lower-bound this quantity for an arbitrary $f$.

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Take any $0<\epsilon<1$ and define: $$ f(x)=\cases{ \epsilon/t-\epsilon, & if $0\le x<t$;\\ \epsilon/x-\epsilon,& if $t\le x\le1$,\\ } $$ where $t=e^{-1/\epsilon}$ is chosen so that $\int_0^1 f(x)\,dx=1$.

$f$ satisfies your requirements, but any rectangle having bottom-left corner at the origin and bottom-right corner at $(x, f(x))$ has area $\epsilon$ at most.

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  • $\begingroup$ You seem to have $f(1)=\epsilon$ rather than $0$, though this will not make much difference $\endgroup$ – Henry Jun 20 '16 at 17:31
  • $\begingroup$ Ouch... you are right! I should adjust that. $\endgroup$ – Aretino Jun 20 '16 at 17:35
  • $\begingroup$ I've updated my answer (thanks to Henry!). Hope it works now. $\endgroup$ – Aretino Jun 20 '16 at 17:44
  • $\begingroup$ This function is not continuous at $t$, is it? $\endgroup$ – pi66 Jun 20 '16 at 21:00
  • $\begingroup$ @pi66 Right: I fixed it now. $\endgroup$ – Aretino Jun 20 '16 at 21:25
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For sure, $f(x)=x$ for some $\alpha\in(0,1)$. Now we may prove that $\alpha$ cannot be too small. We have: $$ 1=\int_{0}^{1}f(x)\,dx = \alpha^2+\int_{\alpha}^{1}f(x)\,dx+\int_{\alpha}^{f(0)}f^{-1}(x)\,dx\leq \alpha^2+\alpha(1+f(0)-2\alpha) $$ hence $$ \alpha \geq \frac{1+f(0)-\sqrt{(1+f(0))^2-4}}{2} $$ where $f(0)\geq 1$. However, if $f(0)$ is unbounded, that inequality is ineffective.

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