I made the following claim, which I now see that I don't know how to prove. Can anyone prove it?

Claim. Let $f$ be a concave and non-negative function on $[0,1]$ with $$\int_0^1 f = 1,$$ $$\int_0^1 |f-1| > 2\varepsilon.$$ Then $$\min\left(\int_0^{1/4} f , \int_{3/4}^1 f\right) < \frac14 - \frac\varepsilon8.$$

In principle, this claim can be settled via computer algebra. For any $f$ satisfying the conditions, there is a piecewise-linear function $g$ of 10 pieces which satisfies the same conditions and has the same integrals as $f$ and $|f-1|$ on $[0,\frac14]$, $[\frac14,\frac34]$ and $[\frac34,1]$. The existence of such $g$ is just a matter of the existence of 9 $x$'s, 11 $y$'s, and 1 $\varepsilon$ satisfying some multilinear equalities and inequalities, and therefore is algorithmically solvable. In practice, my simplifications in Mathematica get overwhelmed before they finish.

Does anyone see a proof or a counterexample? I'd be happy to see a proof with $8$ replaced by any other small number.

Update: Distributions like $f(x)=\min(cx,c(1-x),\frac12(c-\sqrt{c^2-4c}))$ may provide a good source of extreme examples; for high $c$, they satisfy only $\min(\int_0^{1/4}f,\int_{3/4}^1f)<\frac14-\frac\epsilon4$.

I made the following claim, which I now see that I don't know how to prove. Can anyone prove it?

Claim. Let $f$ be a concave and non-negative function on $[0,1]$ with $$\int_0^1 f = 1,$$ $$\int_0^1 |f-1| > 2\varepsilon.$$ Then $$\min\left(\int_0^{1/4} f , \int_{3/4}^1 f\right) < \frac14 - \frac\varepsilon8.$$

In principle, this claim can be settled via computer algebra. For any $f$ satisfying the conditions, there is a piecewise-linear function $g$ of 10 pieces which satisfies the same conditions and has the same integrals as $f$ and $|f-1|$ on $[0,\frac14]$, $[\frac14,\frac34]$ and $[\frac34,1]$. The existence of such $g$ is just a matter of the existence of 9 $x$'s, 11 $y$'s, and 1 $\varepsilon$ satisfying some multilinear equalities and inequalities, and therefore is algorithmically solvable. In practice, my simplifications in Mathematica get overwhelmed before they finish.

Does anyone see a proof or a counterexample? I'd be happy to see a proof with $8$ replaced by any other small number.

Update: Distributions like $f(x)=\min(cx,c(1-x),\frac12(c-\sqrt{c^2-4c}))$ may provide extreme examples. The graph below shows $c=24$, with a maximum of $12-\sqrt{120}$, and a bound of $\frac14-\varepsilon/3.66$. For $f$ of this form with high $c$, the bound is $\min(\int_0^{1/4}f,\int_{3/4}^1f)<\frac14-\varepsilon/4$.

I made the following claim, which I now see that I don't know how to prove. Can anyone prove it?

Claim. Let $f$ be a concave and non-negative function on $[0,1]$ with $$\int_0^1 f = 1,$$ $$\int_0^1 |f-1| > 2\varepsilon.$$ Then $$\min\left(\int_0^{1/4} f , \int_{3/4}^1 f\right) < \frac14 - \frac\varepsilon8.$$

In principle, this claim can be settled via computer algebra. For any $f$ satisfying the conditions, there is a piecewise-linear function $g$ of 10 pieces which satisfies the same conditions and has the same integrals as $f$ and $|f-1|$ on $[0,\frac14]$, $[\frac14,\frac34]$ and $[\frac34,1]$. The existence of such $g$ is just a matter of the existence of 9 $x$'s, 11 $y$'s, and 1 $\varepsilon$ satisfying some multilinear equalities and inequalities, and therefore is algorithmically solvable. In practice, my simplifications in Mathematica get overwhelmed before they finish.

Does anyone see a proof or a counterexample? I'd be happy to see a proof with $8$ replaced by any other small number.

I made the following claim, which I now see that I don't know how to prove. Can anyone prove it?

Claim. Let $f$ be a concave and non-negative function on $[0,1]$ with $$\int_0^1 f = 1,$$ $$\int_0^1 |f-1| > 2\varepsilon.$$ Then $$\min\left(\int_0^{1/4} f , \int_{3/4}^1 f\right) < \frac14 - \frac\varepsilon8.$$

In principle, this claim can be settled via computer algebra. For any $f$ satisfying the conditions, there is a piecewise-linear function $g$ of 10 pieces which satisfies the same conditions and has the same integrals as $f$ and $|f-1|$ on $[0,\frac14]$, $[\frac14,\frac34]$ and $[\frac34,1]$. The existence of such $g$ is just a matter of the existence of 9 $x$'s, 11 $y$'s, and 1 $\varepsilon$ satisfying some multilinear equalities and inequalities, and therefore is algorithmically solvable. In practice, my simplifications in Mathematica get overwhelmed before they finish.

Does anyone see a proof or a counterexample? I'd be happy to see a proof with $8$ replaced by any other small number.

Update: Distributions like $f(x)=\min(cx,c(1-x),\frac12(c-\sqrt{c^2-4c}))$ may provide a good source of extreme examples; for high $c$, they satisfy only $\min(\int_0^{1/4}f,\int_{3/4}^1f)<\frac14-\frac\epsilon4$.