$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$Instead of $\int_0^1|f-1|>2\ep$, let us write 
\begin{equation*}
	\ep:=\frac12\int_0^1|f-1|.
\end{equation*}
We want to show that 
\begin{equation*}
	\min\Big(\int_0^{1/4}f,\int_{3/4}^1 f\Big)\le\frac14-\frac\ep8. \tag{!}
\end{equation*}

Without loss of generality (wlog), $\ep>0$. 
By approximation, wlog the function $f\in C^2[0,1]$ with $f''>0$ on $(0,1)$ and the set $\{x\in[0,1]\colon f(x)\ge1\}$ is an interval $[a,b]$ such that $0<a<b<1$, with $f(a)=f(b)=1$. 

So, 
\begin{equation*}
	\ep=\ep_1+\ep_2,
\end{equation*}
where
\begin{equation*}
	\ep_1:=\int_0^a(1-f),\quad\ep_2:=\int_b^1(1-f).
\end{equation*}
Let 
\begin{equation*}
	\de_1:=\frac14-\int_0^{1/4}f=\int_0^{1/4}(1-f),\quad
	\de_2:=\frac14-\int_{3/4}^1 f=\int_{3/4}^1(1-f). 
\end{equation*}
It suffices to show that 
\begin{equation*}
	\max(\de_1,\de_2)\overset{\text{(?)}}\ge\frac{24}{121}\,\ep, \tag{*}
\end{equation*}
which will improve the constant factor in (!) from $1/8=0.125$ to $\frac{24}{121}=0.19834\dots$. 
Moreover, the factor $\frac{24}{121}=0.19834\dots$ in the lower bound $\frac{24}{121}\,\ep$ in (*) is the best possible.

Our main tool here will be the following lemma. 

>**Lemma 1:** Suppose that $f\in C^2[0,1]$ with $g''>0$ on $(0,1)$ and $g(u)=0$ for some $u\in(0,1)$. Then the function $h\colon[0,1]\to\mathbb R$ defined by 
\begin{equation*}
	h(x):=\frac1{(x-u)^2}\,\int_u^x g
\end{equation*}
for $x\in[0,1]\setminus u$, with $h(u):=g'(u)/2$, is nonincreasing. Here, $\int_u^x:=-\int_x^u$ if $x<u$. 

The proof of this lemma consists in a triple application of the special-case l'Hospital-type rule for monotonicity given by [Proposition 4.1][1]. 

In particular, it follows from Lemma 1 (applied to $g:=f-1$) that 
\begin{equation*}
	b-a\ge1/2. \tag{0}
\end{equation*}
Indeed, suppose (0) is false. Then there are $l_1\in(0,a)$ and $l_2\in(0,1-b)$ such that $l_1+l_2=b-a$ (e.g., let $l_1:=\frac{b-a}{a+1-b}\,a$ and $l_1:=\frac{b-a}{a+1-b}\,(1-b)$). By Lemma 1 (it helps to draw a picture), 
\begin{equation*}
	\int_a^{a+l_1}(f-1)\le\frac{l_1^2}{a^2}\,\int_a^0(f-1)=\frac{l_1^2}{a^2}\,\int_0^a(1-f)
	=\frac{l_1^2}{a^2}\,\ep_1\le\ep_1, \tag{0.5}
\end{equation*}
so that $\int_a^{a+l_1}(f-1)\le\ep_1$, and the latter inequality is strict if $\ep_1>0$. Similarly, $\int_{a+l_1}^b(f-1)=\int_{b-l_2}^b(f-1)\le\ep_2$, and the latter inequality is strict if $\ep_2>0$. So, 
\begin{equation*}
	\ep=\int_a^b(f-1)=\int_a^{a+l_1}(f-1)+\int_{a+l_1}^b(f-1)\le\ep_1+\ep_2=\ep,
\end{equation*}
and the latter inequality is strict, since $\ep>0$ and hence either $\ep_1>0$ and $\ep_2>0$. Thus, we have $\ep<\ep$, a contradiction, which proves (0). 

In view of (0) and by symmetry, it is enough to consider the following two cases:
\begin{equation*}
	0<a\le1/4,\quad 3/4\le b<1 \tag{i}
\end{equation*}
and 
\begin{equation*}
	1/4<a\le1/2,\quad 3/4\le b<1 \tag{ii}
\end{equation*}

Consider now case (i). Then, letting 
\begin{equation*}
	\eta_1:=\int_a^{1/4}(f-1),\quad\eta_2:=\int_{3/4}^b(f-1),
\end{equation*}
we have 
\begin{equation*}
	\de_1=\ep_1-\eta_1,\quad\de_2=\ep_2-\eta_2. \tag{1.5}
\end{equation*}
Let now 
\begin{equation*}
	c:=f(1/4)-1[\ge0],\quad d:=f(3/4)-1[\ge0],
\end{equation*}
\begin{equation*}
	p(x):=c+(x-1/4)\frac{d-c}{1/2}=\frac{3c-d}2+2x(d-c). 
\end{equation*}
Then $p(1/4)=c=f(1/4)-1$ and $p(3/4)=d=f(3/4)-1$. So, by the concavity of $f$, we have $p\le f-1$ on $[1/4,3/4]$ and $p\ge f-1$ on $[0,1]\setminus[1/4,3/4]$. So, 
\begin{equation*}
	\eta_1=\int_a^{1/4}(f-1)\le\int_a^{1/4}p=\frac{5c-d}{16};
\end{equation*}
similarly, $\eta_2\le\int_a^{1/4}p=\frac{5d-c}{16}$, whence
\begin{equation*}
	\eta_1+\eta_2\le\frac{5c-d}{16}+\frac{5d-c}{16}=\frac{c+d}4, 
\end{equation*}
whence, by (1.5), 
\begin{equation*}
	\de_1+\de_2=\ep_1+\ep_2-(\eta_1+\eta_2)=\ep-(\eta_1+\eta_2)\ge\ep-(c+d)/4;
\end{equation*}
and 
\begin{equation*}
	\int_{1/4}^{3/4}(f-1)\ge\int_{1/4}^{3/4}p=\frac{c+d}4,
\end{equation*}
so that 
\begin{equation*}
	\ep_1+\ep_2=\ep=\int_a^b(f-1)=\int_a^{1/4}(f-1)+\int_{1/4}^{3/4}(f-1)+\int_{3/4}^b(f-1)
	\ge\eta_1+\frac{c+d}4+\eta_2,
\end{equation*}
whence, by (1.5), $\de_1+\de_2\ge(c+d)/4$. 
So, 
\begin{equation*}
	\max(\de_1,\de_2)\ge\frac12\,(\de_1+\de_2)
	\ge\frac12\,\max((c+d)/4,\ep-(c+d)/4)\ge\frac\ep4.  
\end{equation*}
So, (*) holds if case (i) takes place. 


Consider now case (ii). By Lemma 1 (cf. (0.5)),
\begin{equation*}
	\ep=\int_a^b(f-1)\le\frac{(b-a)^2}{a^2}\,\int_0^a(1-f)\le\frac{(1-a)^2}{a^2}\,\int_0^a(1-f)
	=\frac{(1-a)^2}{a^2}\,\ep_1,
\end{equation*}
\begin{equation*}
	\ep-\eta_2=\int_a^{3/4}(f-1)
\ge\frac{(3/4-a)^2}{(b-a)^2}\,\int_a^b(1-f)=\frac{(3/4-a)^2}{(b-a)^2}\,\ep\ge\frac{(3/4-a)^2}{(1-a)^2}\,\ep,
\end{equation*}
and, similarly, 
\begin{equation*}
 \ep_1-\de_1=\int_a^{1/4}(f-1)\le\frac{(a-1/4)^2}{a^2}\int_a^0(f-1)
 =\frac{(a-1/4)^2}{a^2}\,\ep_1,
\end{equation*}
which implies 
\begin{equation*}
	\de_1\ge n(a)\ep_1\ge m(a)\ep, \tag{4}
\end{equation*}
where 
\begin{equation*}
n(a):=1-\frac{(a-1/4)^2}{a^2},\quad m(a):=\frac{a^2}{(1-a)^2}n(a). 
\end{equation*}
So, 
\begin{equation*}
	\de_2=\ep_2-\eta_2=-\ep_1+(\ep-\eta_2)\ge-\de_1/n(a)+\frac{(3/4-a)^2}{(1-a)^2}\,\ep. 
\tag{5} 
\end{equation*}

Minimizing now $\max(\de_1,\de_2)$ over all $\de_1,\de_2,a$ such that $1/4\le a<1$, $\de_1\ge m(a)\ep$ (cf. (4)), and $\de_2\ge-\de_1/n(a)+\frac{(3/4-a)^2}{(1-a)^2}\,\ep$, we get
$\max(\de_1,\de_2)\ge\frac{24}{121}\,\ep$ in case (ii), so that (*) holds in case (ii) as well.
This calculation takes about 0.11 sec with Mathematica: 

[![enter image description here][2]][2]

$\Box$

**Remark:** Following the lines of the consideration of case (ii) (with $a=5/16$, $\ep\downarrow0$, and $b\uparrow1$), one can see that the factor $\frac{24}{121}=0.19834\dots$ in the lower bound $\frac{24}{121}\,\ep$ in (*) is the best possible. 

More explicitly, for $k\in(0,1)$ and $x\in[0,1]$, let 
\begin{equation}
	f(x):=\left\{
	\begin{aligned}
 k \left(x-5/16\right)+1 & \text{ if }x\le\frac{5 k+16}{11 k+16}  \\
 \frac{(25 k^2+352 k+256) (1-x)}{96 k} & \text{ if }x\ge\frac{5 k+16}{11 k+16}. 
\end{aligned}
\right.
\end{equation}
Then $f$ satisfies all the conditions. Moreover, then 
\begin{equation}
	a=5/16,\quad b=\frac{25 k^2+256 k+256}{25 k^2+352 k+256},
\end{equation}
and
\begin{equation}
	\frac{\de_1}\ep=\frac{\de_2}\ep=\frac{24 \left(25 k^2+352 k+256\right)}{(25 k+176)^2}\to\frac{24}{121}
\end{equation}
as $k\downarrow0$. 

Here is the plot of this particular $f$ for $k=1/5$: 

[![enter image description here][3]][3]


  [1]: https://www.emis.de/journals/JIPAM/images/157_05_JIPAM/157_05.pdf
  [2]: https://i.sstatic.net/Zy862.png
  [3]: https://i.sstatic.net/NolQC.png