Minimize $\min_{f\in E}\left(\int_0^1f(x) dx\right)$ Inspired by this question, I pose this following problem.
Let $E$ be the set of all nonnegative continuous functions $f:[0,1]\to \mathbb{R}$ such that $$f(x)\,f(y)\ge |x-y|\qquad\forall\{x,y\}\subset [0,1]$$
Find $$\min_{f\in E}\left(\int_0^1f(x) \,dx\right)$$
 A: Integrating both sides with respect to x and y we get 
$$\left(\int_0^1f(x) dx\right)^2=
\int_0^1\int_0^1f(x)f(y)dxdy \ge\int_0^1\int_0^1
 |x-y|dxdy= \frac13.$$
that is 
$$\int_0^1f(x)dx \ge  \frac{1}{\sqrt3.}$$
I am still searching whether the min is attain or not.
A: Let $A=A(f)=\int_0^1f(x)\,dx$. By the AGM inequality, 
$$A=\frac12\int_0^1 \big(f(x)+f(1-x)\big) dx \ge \int_0^1 \sqrt{f(x)f(1-x)}\,dx \ge \int_0^1 \sqrt{|1-2x|}dx = \frac23.$$
Similarly, 
$$A=2\int_0^{1/2} \frac{f(x)+f(x+1/2)}2\, dx $$ $$\ge 
2\int_0^{1/2} \sqrt{f(x)f(x+1/2)} \, dx\ge 2\int_0^{1/2}\sqrt{1/2}\,dx=\frac1{\sqrt2}.$$
Added 14 Nov.
This might or might not have been obvious all along, but not explicitly so to me until just now. Note that if $g_1$ and $g_2$ satisfy the constraints, then so does $g(x) = \sqrt{g_1(x)g_2(x)}$, and moreover, by the AGM inequality, $$\int_0^1 \sqrt{g_1(x)g_2(x)}\,dx\le \int_0^1 \frac {g_1(x)+g_2(x)} 2\, dx.$$ Given any admissible $h$, the functions $h_1(x) = h(x)$, $h_2(x)=h(1-x)$, and $f(x)=\sqrt{h_1(x)h_2(x)} = \sqrt{h(x)h(1-x)}$ are admissible, and $\int_0^1 f\le \int_0^1 h$.
Hence one cannot beat functions of the form $\sqrt{h(x)h(1-x)}$, that is, functions that depend only on $|2x-1|$.
Another consequence of the AGM inequality is that if the minimum is attained, it is attained uniquely:  Suppose $A(f)=A(g)$ where $f\ne g$. Then  $A(h)<A(f)$, where $h(x)=\sqrt{f(x)g(x)}$.
Added 17 Nov.
Not only can one not beat functions of form $f(x)=\phi(|2x-1|)$
, but one cannot beat such functions where continuous $\phi:[0,1]\to\mathbb R^+$ is also required to be nondecreasing. The $f(x)f(y)\ge|x-y|$ constraint translates to a $\phi(u)\phi(v)\ge(u+v)/2$ constraint for $u,v\in[0,1]$. If $\phi$ obeys these constraints, so does monotonic $\phi^*(u)=\min\{\phi(t): t\in[u,1]\}$, for which the corresponding $A(f^*)\le A(f)$.
(I'm hoping to use a compactness argument somehow to then prove the minimum $A(f)$ is attained. I think that such an argument succeeds in the restricted problem where one sticks in an additional constraint $f(0)=\gamma$, but don't yet see yet why the restricted optimal value of $A$ depends continuously on $\gamma$.)
A: NOTE: Still not a complete answer, just improving on the lower bound.
EDIT: Added an improvement on the upper bound (a better function which "works"). 
I decided to combine the two ideas in kimchi lover's bounds, by putting a bound on 
$$
\int_0^1 f(x)dx = \int_0^{1/4} \left[f(x) + f(1-x) + f(1/2+x) + f(1/2-x)\right]dx
$$
Considering the four values $(u,v,s,t) = (f(x), f(1-x), f(1/2-x), f(1/2+x))$ as independent, but satisfying the constraints
$$
\begin{align}
&uv \geq 1-2x \\
&ut \geq 1/2 \\
&sv \geq 1/2 \\
&st \geq 2x, 
\end{align}
$$
I found a lower bound for $u+s+v+t$. (There are two more inequalities, of course, but the minimizing set of values turns out to be symmetric, $s = t$ and $u = v$, so the $us$ and $tv$ inequalities are weaker than the $ut$ and $sv$ ones.) The approach is rather brute-force: pick $u$ and $s$ as independent, consider the different cases for which of the inequalities involving $t$ and $v$ are the stronger ones, then minimize an expression in terms of $u$ and $s$. The calculations are tedious, and since this is not a complete answer anyway, I'll omit them. I got the following:
$$
u + s + t + v \geq \frac{3-4x}{\sqrt{1-2x}},
$$
minimum attained at $u = v = \sqrt{1-2x}$ and $s = t = {1\over2u}$ (these are not too surprising, of course). Thus
$$
\int_0^1 f(x) dx \geq \int_0^{1\over4} \frac{(3-4x)dx}{\sqrt{1-2x}} = {5-2\sqrt2\over 3} \approx 0.72386
$$
If we try to piece together a function from the $(u,s,t,v)$'s for $x \in [0,1/4]$, it still doesn't satisfy some of the constraints, such as for example $f(x)f(1) \geq 1-x$.  
A function which does meet the constraints everywhere is
$$
f(x) = \frac{e^{|2x-1|}}{\sqrt{2e}}.
$$
That it does work can be verified using the triangle inequality
$$
f(x)f(y) = \frac{e^{|2x-1|+|2y-1|}}{2e} \geq \frac{e^{2|x-y|}}{2e} \geq |x-y|
$$
where the last one holds because $e^{2t} - 2et$ has a global minimum $0$ at $t = 1/2$. Its integral is
$$
\int_0^1 \frac{e^{|2x-1|}}{\sqrt{2e}}dx = 2\int_{1/2}^1 \frac{e^{2x-1}}{\sqrt{2e}}dx = \frac{e-1}{\sqrt{2e}} \approx 0.73694
$$
EIDT: Following @kimchilover's idea to generalize the exponential function to $\beta e^{\alpha|2x-1|}$, the constraints will be satisfied if and only if the minimum of 
$$
\beta^2 e^{\alpha|2x-1|}e^{\alpha|2y-1|} - |x-y|
$$
over $[0,1]\times[0,1]$ is non-negative. The strongest constraints are obtained using $x \in [0,1/2], y \in [1/2,1]$ so that we need
$$
\beta^2 e^{2\alpha(y-x)} - (y-x) = \beta^2 e^{2\alpha t} - t \geq 0
$$
for $t \in [0,1]$. We can find the minimum, then if the minimum is in $[0,1]$, require that it is non-negative, or if it is outside, require that the closest "cut-off" is non-negative. Long story short, the optimal ones I found to be the root of $e^\alpha(\alpha - 3/2) + 3/2 = 0$, $\alpha \approx 0.874217$ (by Wolfram), $\beta = 1/\sqrt{2e\alpha}$, and for the integral $\approx 0.733001$. 
A: Extending kimchi lover's idea:
$$I = n \int_0^{1/n} \frac{1}{n} \sum_{k=0}^{n-1} f(x+k/n) dx \geq n \int_0^{1/n} \left(\prod_{k=0}^{n-1} f(x+k/n) \right)^{1/n} dx$$ 
Pairing up opposite edges, then we can pick pairs $(x+k/n, x+(n-k+1)/n)$, with $n=2m$, such that 
$$\lim_{n \to \infty} I \geq \lim_{n \to \infty} 2m \int_0^{1/2m} \left( \prod_{k=1}^{m} \left(\frac{2k-1}{2m} \right) \right)^{1/2m} dx = \lim_{m \to \infty} \frac{[(2m-1)!!]^{1/2m}}{(2m)^{1/2}} = \sqrt{\frac{1}{e}}$$
where we compute the limit using wolfram alpha.
This bound isn't the best one found so far, but perhaps my approach will be inspiring. Loved thinking about the problem!
EDIT: More information. Using a continuous approximate (from above) to the step function $f(x)=a^{1/2} \chi_{[0,a]} + a^{-1/2} \chi_{(a,1]}$, you should be able to obtain a value below 1 for the integral if I did by calculation correctly. It is clear that $f=1$ means the minimum is less than or equal to 1.
