Find al continuous functions $f:[0,1]\to \mathbb{R}$ such that $f(x) f(1-x) \ge \int_0^1 (f^2(x)-f(x)f(1-x)+f^2(1-x)) dx, \forall x\in [0,1]$ Find al continuous functions $f:[0,1]\to \mathbb{R}$ such that $f(x) f(1-x)  \ge \int_0^1 (f^2(x)-f(x)f(1-x)+f^2(1-x)) dx, \forall x\in [0,1]$.
I began by integrating this inequality on $[0,1]$ to get that $\int_0^1 (f(x)-f(1-x))^2 dx\le0$.
Since we also have that $\int_0^1 (f(x)-f(1-x))^2 dx\ge 0$ we obtain that $\int_0^1 (f(x)-f(1-x))^2 dx=0$ and because $f$ is continuous it follows that $f(x)=f(1-x), \forall x\in [0,1]$. Now, this functional equation has infinitely many solutions, so we can't determine $f$ only from here.
After substituting back into the original equation we get that $f^2(x) \ge \int_0^1 f^2(x) dx,  \forall x\in [0,1]$. I honestly do not know how to proceed from here.

Starting from $f(x)^2\geq\int_0^1f(x)^2dx$, we apply the mean value theorem to obtain $$f(x)^2\geq\int_0^1f(x)^2dx=f(c)^2$$ for some all $x\in[0,1]$ and for some $c\in[0,1]$. Note that we have used the fact that the length of the interval is $1$. Now, we have $$f(x)^2\geq f(c)^2\implies f(x)^2-f(c)^2\geq 0 \  \ \ \forall x\in[0,1].$$ Integrating this new function, we obtain $$\int_0^1f(x)^2-f(c)^2dx = \int_0^1f(x)^2dx-\int_0^1f(x)^2dx=0.$$ Since $f(x)^2-f(c)^2\geq0$ and $\int_0^1 f(x)^2-f(c)^2dx= 0$, it must be that $f(x)^2-f(c)^2=0$, so $f(x)$ is constant on $[0,1]$.
