Given $f(x)$ is continuous on $[0,1]$ and $f(f(x))=1$ for $x\in[0,1]$. Prove that $\int_0^1 f(x)\,dx > \frac34$.

$f(1)=f(f(f(1)))=(f\circ f) (f(1))=1$

$f([0,1])=[a,1]$ for some $a >0$ since the image is connected hence an interval ending at $1$ and compact hence the interval is closed while obviously $f([a,1])=1$ so $a >0$

But now on $[0,a], f(x) \ge a$ so $\int_0^1f(x)dx=\int_0^af(x)dx+\int_a^1f(x)dx \ge a^2+1-a \ge 3/4$ and we cannot have equality since then $a=1/2$ and because $f(1/2)=1, f(x) \to 1, x \to 1/2, x<1/2$ so $f$ cannot be identically $1/2$ on $[0,1/2)$ and it is bigger on at least a small interval near $1/2$

Note that by choosing that interval very small and making $f$ linear there (and $1/2$ before, $1$ after) we can get the integral $3/4+\epsilon$ so the result is sharp.

Done!