Prove that $\int_0^1 f^2(x)dx > \frac{1}{6}$ , knowing that $f(f(x)) = x^2$. How can I show that $\int_0^1 f^2(x)dx > \frac{1}{6}$, where $f(x)$ is continuous for $x\in[0,\infty)$, and $f(f(x)) = x^2$?  
I figured out that $f(x^2)=f(f(f(x)))=f^2(x)$ then $\int_0^1f^2(x)dx = \int_0^1f(x^2)dx$, but I don't know what to do next.

It can be proven that if $f:[0,\infty)\mapsto [0,\infty)$ is continuous and $f(f(x))=x^2$, then $f(x)$ is bounded (not strictly) between $x$ and $x^2$ for all $x\in (0,1)$. I will not prove this rigorously here, and leave it as an exercise (it can be done using the continuity requirement and the generalized intermediate value theorem for two curves).
This shows that $f^2(x)$ is between $x^2$ and $x^4$, and that $\int_0^1 f^2(x)dx\ge \int_0^1 x^4=1/5\gt 1/6$.
MORE DETAILS FOR THE PROOF: 
Since $f(f(x))=x^2$ is injective, $f(x)$ must be injective, and thus either strictly increasing or strictly decreasing; I’ll let you rule out the latter case.
Suppose it is increasing. If $f(x)\gt x$ for some $x\in (0,1)$, then $f(f(x))\gt f(x)$ and $x^2\gt f(x)\gt x$ which is impossible. If $f(x)\lt x^2$ for some $x\in (0,1)$, then $f(x)\lt f(f(x))$ and $x\lt f(x)\lt x^2$ which is impossible. Thus $x^2\le f(x)\le x$.
