Integral Inequality

Problem. Let $k$ be a positive integer.  $f:[0,\infty)\to[0,\infty)$ is a continuous function such that $f(f(x))=x^k$, $\forall x\in[0,\infty)$. Show that $$\int_0^1f(x)^2\,dx\ge\frac{2k-1}{k^2+6k-3}.$$

This statement is provable by Young's inequality. By assuming $f$ be continuously differentiable (or any kind of continuous such that the integration by parts on $[0,1]$ work, say $f$ is absolutely continuous), then we have a stronger bound: $$\int_0^1f(x)^2\,dx\ge\frac{2k-1}{k^2+4k-2}.$$ Which becomes the previous one by adding $2k-1$ to the denominator.

Solution. There is a few observations. First, $f$ is injective. Second, $f(f(x))=x^k$ implies $f(f(f(x)))=f(x)^k$, therefore  $f(x^k)=f(x)^k$. In particular, if we put $x=0$ and $x=1$, then we have $f(0)=f(0)^k$ and $f(1)=f(1)^k$. Therefore both $f(0),f(1)\in \{0,1\}$.  By injectivity and the fact $f\ge 0$, we have $f(0)=0$ and $f(1)=1$.

By Cauchy-Schwarz inequality we have $I:=\int_0^1f^2\,dx\ge \int_0^1 f\,dx$. We substitute $x=f(y)$ to get $$I=\int_0^1 f(f(y))(f(y))'\,dy=\int_0^1 y^k(f(y))'\,dy=1-k\int_0^1f(y)y^{k-1}.$$ The last equality follows from integration by parts. Now we shrink RHS by Cauchy-Schwarz inequality to get $I\ge1- \frac{k}{\sqrt{2k-1}}\sqrt{I}$, which we rearrange to get $$(\sqrt{I})^2 +\frac{k}{\sqrt{2k-1}} \sqrt{I}-1\ge 0.$$ By completing square we find that this implies $$\sqrt{I}\ge \frac{-\frac{k}{\sqrt{2k-1}}+ \sqrt{\frac{k^2+8k-4}{2k-1}}}{2}=\frac{2\sqrt{2k-1}}{\sqrt{k^2+8k-4}+k}.$$ Now by Cauchy-Schwarz inequality  we also have $$\sqrt{k^2+8k-4}+k\leq \sqrt{(k^2+8k-4+k^2)(1+1)}=2\sqrt{k^2+4k-2},$$ hence we have $$I\ge \frac{2k-1}{k^2+4k-2}.\qed$$