Prove that $\int_0^1\sqrt{f(x)} dx \le \frac{\pi\sqrt 5}{8}$ where $f(x) + f((1 - \sqrt{x})^2) \le 1, \forall x \in [0; 1]$.

Question

Consider function $f$ which is continuous on the closed interval $[0; 1]$ such that $f(x) > 0, \forall x \in [0; 1]$ and $f(x) + f\left((1 - \sqrt{x})^2\right) \le 1, \forall x \in [0; 1]$. Prove that $\displaystyle \int_0^1\sqrt{f(x)}\, \mathrm dx \le \dfrac{\pi\sqrt 5}{8}$.

Below is an attempt of mine at solving this problem. One question, why is $\pi$ there, just whhyyy~?

(Welp, restarting in 3... 2... 1...)

We have that $$\begin{aligned} \left(\sqrt{f(x)} + \sqrt{f\left((1 - \sqrt{x})^2\right)}\right)^2 \le 2\left[f(x) + f\left((1 - \sqrt{x})^2\right)\right] &\le 2\\ \iff \sqrt{f(x)} + \sqrt{f\left((1 - \sqrt{x})^2\right)} &\le \sqrt 2, \forall x \in [0; 1]\\ \iff \int_0^1\sqrt{f(x)}\, \mathrm dx + \int_0^1\sqrt{f\left((1 - \sqrt{x})^2\right)}\, \mathrm dx &\le \sqrt 2\\ \iff \int_0^1\sqrt{f(x)}\, \mathrm dx + \int_1^0\dfrac{1 - \sqrt{x}}{\sqrt{x}}\sqrt{f(x)}\, \mathrm dx &\le \sqrt 2\\ \iff \int_0^1\dfrac{2\sqrt{x} - 1}{\sqrt{x}}\sqrt{f(x)}\, \mathrm dx &\le \sqrt 2\end{aligned}$$

I'll add more thoughts as time goes on, but this is all for now, thanks for reading (and more if you could help~)!

Answer 1

First, substitute $x=s^2$, so that $$ I=2\int_0^1\sqrt{f(s^2)}s\,ds $$ Then split into two copies and reverse the direction of integration in one $$ I=\int_0^1\left[\sqrt{f(s^2)}s+\sqrt{f((1-s)^2)}(1-s)\right]\,ds $$ Apply Cauchy-Schwarz $$ I\le\int_0^1\sqrt{f(s^2)+f((1-s)^2)}\sqrt{s^2+(1-s)^2}\,ds\le\int_0^1\sqrt{1-2s+2s^2}\,ds $$ and if all goes well that integral evaluates to the claimed bound.