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

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~)!

• First line, setting $x=\sin^2 t$ yields $f(\sin^2t) + f((1-\sin t)^2) \le 1$ Apr 24, 2022 at 5:21
• Oh, ohhh, ohhhh~ Well, that sucks. How about setting $x = \sin^4t$? I'll have to rewrite this. Apr 24, 2022 at 5:45

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.
• An antiderivative of the last integrand is $\frac{1}{8} \left(2 (2 x-1) \sqrt{2 x^2-2 x+1}-\sqrt{2} \log \left(\sqrt{4 x^2-4 x+2}-2 x+1\right)\right)$, and so the last integral evaluates to $\frac{1}{2}+\frac{\log \left(1+\sqrt{2}\right)}{2 \sqrt{2}} \approx 0.811613$, better than the proposed $\frac{\pi\sqrt5}8 \approx 0.878102$. Apr 24, 2022 at 6:34