IMC 2001 · Day 2 · P12

We prove that $g(\vartheta) = |\sin\vartheta| |\sin(2\vartheta)|^{1/2}$ attains its maximum value $(\sqrt{3}/2)^{3/2}$ at points $2k\pi/3$ (where $k$ is a positive integer). This can be seen by using derivatives or a classical bound like $$\begin{align*} g(\vartheta) &= |\sin\vartheta| |\sin(2\vartheta)|^{1/2} = \frac{\sqrt{2}}{\sqrt[4]{3}} \left( \sqrt[4]{|\sin\vartheta| \cdot |\sin\vartheta| \cdot |\sin\vartheta| \cdot |\sqrt{3} \cos\vartheta|} \right)^2 \\ &\le \frac{\sqrt{2}}{\sqrt[4]{3}} \cdot \frac{3 \sin^2\vartheta + 3 \cos^2\vartheta}{4} = \left( \frac{\sqrt{3}}{2} \right)^{3/2}. \end{align*}$$ Hence $$\begin{align*} \left| \frac{f_n(\vartheta)}{f_n(\pi/3)} \right| &= \left| \frac{g(\vartheta) \cdot g(2\vartheta)^{1/2} \cdot g(4\vartheta)^{3/4} \cdots g(2^{n-1}\vartheta)^{E}} {g(\pi/3) \cdot g(2\pi/3)^{1/2} \cdot g(4\pi/3)^{3/4} \cdots g(2^{n-1}\pi/3)^{E}} \right| \cdot \left| \frac{\sin(2^n \vartheta)}{\sin(2^n \pi/3)} \right|^{1 - E/2} \\ &\le \left| \frac{\sin(2^n \vartheta)}{\sin(2^n \pi/3)} \right|^{1 - E/2} \le \left( \frac{1}{\sqrt{3}/2} \right)^{1 - E/2} \le \frac{2}{\sqrt{3}}. \end{align*}$$ where $E = \frac{2}{3} (1 - (-1/2)^n)$. This is exactly the bound we had to prove.