Proving that $f(x)=x\arccos\left(1-p+p\cos\left(\frac{2\pi}{x}\right)\right)$ is increasing on $[2,\infty)$ with $0<p<1$

Question

The function$$f(x)=x\arccos\left(1-p+p\cos\left(\frac{2\pi}{x}\right)\right)$$on $[2,\infty)$ with $0<p<1$ is clearly increasing by looking at its graph.

I calculated its derivative to be $$f'(x)=\arccos\left(1-p+p\cos\left(\frac{2\pi}{x}\right)\right)-\frac{2\pi p}{x}\frac{\sin\left(\frac{2\pi}{x}\right)}{\sqrt{1-\left(1-p+p\cos\left(\frac{2\pi}{x}\right)\right)^{2}}}$$ but couldn't prove that it is positive on the interval.

I tried to prove the equivalent result $$\frac{x}{x+h} > \frac{\arccos\left(1-p+p\cos\left(\frac{2\pi}{x+h}\right)\right)}{\arccos\left(1-p+p\cos\left(\frac{2\pi}{x}\right)\right)} $$ for all $x\ge2$ and $h>0$ but couldn't deal with the $\arccos$ terms.

I also calculated the limit of $f(x)$ at infinity to be $2\pi \sqrt{p}$ if this can help somehow.

Answer 1

$\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$Define $g(u) = \arccos u$. Starting from the well-known fact that $\dfrac{\tan v}{v}$ is strictly increasing for $v \in \left(0, \dfrac{π}{2}\right)$ and letting $u = \cos 2v$, it shows that$$ (1 - u) \frac{g'(u)}{g(u)} = -\frac{1}{\arccos u} \sqrt{\frac{1 - u}{1 + u}} = -\frac{\tan v}{2v} $$ is strictly increasing for $u \in (-1, 1)$. For any $u \in (-1, 1)$, since $1 - p + pu > u$, then\begin{align*} &\peq \frac{\d}{\d u}\left(\frac{\arccos(1 - p + pu)}{\arccos u}\right) = \frac{\d}{\d u}\left(\frac{g(1 - p + pu)}{g(u)}\right)\\ &= \frac{1}{(g(u))^2} (pg'(1 - p + pu) g(u) - g(1 - p + pu) g'(u))\\ &= \frac{g(1 - p + pu)} {(1 - u) g(u)} \left((1 - (1 - p + pu)) \frac{g'(1 - p + pu)}{g(1 - p + pu)} - (1 - u) \frac{g'(u)}{g(u)}\right) > 0, \end{align*} which implies that $\dfrac{\arccos(1 - p + pu)}{\arccos u}$ is strictly increasing for $u \in [-1, 1)$. Finally, letting $x = \dfrac{2π}{\arccos u}$ shows that$$ f(x) = x \arccos\left(1 - p + p\cos\frac{2π}{x}\right) = 2π \frac{\arccos(1 - p + pu)}{\arccos u} $$ is strictly increasing for $x \in [2, +∞)$.

Answer 2

Alert: just a long comment

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Let $f_1(x)=\frac{2\pi}{x}$, $f_1$ is decreasing because $\displaystyle\lim_{x\to\infty}f_1(x)=0$. We all agree here,but just in case.

Let $f_2=\cos(x)_{|I}\to [-1,1]\; \\I\subseteq \langle 0,2\pi\rangle$

As a composition of two (strictly) increasing functions (remember, we restricted $\cos$), $$g(x)=f_2(f_1(x))=\cos\left(\frac{2\pi}{x}\right)\;\text{is a strictly increasing function},\lim_{x\to\infty}\cos\left(\frac{2\pi}{x}\right)=1$$ $$\implies 1-p+p\cos\left(\frac{2\pi}{x}\right)\;\text{is strictly decreasing}$$ This function can be strictly increasing and converging for any choice of $p$ on the interval $[2,\infty\rangle$, as can be seen in the post and the OP's approach.

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