Proving $ \cos x + \frac{2p}{\pi}\sin x \geq 1 - 2 \left(\frac{x}{\pi}\right)^p$ for $p>0$

Question

I came across the following inequality: $$ \cos x + \frac{2p}{\pi}\sin x \geq 1 - 2 \left(\frac{x}{\pi}\right)^p, $$ which should be valid for any $0\leq x \leq \pi$ and $p\in\mathbb R$ according to the graphs.

I managed to prove it myself for $p\leq0$ as follows:

If $p=0$ it's trivial, and if $p<0$, I look at the difference $$f(x) = \cos x + \frac{2p}{\pi}\sin x - 1 + 2\left(\frac{x}{\pi}\right)^p$$ Noticing that $f(\pi)=0$, it is enough to show $-f'(x)\geq0$. Then, $$-f'(x) = \sin x - \frac{2p}{\pi}\left(\cos x + \left(\frac{x}{\pi}\right)^{p - 1}\right)$$ and all terms are nonnegative for $x\in[0,\frac{\pi}{2}]$. If $x\in[\frac{\pi}{2},\pi]$ then still $\cos x + \left(\frac{x}{\pi}\right)^{p - 1}\geq0$ as the latter term is at least $1$.

This method doesn't help me for the case $p>0$, though, as the function is no longer monotone. I tried to calculate (among other things) a taylor and fourier series, but found it wasn't much use for general $p$.

Any help or hints will be appreciated.

Answer 1

This is a partial solution or rather a long comment to give some ideas. It is probably better to compare the function $\phi_p(x)=\cos x+\frac{2p}{\pi}\sin x$ and $f_p(x)=1-2(x/\pi)^p$.

$\phi'_p(x)=-\sin x+\frac{2p}{\pi}\cos x=0$ on $[0,\pi]$ iff $\tan(x)=\frac{2p}{\pi}$. This has a unique solution $\xi^\phi\in (0,\pi/2)$, namely $$\xi^\phi(p)=\arctan\big(\frac{2p}{\pi}\big)\nearrow\frac{\pi}{2}\quad\text{as $p\nearrow\infty$}$$ $\phi''_p(x)=-\phi_p(x)=0$ on $[0,\pi]$ iff $\tan(x)=-\frac{\pi}{2p}$. This has a unique solution $\xi_\phi\in(\pi/2,\pi)$, namely $$\zeta_\phi(p)=\pi-\arctan\big(\frac{\pi}{2p}\big)\nearrow\pi\quad\text{as $p\nearrow\infty$}$$ Furthermore, $\phi_p$ attains its maximum in $[0,\pi]$ at $\xi^\phi(p)$, and it has a unique zero and inflection point at $\zeta_\phi(p)$.

The function $f_p$ is strictly decreasing on $[0,\pi]$, and has a unique zero on $[0,\pi$, namely $$\zeta_f(p)=2^{-1/p}\pi\nearrow\pi\quad\text{as $p\nearrow\infty$}$$ For $0<p\leq1$, $\zeta_f(p)\in(0,\pi/2]$ whereas for $p>1$, $\zeta_f(p)\in(\pi/2,\pi)$.
Since $f''(x)=-\frac{2p(p-1)}{\pi^2}(x/\pi)^{p-2}$ for $p\neq1$, $f$ is concave if $p>1$ and convex if $p<1$.

• For $0<p\leq1$, $g=\phi-f$ is concave and $g(0)=g(1)$, monotone increasing near $0$ and monotone decreasing near $\pi=0$. Thus there cannot be solutions to $g=0$ inside $(0,\pi)$ and so $g\geq0$.

• The case $p>1$ involves more work. The following seems to hold $$\frac{\pi}{2}<\zeta_f(p)<\zeta_\phi(p)$$ The left-hand-side is obvious; the second should follow by showing that $$G(p):=\arctan\big(\frac{\pi}{2p}\big)+2^{-1/p}\pi<\pi$$ Clearly $G(p)\xrightarrow{p\rightarrow\infty}\pi$. Taking derivatives on the left-hand-side, for example, yields $$ G'(p)=-\frac{2\pi}{\pi^2+4p^2}+\frac{\pi 2^{-1/p}\log 2}{p^2} $$ I have not checked this carefully, but it seems that indeed $G'(p)>0$. Putting all this together will imply that $g_p>f_p$ on $[0,\zeta\phi(p)]$.

For the interval $[\zeta_\phi(p),\pi]$ we notice that $\phi'_p(\pi)=-\frac{2p}{\pi}=f'(\pi)$. Since $\phi_p$ is convex on this interval and $f_p$ is concave, it must be the $g_p>f_p$ on $[\zeta_\phi(p),\pi)$. This would be the end of the proof.

There might be more clever ways to prove this inequality and I am looking forward to see other attempts.

Answer 2

Problem: Prove that, for all $x \in (0, \pi)$ and $p > 0$, $$\cos x + \frac{2p}{\pi}\sin x + 2\left(\frac{x}{\pi}\right)^p - 1 \ge 0.$$

Proof.

We split into two cases.

Case 1: $x\in (0, \pi/2]$

Using the known inequalities $\cos x \ge 1 - \frac{x^2}{2}$ and $\sin x \ge \frac{2}{\pi}x $ on $(0, \pi/2]$, it suffices to prove that $$1 - \frac{x^2}{2} + \frac{2p}{\pi}\cdot \frac{2}{\pi}x + 2\left(\frac{x}{\pi}\right)^p - 1 \ge 0$$ or $$ - \frac12x^2 + \frac{4px}{\pi^2} + 2\left(\frac{x}{\pi}\right)^p \ge 0 \tag{1}$$ which is true. The proof of (1) is given at the end.

Case 2: $x\in (\pi/2, \pi)$

Letting $x = \pi - y$, it suffices to prove that, for all $y \in (0, \pi/2)$, $$- \cos y + \frac{2p}{\pi}\sin y + 2\left(\frac{\pi - y}{\pi}\right)^p - 1 \ge 0$$

Using the know inequalities $\cos y \le 1 - \frac{y^2}{2} + \frac{y^4}{24}$ and $\sin y \ge y - \frac{y^3}{6}$ on $(0, \infty)$, it suffices to prove that $$- \left(1 - \frac{y^2}{2} + \frac{y^4}{24}\right) + \frac{2p}{\pi}\left(y - \frac{y^3}{6}\right) + 2\left(\frac{\pi - y}{\pi}\right)^p - 1 \ge 0 \tag{2}$$ which is true. The proof of (2) is similar to that of (1) ($0 < p < 1$, $1\le p \le 2$, $p> 2$, Bernoulli inequality etc).

We are done.

$\phantom{2}$

---

Proof of (1):

(1) $p \ge 2$

We have $$\mathrm{LHS} \ge - \frac12x^2 + \frac{8x}{\pi^2} \ge 0.$$

(2) $1\le p < 2$

Using $u^v \ge \frac{u}{u + v - uv}$ for all $u > 0$ and $v\in (0, 1]$ (equivalently $u^{1-v} \le 1 + (u- 1)(1 - v)$ which is Bernoulli inequality), we have $$\left(\frac{x}{\pi}\right)^{p-1} \ge \frac{x/\pi}{x/\pi + (p-1) - (p-1)x/\pi}.$$

It suffices to prove that $$- \frac12x^2 + \frac{4px}{\pi^2} + 2\left(\frac{x}{\pi}\right) \cdot \frac{x/\pi}{x/\pi + (p-1) - (p-1)x/\pi} \ge 0$$ or $$- \frac12x + \frac{4p}{\pi^2} + 2\left(\frac{x}{\pi}\right) \cdot \frac{1}{x + (p- 1)(\pi - x)} \ge 0$$ or \begin{align*} &- \frac12x + \frac{4}{\pi^2} - \frac{4x}{\pi^2(\pi - x)}\\[6pt] &\quad + \frac{4}{\pi^2(\pi - x)}[x + (p- 1)(\pi - x)] + 2\left(\frac{x}{\pi}\right) \cdot \frac{1}{x + (p- 1)(\pi - x)} \ge 0. \end{align*}

By AM-GM, it suffices to prove that \begin{align*} &- \frac12x + \frac{4}{\pi^2} - \frac{4x}{\pi^2(\pi - x)}\\[6pt] &\quad + 2\sqrt{\frac{4}{\pi^2(\pi - x)}[x + (p- 1)(\pi - x)] \cdot 2\left(\frac{x}{\pi}\right) \cdot \frac{1}{x + (p- 1)(\pi - x)}} \ge 0 \end{align*} or $$- \frac12x + \frac{4}{\pi^2} - \frac{4x}{\pi^2(\pi - x)} + \frac{4}{\pi}\sqrt{\frac{2x}{\pi(\pi - x)}} \ge 0$$ which is true.

(3) $0 < p < 1$

Using $u^v \ge \frac{u}{u + v - uv}$ for all $u > 0$ and $v\in (0, 1]$, we have $$\left(\frac{x}{\pi}\right)^p \ge \frac{x/\pi}{x/\pi + p - px/\pi}.$$

It suffices to prove that $$- \frac12x^2 + \frac{4px}{\pi^2} + 2\cdot \frac{x/\pi}{x/\pi + p - px/\pi} \ge 0 $$ or $$- \frac12x + \frac{4p}{\pi^2} + 2\cdot \frac{1}{x + p(\pi - x)} \ge 0 $$ or $$-\frac12 x - \frac{4x}{\pi^2(\pi - x)} + \frac{4}{\pi^2(\pi - x)}[x + p(\pi - x)] + \frac{2}{x + p(\pi - x)} \ge 0.$$

By AM-GM, it suffices to prove that $$-\frac12 x - \frac{4x}{\pi^2(\pi - x)} + 2\sqrt{\frac{4}{\pi^2(\pi - x)}[x + p(\pi - x)] \cdot \frac{2}{x + p(\pi - x)}} \ge 0$$ or $$-\frac12 x - \frac{4x}{\pi^2(\pi - x)} + \frac{4}{\pi}\sqrt{\frac{2}{\pi - x}} \ge 0$$ which is true.

We are done.