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I am trying to prove the following result.

Prove that $(\cos \theta)^p \leq \cos(p \theta)$ for $0 \leq \theta < \pi/2$ and $0 <p < 1$.

Here is my attempt.

Let $f:\left[0, \frac{\pi}{2}\right) \to \mathbb{R}$ be defined by $f(\theta) = \cos (p \theta) - (\cos \theta)^p$. Then, for any $\theta$, we have \begin{align*} f'(\theta) & = - p \sin (p \theta) + p \cos^{p-1} \theta \sin \theta \\ & = p \left(\cos^{p-1}\theta \sin \theta - \sin(p \theta) \right) \\ & = p \left(\frac{\sin \theta}{\cos^{1-p} \theta} - \sin (p\theta) \right) \end{align*} As $0 \leq \theta < \frac{\pi}{2}$, we have $\cos \theta > 0$; in particular, $0 < \cos \theta < 1$, so $0 < \cos^{1-p} \theta < 1$ and hence $\frac{\sin \theta}{\cos^{1-p} \theta} > \sin \theta$, so we have: $$ f'(\theta) \geq p \left(\sin \theta - \sin (p\theta)\right). $$ As $0 < p < 1$, $p\theta < \theta$. As $\cos x > 0$ for any $x \in \left[0, \frac{\pi}{2}\right)$, $\sin x$ is monotonically increasing on $\left[0, \frac{\pi}{2}\right)$, so $\sin \theta \geq \sin (p\theta)$, i.e., $\sin \theta - \sin (p\theta) \geq 0$. Therefore, we have $p \left(\sin \theta - \sin (p\theta)\right) \geq 0$ since $p \left(\sin \theta - \sin (p\theta)\right)$ is a product of two nonnegative quantities and hence nonnegative. Therefore, for all $\theta \in \left[0, \frac{\pi}{2}\right)$, $f'(\theta) \geq 0$. But $f(0) = \cos (p(0)) - (\cos (0))^p = 0$. As $f'(\theta) \geq 0$ for any $\theta$, we have $f(\theta) \geq 0$ for any $\theta \in \left[0, \frac{\pi}{2}\right)$, which implies that $(\cos \theta)^p \leq \cos(p \theta)$, as required.

How does this look?

Cardinality
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    $0 < \cos^{1-p} \theta < 1$ does not imply $\frac{\sin \theta}{\cos^{1-p} \theta} < \sin \theta$. E.g. let $p$ be very small and take $\theta=\pi/4$. I think you meant $\gt$. There are also several places where $lt$ should $\le$. – NickD Jun 18 '22 at 01:53
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    Simplify NickD comment. $cos^{1-p}\theta \lt 1$ implies$\frac{sin \theta}{cos^{1-p}\theta} \gt sin\theta$ – herb steinberg Jun 18 '22 at 02:36
  • @herbsteinberg Thank you, I fixed it. Does the proof look correct otherwise? – Cardinality Jun 18 '22 at 02:49
  • @NickD Thank you for catching this. Is the inequality incorrect elsewhere? – Cardinality Jun 18 '22 at 02:49
  • I think the rest of it is correct, modulo the strict vs non-strict inequalities in various places. I don't think that those would invalidate the proof, but I think you should go through and make sure each one is correctly denoted. – NickD Jun 18 '22 at 13:54

2 Answers2

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I don't have the patience to perform so much calculation. We consider the function $f(x) = \log \cos(\theta)$ over $[0,\pi/2)$. Note that $f'(x) = -\tan(\theta)$ is decreasing on the fixed interval, we know $f(x)$ is concave on the interval, therefore $$f((1-p)0+p\theta)\ge (1-p) f(0) + p f(\theta)$$ which evaluates as $$\log(\cos(p\theta))\ge p\log(\cos\theta)=\log(\cos^p(\theta))$$ Take exponential on both sides, we get the desired $\cos(p\theta)\ge \cos^p(\theta)$.

Just a user
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As @Just a user answered, instead of considering the function $$f(t)=\cos(p\theta)-\cos^p(\theta)$$ consider $$g(\theta)=\log (\cos (p \theta))-p \log (\cos (\theta))$$ and use series expansion around $\theta=0$.

This is $$g(\theta)=p \sum_{n=1}^\infty \frac{2^{2 n-3} \left|E_{2 n-1}(0)-E_{2 n-1}(1)\right|}{n (2 n-1)!} \left(1-p^{2 n-1}\right)\, \theta^{2n}$$

All terms being positive, then $$\log (\cos (p \theta))-p \log (\cos (\theta)) \geq0 \implies \cos(p\theta)\geq \cos^p(\theta)$$

Claude Leibovici
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