TITLE: Proof: $\cos^p (\theta) \le \cos(p\theta)$
QUESTION [15 upvotes]: I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!
Enough backstory. Time for math. The problem is as follows:
Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.
I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.
Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions!

REPLY [2 votes]: $$\color{blue}{\cos^p(\theta)}\le\color{green}{\cos(p\theta)}.$$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
For fixed $p$ ($0.5$ in the figure), take the logarithm
$$p\log(\cos(\theta))\le\log(\cos(p\theta)),$$
and derive on $\theta$
$$-p\tan(\theta)\le-p\tan(p\theta).$$
The latter inequality is obviously true as the $\tan$ function is increasing.
This shows that the $\color{blue}{LHS}$ of the original inequality decreases faster than the $\color{green}{RHS}$, and they are equal at $\theta=0$.