Given $0 < \Theta < \frac{\pi}{2}$ and $0 < p < 1$, show that $$\cos^p{\Theta} \le \cos{p\Theta}$$
Can you please check if my proof is correct? Would also love to know if there're other ways of proving this.
Proof. Let $$f(\Theta) = \frac{\cos^p{\Theta}}{\cos{p\Theta}}$$considering $p$ fixed. The derivative is
$$f'(\Theta) = \frac{p*\cos^{p-1}{\Theta}*{(-\sin{\Theta})*\cos{p\Theta} -\cos^p{\Theta}*(-\sin{p\Theta}*p) } } {\cos^2{p\Theta}} = \frac{p*\cos^{p-1}{\Theta}*\sin{(p-1)\Theta}}{\cos^2{p\Theta}}$$
We have $f(0)=1$, and because $0<p<1$, the term $\sin(p-1)\Theta$ makes the derivative negative: $f'(\Theta) < 0$ on $(0,\frac{\pi}{2})$.
At $0$ we have $f'(0)=0$, so we cannot immediately deduce that $f$ is decreasing on all of $[0,\frac{\pi}{2})$.
However, considering that $f(0)=1, f(\frac{\pi}{2})=0$, and both $f$ and $f'$ are continuous on $[0,\frac{\pi}{2})$, if it were the case that $f$ rises above $1$ at some $\Theta \in (0,\frac{\pi}{2})$, $f$ would have to have a local maximum somewhere in $(0,\frac{\pi}{2})$, with the derivative vanishing at that point, which cannot happen. Therefore $f(\Theta) \le 1$ throughout $[0,\frac{\pi}{2}]$, Q.E.D.
Use Mean Value Theorem you can prove the following lemma:
suppose $f(x)$ is continuous on $[a,b]$ and has derivative in $(a,b)$. If $f'(x)<0$ for $x\in(a,b)$, then $f(x)$ is strictly decreasing on $[a,b]$.
So you need only $f'(\Theta) < 0$ for $\Theta\in(0,\pi/2)$ in your situation!
Also we can deal this one as follows: For fix $p\in(0,1)$, let $$f(x)=\cos px-\cos^px,x\in[0,\pi/2],$$ then $f$ is continuous on $[0,\pi/2]$ and is differentiable in $(0,\pi/2)$. Its derivative is $$f'(x)=p\sin x\left(\cos^{p-1}x-\frac{\sin px}{\sin x}\right),x\in(0,\pi/2).$$ Because $p\in(0,1)$, $0<px<x<\pi/2$, we have $$0<\frac{\sin px}{\sin x}<1<\frac{1}{\cos^{1-p}x}=\cos^{p-1}x.$$ So $$f'(x)>0,\ \forall x\in(0,\pi/2).$$ This implies $$f(x)>f(0)=0,\forall x\in(0,\pi/2).$$