Proving that $(\cos \theta )^p\leq (\cos p\, \theta )$ for $0\leq\theta\leq \pi/2$ and $0<p<1$ through an alternative method?

Question

I'm reading the Berkeley Problems in Mathematics book:

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

I could find other ways to prove it, but this time I wanted to probe something: I've treated them as functions and made some plots of them with Mathematica - I'm looking for a proof envolving the differences of both sets.

   ListLinePlot[
   Table[Cos[x]^p, {x, 0, \[Pi]/2, 0.01}, {p, 0, 1, 0.01}], 
   ImageSize -> 400, PlotRange -> {{0, 100}, {0, 1}}];

$ $

   ListLinePlot[
   Table[Cos[p x], {x, 0, \[Pi]/2, 0.01}, {p, 0, 1, 0.01}], 
   ImageSize -> 400, PlotRange -> {{0, 100}, {0, 1}}];

Respectively, the results were:

I can see that the second plot takes the some of the values of the first plot, but the first plot takes some smaller values that are not present on the second. From here, I can conclude that they would be equal if the plots were equal, they would be different $(<,>)$ if the plots had no intersection, such as this sketch:

And they would be $(\leq,\geq)$ if there is some intersection between then but $A\cap C \neq A$ and $A\cap C \neq C$, such as in this sketch:

My question is: Could this be used as a proof? I am aware of some pitfalls that may happen with visual reasoning, I'm just not sure if I could fall in one of these here. Notice I took the care of making the plots visually proportional with PlotRange -> {{0, 100}, {0, 1}}.

Note: I'm not looking for some traditional way of proving it, I guess I have an idea on how to do it via the traditional way, I'm curious about possible flaws in the reasoning I've provided.

Sorry if it's too naive/confusing, I'm trying to think and It's a complete mess.

Answer 1

I think that is a nice idea. Translated in the analytical language, your argument could base on convexity:

We fix a $\theta \in (0,\frac{\pi}{2})$ and write both functions as functions of $p$: $$f_\theta (p) = cos(\theta)^p\\g_\theta(p) = cos(\theta p).$$

• For $p=0,1$ we have $f_\theta(p) = g_\theta(p)$.

• $f_\theta$ is a convex function in $p$. You can check that by $$\frac{\partial^2 f_\theta}{(\partial p)^2}(p) = \cos(t)^p\log(\cos(t))^2>0$$.

• $g_\theta$ is a concave function because $$\frac{\partial^2 g_\theta}{(\partial p)^2}(p)=-t^2\cos(tp)<0$$.

In other words:

$f_\theta$ is convex and thus always below the straight line through $(0,f_\theta(0))=(0,1)$ and $(1,f_\theta(1))=(1,\cos(\theta))$ while $g_\theta$ is concave and thus always above. So clearly the two never intersect and $f_\theta$ is always below $g_\theta.$