Prove that \[ (\cos\theta)^p\le \cos(p\theta) \] for \(0\le\theta\le\pi/2\) and \(0<p<1\).
Solution
1. The function\[ \phi(x)=\log(\cos x) \]is strictly concave on \([0,\pi/2)\) and satisfies \(\phi(0)=0\).
For \(0\leq x<\pi/2\), \[ \phi''(x)=-\sec^2x<0. \] Hence \(\phi\) is strictly concave on this interval. Also \[ \phi(0)=\log1=0. \]
2. If \(0\leq\theta<\pi/2\) and \(0<p<1\), then\[ p\log(\cos\theta)\leq\log(\cos(p\theta)). \]
Since \[ p\theta=(1-p)\cdot0+p\theta, \] concavity from step 1 gives \[ \begin{aligned} \phi(p\theta) &\geq (1-p)\phi(0)+p\phi(\theta)\\ &= p\phi(\theta). \end{aligned} \] Substituting the definition of \(\phi\) gives the claim.
3. If \(0\leq\theta<\pi/2\), then\[ (\cos\theta)^p\leq\cos(p\theta). \]
Exponentiating the inequality in step 2 gives \[ \exp\bigl(p\log(\cos\theta)\bigr) \leq \exp\bigl(\log(\cos(p\theta))\bigr), \] which is exactly the asserted inequality.
4. The inequality also holds when \(\theta=\pi/2\).
Since \(0<p<1\), \[ 0<\frac{p\pi}{2}<\frac{\pi}{2}, \] and therefore \[ (\cos(\pi/2))^p=0 \leq \cos(p\pi/2). \]
5. Q.E.D.
Steps 3 and 4 cover the entire interval \(0\leq\theta\leq\pi/2\).