问题及解答

证明: $(\cos\theta)^p\leqslant\cos(p\theta)$, 这里 $\theta\in[0,2\pi]$, $0 < p <1$.

Berkeley Problems in Mathematics, Third Edition. (伯克利数学问题集)

令 $f(\theta)=\cos(p\theta)-(\cos\theta)^p$, 则 $f(0)=0$. 设 $\theta\in[0,\frac{\pi}{2})$.

\[
\begin{split}
f'(\theta)&=-\sin(p\theta)\cdot p-p(\cos\theta)^{p-1}\cdot(-1)\sin\theta\\
&=p\Bigl(-\sin(p\theta)+\frac{\sin\theta}{\cos^{1-p}\theta}\Bigr).
\end{split}
\]

由于 $0 < p < 1$, 对于 $\theta\in(0,\frac{\pi}{2})$, $0 < \cos^{1-p}\theta < 1$. $\sin\theta > \sin(p\theta)$. 故

\[
-\sin(p\theta)+\frac{\sin\theta}{\cos^{1-p}\theta} > 0.
\]

即 $f'(\theta) > 0$. 因此, $f(\theta)\geqslant 0$, 对任意 $\theta\in[0,\frac{\pi}{2}]$.