Prove the following inequality without using induction: $\frac{1}{2^k-1}\leq \sin^{2k}\theta+\cos^{2k}\theta\leq 1$ 
How to prove the following inequality (without using induction)? 
  $$\frac{1}{2^k-1}\leq \sin^{2k}\theta+\cos^{2k}\theta\leq 1,\quad k\in\Bbb N.$$

 A: We have to assume $k>0$ or the term $1/(2^k-1)$ would be meaningless. The case $k=1$ is trivial, so I'll assume $k>1$.
Consider, for $k\ge0$, the function
$$
f_k(\theta)=\sin^{2k}\theta+\cos^{2k}\theta
$$
which has period $\pi/2$. Its derivative is
\begin{align}
f_k'(\theta)
&=2k\sin^{2k-1}\theta\cos\theta-2k\cos^{2k-1}\theta\sin\theta\\[3px]
&=2k\sin\theta\cos\theta(\sin^{2k-2}\theta-\cos^{2k-2}\theta)\\[3px]
&=k\sin(2\theta)(\sin^{2k-2}\theta-\cos^{2k-2}\theta)
\end{align}
We need to find where the derivative vanishes. It happens at $\theta=0$, and $\theta=\pi/2$ or at points where
$$
\sin^{2k-2}\theta-\cos^{2k-2}\theta=0
$$
that is, $\tan\theta=\pm1$, that is $\theta=\pi/4$ (in the interval of periodicity).
So we have the critical points $0$, $\pi/4$ and $\pi/2$. We have
\begin{align}
f_k(0)&=1\\
f_k(\pi/4)&=\frac{1}{2^k}+\frac{1}{2^k}=\frac{1}{2^{k-1}}
\end{align}
Since
$$
\frac{1}{2^k-1}<\frac{1}{2^{k-1}}
$$
for $k>1$, the result is proved.
A: for Left hand ,we  use Holder inequality
$$(\sin^{2k}{\theta}+\cos^{2k}{\theta})(1+1)^{k-1}\ge (\sin^2{\theta}+\cos^2{\theta})^k=1$$
so
$$\sin^{2k}{\theta}+\cos^{2k}{\theta}\ge\dfrac{1}{2^{k-1}}$$
not $\dfrac{1}{2^k-1}$
