My problem is to prove
$$
\left(\cos\frac{m}{2n}\pi\right)^{4n}\ge \left(\cos\frac{m+1}{2n}\pi\right)^{2n-1}
\left(\cos\frac{m-1}{2n}\pi\right)^{2n+1}
$$
holds for any positive integer $n$ and $m = 1, 2, \dots, n-1$.



An inequality involving the power of cosine
A bit more general inequality is
\begin{equation\*}
 g(h):=g\_x(h):=\ln\frac{\cos ^{1+h}(\pi (x-h)) \cos ^{1-h}(\pi (x+h))}{\cos ^2(\pi x) }\le0 \tag{1}
\end{equation\*}
for $x\in(0,1/2)$ and $h\in(0,\min(x,1/2-x))$.


Suppose that indeed $x\in(0,1/2)$ and $h\in(0,\min(x,1/2-x))$, so that $h\in(0,1/4)$.
Note that
\begin{equation\*}
 \frac{g''(h)}\pi=f(\pi (x-h))+f(\pi (x+h))-\frac\pi2 c,
\end{equation\*}
where
\begin{equation\*}
 f(u):=2 \tan u-\frac\pi2 \sec ^2u=(\sin 2u-\frac\pi2)\sec ^2u
\end{equation\*}
and
\begin{equation\*}
 c:=(1+2 h) \sec ^2(\pi (x-h))+(1-2 h) \sec ^2(\pi (x+h)).
\end{equation\*}
Obviously, $f(u)<0$ for $u\in(0,1/2)$ and $c>0$. So, $g''<0$ and hence the function $g$ is concave. Also, $g(0+)=g'(0+)=0$. So, (1) follows.


