Minimum of $\cos^2(x) (\csc^2(\frac{\pi}{2n} - \frac{x}{n}) + \csc^2(\frac{\pi}{2n} + \frac{x}{n}))$ for $0 \leq x < \pi/2$ and $n \geq 3$ I am trying to proove that $x=0$ is the minimum point of the function
$$f(x) = \cos^2(x)\left( \csc^2(x_+) + \csc^2(x_-) \right)$$
in the interval $0 \leq x < \pi/2$ where $x_\pm = \frac\pi{2n} \pm \frac x n$,
and $n \in \mathbb{N}$ such that $n \geq 3$.
I have unsuccessfully tried the following approaches.

*

*Attempted to show that $f(x) - f(0) \geq 0$, i.e.
$$\cos^2(x)\left(
    \csc^2\left(\frac\pi{2n} - \frac x n\right) +
    \csc^2\left(\frac\pi{2n} + \frac x n\right)
\right) - 2\csc^2\left(\frac\pi{2n}\right) \geq 0;$$

*Computed $f'(x) = \frac\partial{\partial x} f(x)$ and tried to show that $f'(x) \geq 0$ in the interval, where
\begin{align}
\frac\partial{\partial x} f(x) = 2\cos^2(x) \left(
    \frac{1}{n} \cot(x_-)\csc^2(x_-) - 
    \frac{1}{n} \cot(x_+)\csc^2(x_+) -
    \tan(x)\left(\csc^2(x_-) + \csc^2(x_+)\right)
\right);
\end{align}

*Used Taylor Series Expansion in $f(x)$ attempting to find a tight lower bound for $f(x)$.

Any help or hint would be appreciated.

Trying to proove that $x=0$ corresponds to $\large \color{red}{a}$ minimum point is not so difficult if you make a Taylor expansion of
$$f_n(x)=\cos ^2(x) \left(\csc ^2\left(\frac{\pi -2 x}{2 n}\right)+\csc ^2\left(\frac{\pi+2
   x }{2 n}\right)\right)$$
$$f_n(x)=2 \csc ^2\left(\frac{\pi }{2 n}\right)\Bigg[1+ \left(\frac{1+3 \cot ^2\left(\frac{\pi }{2
   n}\right)}{n^2}-1\right)x^2+O\left(x^4\right) \Bigg]$$ So, we have
$$f_n(0)=2 \csc ^2\left(\frac{\pi }{2 n}\right)\qquad f'_n(0)=0\qquad f''_n(0)=\frac{2 \left(\left(n^2+2\right) \cos \left(\frac{\pi }{n}\right)+4-n^2\right)
   \csc ^4\left(\frac{\pi }{2 n}\right)}{n^2}$$ and
$$f''_n(0) > 0 \qquad \text{     } \qquad \forall n \quad \geq 3$$
