Analytic proof: $\frac{2}{t^2}-\frac{\frac{1}{2}\cos(\frac{t}{2})}{\sin^2(\frac{t}{2})} \geq 0$ for $t \in [0,\pi]$? I have a relatively simple question:

How can one show that $\frac{2}{t^2}-\frac{\frac{1}{2}\cos(\frac{t}{2})}{\sin^2(\frac{t}{2})} \geq 0$ $\forall t \in [0,\pi]$ or equivalently why $\frac{1}{\sin(\frac{t}{2})}-\frac{2}{t}$ (with value $0$ at $0$) is an increasing function on $[0,\pi]$ (the former being the derivative of the latter)?

This question came up while showing that the the (absolute value) integral of any subinterval in $[-\pi,\pi]$ of the Dirichlet kernel is bounded by an independent constant (later shown to be e.g. $\frac{\pi}{2}+8$). My attempts included pretty much every known inequality of the type $f(\frac{t}{2}) \leq \sin(\frac{t}{2})$ to bound the function from below and obtain an expression $\geq 0$ after some simplification, but it always seems to be just a tiny bit too coarse.
Thank you for reading and have a good day!
 A: To consider the increasing property of: $$\csc(t/2)-2t^{-1}$$On $[0,\pi]$ Is equivalent to consider the increasing property of $\csc(t)-t^{-1}$ on $[0,\pi/2]$. However, we know that: $$\sin t=t-\frac{1}{6}t^3+O(t^5)$$So that: $$\csc t=\frac{1}{t-\frac{1}{6}t^3+O(t^5)}=t^{-1}+\frac{1}{6}t+\frac{7}{360}t^3+O(t^5)$$From some polynomial division, leaving us with $$\csc t-t^{-1}=\frac{1}{6}t+\frac{7}{360}t^3+O(t^5)$$And the terms wrapped up in $O(t^5)$ are all positive. So, the function is clearly increasing!
See this for more.
A: Let $u = t/2$. It suffices to prove that, for all $u \in [0, \pi/2]$,
$$-u^2 \cos u + \sin^2 u \ge 0$$
or
$$-u^2\cos u + 1 - \cos^2 u \ge 0$$
or
$$u^4 + 4 \ge (2 \cos u + u^2)^2$$
or
$$\sqrt{u^4 + 4} - u^2 \ge 2\cos u.$$
It is easy to prove that, for all $u\in [0, \pi/2]$,
$$1 - \frac{1}{2}u^2 + \frac{1}{24}u^4 \ge \cos u.$$
(Note: Use derivatives. LHS is the 4-th degree Taylor approximation of $\cos u$ around $u = 0$.)
Thus, it suffices to prove that
$$\sqrt{u^4 + 4} - u^2
\ge 2\left(1 - \frac{1}{2}u^2 + \frac{1}{24}u^4\right)$$
or
$$\sqrt{u^4 + 4} \ge 2 + \frac{1}{12}u^4$$
or
$$u^4 + 4 \ge \left(2 + \frac{1}{12}u^4\right)^2$$
or
$$96u^4 \ge u^8$$
which is true.
We are done.
