Showing $\sin^s\theta+\cos^s\theta\le1-K_s\cos^2\theta\sin^2\theta$, for $s>2$ and $K_s=\min\{\frac14s(s-2),2\}$ I was reading through an article (X. Lu, B. Wennberg, Solutions with increasing energy for the spatially homogeneous Boltzmann equation (2002)) and this elementary inequality is used in a proof:

For $s>2$ and $\theta\in [0,\pi/2]$,
$$\sin^s(\theta) + \cos^s(\theta) \le 1- K_s \cos^2(\theta)\sin^2(\theta)$$
where $K_s=\min\left\{\frac{1}{4}s(s-2), 2 \right\}$.

I'm trying to show it but I'm not succeeding.
I tried to use another another elementary inequality found in the same proof: for $s>2$ and $a,b \ge 0$
$$a^s+b^s \le (a+b)^s \le a^s + b^s + 2^s(a^{s-1}b+ab^{s-1}),$$
but I didn't succeed as well. Do you have any suggestions? Thank you in advance

$\newcommand{\d}{\mathrm{d}}$
Let $x=\cos^2 \theta$. Over $0\leq x \leq 1$, we want to uniformly bound
$$\begin{split}
\frac{1 - \cos^s \theta - \sin^s\theta}{\cos^2 \theta \sin^2 \theta}
&= \frac{1 - x^{s/2} - (1-x)^{s/2}}{x(1-x)} \\
&= \frac{1 - x^{s/2 -1}}{1 - x} + \frac{1 - (1-x)^{s/2-1}}{x} \\
&= (\tfrac{s}{2}-1)\int_0^1\left((1-tx)^{s/2-2}+(1-t(1-x))^{s/2-2}\right)\d t
\end{split}$$
If $2<s\leq 4$ or $6\leq s$, then $u^{s/2 -2}$ is convex, so that by Jensen's inequality we have
$$\begin{split}
(\tfrac{s}{2}-1)\int_0^1\left((1-tx)^{s/2-2}+(1-t(1-x))^{s/2-2}\right)\d t
&\geq 2(\tfrac{s}{2}-1)\int_0^1 (1-\tfrac{t}{2})^{s/2 - 2} \d t \\
&= 4 - 2^{3-s/2}\text{.}
\end{split}$$
In this case, the inequality is an equality at $x=\tfrac{1}{2}$.
If $4\leq s \leq 6$, then $u^{s/2 -2}$ is concave, so it is bounded below by the secant from $1$ to $u$:
$$\begin{split}
(\tfrac{s}{2}-1)\int_0^1\left((1-tx)^{s/2-2}+(1-t(1-x))^{s/2-2}\right)\d t
&\geq (\tfrac{s}{2}-1)\int_0^1 (1 + (1-t)^{s/2 - 2}) \d t \\
&= \tfrac{s}{2}\text{.}
\end{split}$$
In this case, the inequality is an equality at $x\in\{0,1\}$.
We can summarize the established minimum as
$$\begin{split}\min\frac{1-\cos^s\theta -\sin^s\theta}{\cos^2 \theta \sin^2\theta} &=
\begin{cases}
4-2^{3-s/2} & 2 < s\leq 4 \\
\tfrac{s}{2} & 4 \leq s \leq 6 \\
4 - 2^{3-s/2} & 6\leq s
\end{cases}\\
&=\min\{4-2^{3-s/2},\tfrac{s}{2}\}\text{,}\end{split}$$
and one can show that $\min\{\tfrac{s(s-2)}{4},2\}\leq \min\{4-2^{3-s/2},\tfrac{s}{2}\}$.
