<s>[INST] 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
 [/INST] We split into two cases:
Case 1: $s \ge 4$
We have $K_s = 2$.
We have
$$\sin^s\theta + \cos^s\theta + K_s \cos^2\theta \sin^2\theta
\le \sin^4\theta + \cos^4\theta + 2 \cos^2\theta \sin^2\theta
= 1.$$
$\phantom{2}$
Case 2: $2 < s < 4$
We have $K_s = \frac14 s(s - 2)$.
Using Bernoulli inequality
$(1 + x)^r \le 1 + rx$ for all $x \ge -1, \, 0 < r \le 1$, we have
$$(\sin^2\theta)^{s/2 - 1} = (1 - \cos^2 \theta)^{s/2 - 1}\le 1 - \cos^2 \theta\cdot (s/2 - 1)$$
and
$$(\cos^2\theta )^{s/2 - 1} = (1 - \sin^2 \theta)^{s/2 - 1} \le 1 - \sin^2\theta \cdot (s/2 - 1).$$
Thus, we have
\begin{align*}
 &\sin^s\theta + \cos^s\theta + K_s \cos^2\theta \sin^2\theta \\
 =\,& \sin^2\theta\, (\sin^2\theta)^{s/2 - 1}
 + \cos^2 \theta\, (\cos^2\theta )^{s/2 - 1} + \frac14 s(s - 2)\cos^2\theta \sin^2\theta\\
 \le\,& \sin^2\theta \left[ 1 -\cos^2\theta\cdot (s/2 - 1)\right] + \cos^2\theta \left[1 - \sin^2\theta\cdot (s/2 - 1)\right] + \frac14 s(s - 2)\cos^2\theta \sin^2\theta\\
 =\,& 1 - \frac{(s - 2)(4 - s)}{4}\cos^2\theta \sin^2\theta\\
 \le\,& 1.
\end{align*}
We are done.
</s>