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
 A: Not an answer.
I cannot access the paper but it seems to me that, if $s >2$ is an integer
$$\sin^s(\theta) + \cos^s(\theta) \le 1- K_s \cos^2(\theta)\sin^2(\theta)$$ is satisfied for
$$K_s=4-2^{3-\frac{s}{2}}$$ For $s=4$ and $s=6$, it is an equality but there is one exception (that I do not understand) for $s=5$.
Trying to confirm, searching for the minimum value of
$$f(s)=\frac{1-\big[\sin^s(\theta) + \cos^s(\theta) \big]}{\cos^2(\theta)\sin^2(\theta)}$$ leads to the same conclusion (the minimum occuring most of the time for $\theta=\frac \pi 4$).

A: Two cases:

*

*case $s \ge 4$.
In this case, we have $K_s=2$ which means that the inequality to be established is
$$\sin^s(\theta) + \cos^s(\theta) \le 1- 2 \cos^2(\theta)\sin^2(\theta)\tag{1}$$
(1) is a consequence of the following identity:
$$\sin^4(\theta) + \cos^4(\theta) = (\underbrace{\sin^2(\theta) + \cos^2(\theta)}_1)^2 - 2\cos^2(\theta)\sin^2(\theta), \tag{2}$$
taking into account the fact that $\sin^s(\theta) \le sin^4(\theta)$ and  $\cos^s(\theta) \le cos^4(\theta)$ for $s \ge 4$.

2) Case $2 \le s \le 4$
where the inequation to be established is
$$\sin^s(\theta) + \cos^s(\theta) \le 1- \frac{s(s-2)}{4} \cos^2(\theta)\sin^2(\theta)\tag{3}$$
First of all, inequality (3) is tight as attested by this figure obtained with Geogebra:

Fig 1: Polar plot representation of (2) for $s=3.75$. Green curve for the LHS ; red curve for the RHS.
Let us write the LHS of (3) under the following form (using generalized binomial expansion $(1+X)^{\alpha}=1+\alpha X+...$):
$$\underbrace{(1-\cos^2 \theta)^{s/2}+(1-\cos^2 \theta)^{s/2}}_{LHS}=$$
$$(1-\tfrac{s}{2}\cos^2 \theta+\tfrac12 \tfrac{s}{2}(\tfrac{s}{2}-1)\cos^4 \theta + ...)+(1-\tfrac{s}{2}\sin^2 \theta+\tfrac12 \tfrac{s}{2}(\tfrac{s}{2}-1)\sin^4 \theta + ...)=$$
$$=2-\tfrac{s}{2}+\tfrac12 \tfrac{s}{2}(\tfrac{s}{2}-1)(\cos^4 \theta+\sin^4 \theta) +...=$$
which is equal, using (2), to:
$$=2-\tfrac{s}{2}+\tfrac12 \tfrac{s}{2}\tfrac{s-2}{2}(1- 2\cos^2(\theta)\sin^2(\theta))+...=$$
$$=2-\tfrac{s}{2}+\tfrac12 \tfrac{s}{2}\tfrac{s-2}{2}- \tfrac{s}{2}(\tfrac{s}{2}-1)\cos^2(\theta)\sin^2(\theta))+...=$$
$$=\underbrace{1- \tfrac{s}{2}(\tfrac{s}{2}-1)\cos^2(\theta)\sin^2(\theta))}_{RHS}+\underbrace{1-\tfrac{s}{2}+\tfrac12 \tfrac{s}{2}\tfrac{s-2}{2}+...}_{R_s}\tag{4}$$
The structure of (4):
$$\text{LHS=RHS + R}_s$$
can be seen at two levels. If we don't take into account the dots (i.e., remain  with a second order approximation), we have to show that $R_s=\tfrac18 (s^2-6s+8) \ge 0$, which is true in the domain $2 \le s \le 4$.
But this is not a rigorous proof: we have to consider all the terms at any order...
One thing is very likely : inequation (3) has been found in this way (2nd order approximation).
A: $\newcommand{\d}{\mathrm{d}}$
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 $\min\{\tfrac{s(s-2)}{4},2\}\leq \min\{4-2^{3-s/2},\tfrac{s}{2}\}$.
A: 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.
