<s>[INST] Prove that $(\sin x + \cos x)(6 - \sin x)<9$ Is there any elementary way to prove that $(\sin x + \cos x)(6 - \sin x)<9$?
I've noticed that $(\sin x + \cos x)$ has to be positive so $x \in\left(-\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right)$ and then $(6 - \sin x)\in\left(6-\dfrac{\sqrt2}{2},6+ \dfrac{\sqrt2}{2}\right)$ but since $(\sin x + \cos x) \in\left(0, \sqrt2\right]$ the maximum possible value of $(\sin x + \cos x)(6 - \sin x)$ can be $(6+ \dfrac{\sqrt2}{2}) * \sqrt2 = 1+6\sqrt2$ which is greater than $9$.
My second attempt was to find the derivative, but I couldn't find its roots.
 [/INST] Proof 1: Note that $\sin x + \cos x = \sqrt 2 \, \sin (x + \pi/4) \le \sqrt 2$
and $6 - \sin x > 0$.
If $\sin x \ge 0$, we have
$(\sin x + \cos x)(6 - \sin x) \le \sqrt 2 \cdot (6 - \sin x)
\le \sqrt 2 \cdot 6 < 9$.
If $\sin x < 0$, we have
$\sin x + \cos x \le \cos x \le 1$ and thus $(\sin x + \cos x)(6 - \sin x) 
\le 6 - \sin x \le 7 < 9$.
We are done.
$\phantom{2}$
Proof 2: $(a - b)^2 \ge 0$ yields $ab \le \frac{(a + b)^2}{4}$ for all real numbers $a, b$.
We have
\begin{align*}
 (\sin x + \cos x)(6 - \sin x) &= \frac14 \cdot 4(\sin x + \cos x)\cdot (6 - \sin x)\\
  &\le \frac14 \cdot \frac{(4\sin x + 4\cos x + 6 - \sin x)^2}{4}\\
  &=\frac{(3\sin x + 4\cos x + 6)^2}{16} \\
  &\le \frac{(5 + 6)^2}{16}\\
  & = \frac{121}{16} = 7.5625
\end{align*}
where we have used $(3\sin x + 4\cos x)^2
\le (3^2 + 4^2)(\sin^2 x + \cos^2 x) = 25$ (C-S inequality)
to get $-5 \le 3 \sin x + 4\cos x \le 5$.
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