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.

$\begin{equation} \begin{split}
(\sin x + \cos x)(6 - \sin x) & = 6 \cos x + 6 \sin x - \sin^2 x - \sin x \cos x \\ &= 6 \cos x + 6 \sin x - \dfrac{1}{2} (1 - \cos(2x) ) - \dfrac{1}{2} \sin(2x)\\
&= 6 \cos x + 6 \sin x + \dfrac{1}{2} (\cos(2x) - \sin(2x) ) - \dfrac{1}{2} \\
&\le 6 \sqrt{2} + \dfrac{1}{2} \sqrt{2} - \dfrac{1}{2} = 8.692388 \\ &\lt 9 \end{split} \end{equation} $
