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.

Using the product-to-sum identity $\sin\theta\sin\phi = \frac{1}{2}(\cos(\theta - \phi) - \cos(\theta + \phi))$ to simplify:
\begin{align*}
  & (\sin x + \cos x)(6 - \sin x) \\
= & \sqrt{2}\sin(x + \pi/4)(6 - \sin x) \\
= & 6\sqrt{2}\sin(x + \pi/4) - \sqrt{2}\sin x\sin(x + \pi/4) \\
= & 6\sqrt{2}\sin(x + \pi/4) - \sqrt2\times \frac{\cos(\pi/4) - \cos(2x+\pi/4)}{2} \\
= & 6\sqrt{2}\sin(x + \pi/4) + \frac{\sqrt{2}}{2}\cos(2x + \pi/4) - \frac{1}{2} \\
\leq & 6\sqrt{2} + \frac{\sqrt{2}}{2} - \frac{1}{2} \\
= & 8.69\ldots < 9.  
\end{align*}
