I want to show that $\cos(2y + z) > 0 $ for all $y,z \in \mathbb{R}$ which satisfies $y^2 + z^2 < \left( \frac{\pi}{6} \right)^2$. To do this I thought of solving the max problem \begin{align*} & \max_{y, z} \cos(2y + z) \\ & \text{subject to} \\ & y^2 + z^2 < \left( \frac{\pi}{6} \right)^2 \end{align*} It is easy to see that the maximum value $y$ can attain is when $z = 0$, i.e. $y = \pm \pi/6$. Simairly for $z$ when $y = 0$, i.e. $z = \pm \pi/6$. Therefore, the highest possible value $cos(2y + z)$ can attain is $\cos(2 \pi/6 + \pi/6) = \cos(\pi/2) = 0$, however then our constraint will not be satisfied. Therefore, we need to decrease either $y, z$ or both in order for the constraint to be satisfied, but then $2y+z$ will only attain possible values in some interval $[-k,k] \subseteq [-\pi/2, \pi/2]$ where $\cos(2y + z ) \geq 0$.
Does this proof make sense/is suifficient? I guess I could also do it with Lagrange mulitpliers.
Thanks in advance for any help.