Let U be the following set $$U=\{\left(x, y, z\right)\in\mathbb R^3 \ | \ x^2+y^2+z^2<\left(\frac{\pi}{6}\right)^2\} $$ we define the function $f : U\rightarrow \mathbb R$ as $$f\left(x, y, z\right)=x^2+y^2-\cos\left(2y+z\right)$$
Is $f$ strictly convex?
The way I want to show it, is by showing the following $$v^t Av>0$$ for every $v\in\mathbb R^n \backslash \{0\}$ where $A$ is a symmetric matrix. If the above hold, the matrix $A$ is positive definite and we then have a theorem that says that $f$ is strictly convex.
I set A to be the Hessian matrix, so we get the following: $$\begin{pmatrix} x \\ y \\ z \end{pmatrix}^t \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2+4\cos\left(2y+z\right) & 2\cos\left(2y+z\right) \\ 0 & 2\cos\left(2y+z\right) & \cos\left(2y+z\right) \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}$$ $$=2x^2+4y^2\cos\left(2y+z\right)+2y^2+4yz\cos\left(2y+z\right)+z^2\cos\left(2y+z\right)$$ $$=2x^2+2y^2+\cos\left(2y+z\right)\left(4y^2+4yz+z^2\right)$$ $$=2x^2+2y^2+\cos\left(2y+z\right)\left(2y+z\right)^2$$ it is clear to see that $2x^2, 2y^2, \left(2y+z\right)^2>0$ for any value of $x, y, z$. I know that for any value $x\in\left(-\frac{\pi}{2}+2\pi n,\frac{\pi}{2}+2\pi n\right)$, then $\cos\left(x\right)>0$. Im struggling to find out whether $\cos(2y+z)$ is positve or negative when we have the set $U$.
Thanks in advance