Optimize $x^2 + y^2 +2z^2 +z(x^2-y^2)$ subject to $x+y=2$ $$x^{2}+y^{2}+2z^{2}+zx^{2}-zy^{2}\overset{\left(x=2-y\right)}{\longrightarrow}4-4y+2y^{2}+2z^{2}+4z-4yz\rightarrow FOC: \; \begin{cases}
-4+4y-4z=0\\
4z+4-4y=0
\end{cases}\rightarrow y=1+z\rightarrow f\left(x,y,z\right)=2$$
at any point that satisfies
$$\begin{cases}
x=1-z\\
y=1+z
\end{cases}$$
Hessian $4\begin{pmatrix}1 & -1\\
-1 & 1
\end{pmatrix}$ is positive semidefinite, so I cannot claim that these points are points of strict local minimum.
Question: how to show that these points are points of (not only local but) global minimum? As far as I understand I need to prove $f(x,y,z)\geq 2$, but I cannot imagine how to deal with inequality of 3 variables.

Hint : $4-4y+2y^{2}+2z^{2}+4z-4yz$ can be rewritten as $2(1+(z-y+1)^2)$
