Finding the global minimum of the multivariable function using only algebraic tools Problem says:

Find the global minimum of
$$\begin{align}f(x,y): &= x^2 + y^2 + \alpha xy + x + 2y\end{align}$$
where, $\alpha\in\mathbb R$.


The things I have done:
Let,
$$f(x,y):=x^2+x(\alpha y+1)+2y+y^2$$

*

*Algebraic tool I will use along the way:

$$ax^2+bx+c=a(x-m)^2+n$$
$$m=-\frac{b}{2a},~n=-\frac{\Delta}{4a}$$

In this case, we have
$$\begin{align}m:&=-\frac{\alpha y+1}{2} \\
 n:&=\frac 14 \left(y^2(4-\alpha ^2) + 2y(4-
\alpha)-1\right)&\end{align}$$
Then,
$$\begin{align}f(x,y):=\left(x+\frac{\alpha y+1}{2}\right)^2+\frac 14 \left(y^2(4-\alpha ^2) + 2y(4-\alpha)-1\right)\end{align}$$
If $\alpha =±2$ , putting
$$x=-\frac{\alpha y+1}{2}$$
we get
$$f(x,y):=\frac{y(4-\alpha)}{2}-\frac 14$$
Since $4-\alpha>0$ for $\alpha=±2$, we observe that the lower bound of  $\frac{y(4-\alpha)}{2}-\frac 14$ doesn't exist (applying $y\to -\infty$).
If $|\alpha|>2$, then applying the same method, we have
$$\begin{align}\frac 14 \left(y^2(4-\alpha ^2) + 2y(4-\alpha)-1\right)=y^2\left(1-\frac{\alpha ^2}{4}\right)+y\left(2-\frac{\alpha}{2}\right)-\frac 14=\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
So, we obtain
$$\begin{align}f(x,y):=\left(x+\frac{\alpha y+1}{2}\right)^2+\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
Likewise, if $|\alpha|>2$ then putting $$x=-\frac{\alpha y+1}{2}$$ we get,
$$\begin{align}f(x,y):=\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
Observing that,
$$1-\frac{\alpha^2}{4}<0$$ where $|\alpha|>2$. This means, if $|\alpha|>2$, then the lower bound of
$$\begin{align}\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
doesn't exist. (applying $y\to +\infty$)
Finally, if $|\alpha|<2\iff -2<\alpha<2$, then $1-\frac{\alpha^2}{4}>0$ and putting
$$x=-\frac{\alpha y+1}{2}, ~~y=\frac{\alpha-4}{4- \alpha^2}$$
in the original function
$$\begin{align}f(x,y):=\left(x+\frac{\alpha y+1}{2}\right)^2+\left(1-\frac{\alpha^2}{4}\right)\left(y+\frac{4-\alpha}{4-\alpha ^2}\right)^2+\frac{2 \alpha - 5}{4 - \alpha^2}\end{align}$$
we conclude
$$\begin{align}\min \left\{x^2 + y^2 + \alpha xy + x + 2y {\large{\mid}} -2<\alpha<2\right\}=\frac{2\alpha-5}{4-\alpha^2} ~\text {at}~ x=\frac{2-2\alpha}{\alpha^2-4} ~\text{and}~ y=\frac{\alpha-4}{4- \alpha^2}. \end{align}$$

Questions:

*

*Is the algebraic method I use rigorous enough?  Are there still non-rigorous points in my steps?


*What I'm doing is just finding the minimum of the given function.  How can I show that the minimum I found is a global minimum?
Thank you for reviewing.

Hint.
Another approach is to include $f(x,y)$ in the quadratic class as
$$
f(x,y) = (x-x_0,y-y_0)'M(x-x_0,y-y_0)+c
$$
Choosing $M = \left(
\begin{array}{cc}
 1 & \frac{\alpha }{2} \\
 \frac{\alpha }{2} & 1 \\
\end{array}
\right)$ and equating coefficients we arrive at
$$
\cases{
x_0 = -\frac{2 (\alpha -1)}{\alpha ^2-4}\\
y_0 = \frac{4-\alpha }{\alpha ^2-4}\\
c = \frac{5-2 \alpha }{\alpha ^2-4}
}
$$
and now follow the considerations about $M(\alpha)$ avoiding the case with $\alpha = \pm 2$
