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.
There are a few difficulties in figuring out what is meant in the given solution.
A tool is given $$ ax^2+bx+c=a(x-m)^2+n $$ $$ m=-\frac{b}{2a},~n=-\frac{\Delta}{4a} $$ It is left to the reader to figure out that $\Delta=b^2-4ac$. It was not clear to me, at first, that this would be used on the previous line $$ f(x,y):=\!\!\!\!\overset{\substack{a=1\\\downarrow}}{\vphantom{\sum}}\!\!\!\!x^2+x\overbrace{(\alpha y+1)\vphantom{y^2}}^b+\overbrace{\ 2y+y^2\ }^c $$ For an exposition, it would be nice to
\tagthese lines and reference them as they are used.Other than a typo or two, I think the method is sound.
When $|\alpha|\lt2$, the equation has been reduced to a square plus a constant. The square is minimized at $0$. What else needs to be shown?
That being said, here is how I would proceed.
Expand $$ \begin{align} &(x+dy+e)^2+(y+dx+f)^2\\[6pt] &=\left(1+d^2\right)\left(x^2+y^2\right)+4dxy+2(e+df)x+2(f+de)y+e^2+f^2\tag1 \end{align} $$ Since $$ d=\frac{2-\sqrt{4-a^2}}a\implies\frac{4d}{1+d^2}=a\tag2 $$ we have $$ \begin{align} &\frac{(x+dy+e)^2+(y+dx+f)^2}{1+d^2}\\[6pt] &=x^2+y^2+axy+\frac{2(e+df)}{1+d^2}x+\frac{2(f+de)}{1+d^2}y+\frac{e^2+f^2}{1+d^2}\tag3 \end{align} $$ Since $$ \left. \begin{array}{} \displaystyle e=\frac{b-cd}2\frac{1+d^2}{1-d^2}\\ \displaystyle f=\frac{c-bd}2\frac{1+d^2}{1-d^2} \end{array} \right\}\implies \left\{ \begin{array}{} \displaystyle \frac{2(e+df)}{1+d^2}=b\\ \displaystyle \frac{2(f+de)}{1+d^2}=c\\ \displaystyle \frac{e^2+f^2}{1+d^2}=\frac{b^2+c^2-abc}{4-a^2} \end{array} \right.\tag4 $$ we have $$ \frac{(x+dy+e)^2+(y+dx+f)^2}{1+d^2}=x^2+y^2+axy+bx+cy+\frac{b^2+c^2-abc}{4-a^2}\tag5 $$ Therefore, $$ \bbox[5px,border:2px solid #C0A000]{x^2+y^2+axy+bx+cy\ge\frac{abc-b^2-c^2}{4-a^2}}\tag6 $$ where equality is achieved at $$ \begin{align} x &=\frac{fd-e}{1-d^2}\tag{7a}\\ &=\frac{2cd-b\left(1+d^2\right)}{2\left(1-d^2\right)}\frac{1+d^2}{1-d^2}\tag{7b}\\ &=\frac{ac-2b}{4}\left(\frac{1+d^2}{1-d^2}\right)^2\tag{7c}\\[6pt] &=\frac{ac-2b}{4-a^2}\tag{7d} \end{align} $$ and $$ \begin{align} y &=\frac{de-f}{1-d^2}\tag{8a}\\ &=\frac{2bd-c\left(1+d^2\right)}{2\left(1-d^2\right)}\frac{1+d^2}{1-d^2}\tag{8b}\\ &=\frac{ab-2c}{4}\left(\frac{1+d^2}{1-d^2}\right)^2\tag{8c}\\[6pt] &=\frac{ab-2c}{4-a^2}\tag{8d} \end{align} $$ Explanation:
$\text{(7a) and (8a)}$: set the left side of $(5)$ to $0$ and solve
$\text{(7b) and (8b)}$: apply the left side of $(4)$
$\text{(7c) and (8c)}$: apply the right side of $(2)$
$\text{(7d) and (8d)}$: the right side of $(2)$ says that $4-a^2=4\left(\frac{1-d^2}{1+d^2}\right)^2$
Case $\boldsymbol{|a|\lt2}$
Applying $(6)$ to this question for $|a|\lt2$ gives a global minimum of $$ x^2+y^2+axy+bx+cy\ge\frac{abc-b^2-c^2}{4-a^2}=\frac{2a-5}{4-a^2}\tag9 $$ Case $\boldsymbol{|a|\ge2}$
If $|a|\ge2$, then along the line $y=-\frac2ax$, $$ \begin{align} x^2+y^2+axy+bx+cy &=\left(x+\frac a2y\right)^2-\left(\frac{a^2}4-1\right)y^2+bx+cy\tag{10a}\\ &=-\frac{a^2-4}{a^2}x^2+\left(b-\frac{2c}a\right)x\tag{10b} \end{align} $$ which, if $|a|\gt2$, tends to $-\infty$ as $|x|\to\infty$.
If $|a|=2$ and $b\ne\frac{2c}a$ (as in the question), $\text{(10b)}$ is also unbounded below.
If $|a|=2$ and $b=\frac{2c}a$, then $x^2+y^2+axy+bx+cy=0$.