I would like to prove that the following function $f :\mathbb{R}^2\to\mathbb{R}$ has a global minimum:
$f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y=(x^2+y)^2-4(x^2+2x+2y)$
$f$ has strict local minimum at $f(1,3)=-20$
I think that what I need to show is that $-20$ is a lower bound of this function, and then conclude that's a global minimum, but I didn't manage to do so.
Please advise. Thank you.
On
My favorite way,
$$f(x,y)+20=y^2+y(2x^2-8)+x^4-4x^2-8x+20$$
$$\begin{align}\Delta_{\text{half}}&=(x^2-4)^2-(x^4-4x^2-8x+20)\\ &=-4(x-1)^2≤0.\end{align}$$
This means, $f(x,y)+20≥0.$
Hence, for minimum of $f(x,y)+20$, we need to take $x=1$ and $y=4-x^2$, which gives $f(x,y)+20=0.$
Finally, we deduce that
$$\min\left\{f(x,y)+20\right\}=0~ \\ \text {at}~ (x,y)=(1,3)$$
$$\min\left\{f(x,y)\right\}=-20~ \\ \text {at}~ (x,y)=(1,3)$$
where $f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y.$
Small Supplement:
Using the formula
$$ay^2+by+c=a(y-m)^2+n$$
where $m=-\dfrac{b}{2a}, n=-\dfrac{\Delta}{4a}$
$$\begin{align}y^2+y(2x^2-8)+x^4-4x^2-8x+20=(y+x^2-4)^2+4(x-1)^2≥0.\end{align}$$
You can find the stationary points: \begin{align} \frac{\partial f}{\partial x}&=4x^3+4xy-8x-8 \\[6px] \frac{\partial f}{\partial y}&=2x^2+2y-8 \end{align} At a critical point $y=4-x^2$ and also $$ x^3+x(4-x^2)-2x-2=0 $$ that is, $x=1$, that implies $y=3$.
Since clearly the function is upper unbounded on the line $y=0$, we just need to show it is lower bounded. Conjecturing that the stationary point is a minimum, we have $f(1,3)=-20$, we need to see whether $f(x,y)\ge-20$.
Now let's try completing the square in $$ y^2+2(x^2-4)y+x^4-4x^2-8x+20 $$ Since $(x^2-4)^2=x^4-8x^2+16$, we have $$ f(x,y)+20=(y+x^2-4)^2+4x^2-8x+4=(y+x^2-4)^2+4(x-1)^2 $$ which is everywhere nonnegative, so we proved that $f(x,y)\ge-20$.