# Proving global minimum by lower bound of 2-variable function $f(x,y)=x^4+2x^2y+y^2-4x^2-8x-8y$

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.

• $x^2-4x^2$? Are you sure there is no typo? Commented Apr 10, 2021 at 16:38
• @lonestudent you are right, fixed it! Commented Apr 10, 2021 at 16:43

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}

• Thank you. What is $\Delta_\text{half}$? Didn't get you. Commented Apr 10, 2021 at 17:13
• @Denni "half of discriminant" if $$ax^2+2kx+c=0$$ then we can use $\Delta=k^2-ac$ Commented Apr 10, 2021 at 17:18
• @Michael If one proves that $f(x,y)\ge-20$, then the function is lower bounded. The technique used here is a bit contrived, but not wrong. Commented Apr 10, 2021 at 17:42
• +1: (also) to lone student's answer. @Michael I agree that although lone student's analysis is both accurate and valid, I (for one) found it confusing. I would have written it as $$f(x,y) + 20 = y^2 + y(2x^2 - 8) + (x^4 - 4x^2 - 8x + 20)$$ which equals $$[y + (x^2 - 4)]^2 + 4(x-1)^2.$$ Therefore, $[f(x,y) + 20]$ must have a global minimum when both $$y + (x^2 - 4) = 0 ~~\text{and}~~ (x-1) = 0.$$ lone student is welcome to use this comment to clarify his answer, if he wishes. Commented Apr 10, 2021 at 18:05
• looks good, nice. Commented Apr 11, 2021 at 0:06

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 $$f(x,y)\ge-20$$.