I would like to prove that the following function $f :\mathbb{R}^2\to\mathbb{R}$ has a global minimum:


$f$ 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.

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

2 Answers 2


My favorite way,


$$\begin{align}\Delta_{\text{half}}&=(x^2-4)^2-(x^4-4x^2-8x+20)\\ &=-4(x-1)^2≤0.\end{align}$$

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


where $m=-\dfrac{b}{2a}, n=-\dfrac{\Delta}{4a}$


  • $\begingroup$ Thank you. What is $\Delta_\text{half}$? Didn't get you. $\endgroup$
    – Dennis
    Commented Apr 10, 2021 at 17:13
  • $\begingroup$ @Denni "half of discriminant" if $$ax^2+2kx+c=0$$ then we can use $\Delta=k^2-ac$ $\endgroup$ Commented Apr 10, 2021 at 17:18
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    $\begingroup$ @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. $\endgroup$
    – egreg
    Commented Apr 10, 2021 at 17:42
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    $\begingroup$ +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. $\endgroup$ Commented Apr 10, 2021 at 18:05
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    $\begingroup$ looks good, nice. $\endgroup$ 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 $f(x,y)\ge-20$.


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