Simple question - How $x^4 +y^4 -2(x-y)^2$ having a minimum at $(0,0)$

Question

I know this is a silly question, but I need to prove that

$$ f(x,y) = x^4 + y^4 -2 (x-y)^2 $$

is having a minima at $(0,0)$.

I tried doing maxima and minima method to it but the determinant i.e $AC- B^2= 0 $

So, we need to prove it algebraically only.

Answer 1

$f_x = 4x^3 - 4x +4y = 0, f_y = 4y^3 - 4y + 4x = 0\implies 4(x^3+y^3) = 0 \implies x = -y$ or $x^2 - xy + y^2 = 0$ which yields $x = y = 0$. If $x = -y \implies 4x^3 - 8x = 0 \implies 4x(x^2 - 2) = 0\implies x = 0, \pm \sqrt{2}\implies y = 0,\mp \sqrt{2}$. Next $f_{xx}= 12x^2 - 4$. Thus $f_{xx}(\pm \sqrt{2}) = 20 > 0$. So by the second derivative test ( see a calculus book for this test ) the min are at these points. Thus $f_{\text{min}} = f(\pm \sqrt{2}, \mp \sqrt{2})= -8 $ since $f(0,0) = 0 > -8$.

Answer 2

The minimum is not at $(0,0)$. For example, $f(x,0)=x^4-2x^2$ achieves its minimum (i.e. $-1$) when $x^2=1$.

Answer 3

$x = u+v\\y = u-v$

$f(u,v) = 2u^4 + 12u^2v^2 + 2v^4 - 4v^2$

$f(u,0) = 2u^4$ has a minimum at $(0,0)$

$f(0,v) = 2v^4 - 4v^2$ has a local maximum at $(0,v)$

$f(0,0)$ has a saddle at $(0,0)$

Answer 4

If $x=\sqrt2$ and $y=-\sqrt2$ then we get a value $-8$.

We'll prove that it's a minimal value.

Indeed, we need to prove that $$x^4+y^4-2(x-y)^2\geq-8$$ or $$x^4+y^4+8-2(x-y)^2\geq0.$$ Now, by AM-GM $$x^4+y^4+8\geq2\sqrt{8(x^4+y^4)}.$$ Thus, it remains to prove that $$2\sqrt{8(x^4+y^4)}\geq2(x-y)^2$$ or $$8(x^4+y^4)\geq(x-y)^4$$ or $$8((x^2+y^2)^2-2x^2y^2)\geq(x^2+y^2-2xy)^2$$ or $$7(x^2+y^2)^2+4(x^2+y^2)xy-20x^2y^2\geq0$$ or $$7(x^2+y^2)^2+14(x^2+y^2)xy-10(x^2+y^2)xy-20x^2y^2\geq0$$ or $$7(x^2+y^2)(x^2+y^2+2xy)-10xy(x^2+y^2+2xy)\geq0$$ or $$(x+y)^2(7x^2-10xy+7y^2)\geq0,$$ which is obvious.

Done!