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

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!
