Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$? Why does $a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$ imply $(a^2+b^2-c^2)^2-4a^2b^2<0$, if $x, y$ are reals greater than $1$, and $a, b, c$ are positive reals?
A proof with all the math to go from one to the other would be nice.

since you added a clause about $x,y > 1,$ some two minutes after first posting the question, the conclusion you require is false. In fact, take any $a,b$ (not both zero) at all and then take $c = 0,$ then for $x,y > 1,$
$$  a^2 x^2 + (a^2 + b^2) xy + b^2 y^2 > 0 $$
however
$$  (a^2 + b^2)^2 - 4 a^2 b^2 = (a^2 - b^2)^2 \geq 0 $$
EXAMPLE WITH NONZERO  $c...$  take $a= 20, b = 10, c = 1.$ We find
$$ \color{red}{ 400 x^2 + 499 xy + 100 y^2}  $$
is positive whenever both $x,y$ are bigger than one. HOWEVER, the discriminant is
$$ 499^2 - 4 \cdot 400 \cdot 100 = 499^2 - 400^2 = 99 \cdot 899 = 89001 > 0.  $$
Furthermore, when we take $$ x = 1, y = -2,  $$ we get
$$ 400 - 2 \cdot 499 + 4 \cdot 100 = 400 + 400 - 998 = 800 - 998 = -198. $$
The quadratic form $\langle 400, 499, 100 \rangle$ is positive when both arguments are positive, but not always.
