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.

Below is old incorrect answer ignoring $x>1$ condition. Explanation:
The discriminant is less than $0$ iff for all $x$ the quadratic has no roots.
However, we only know the LHS has no roots for all $x>1$ so this cannot be used. In fact, the counterexample pointed out by @WillJagy shows this.

$$a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0$$
$\implies$ (discriminant for LHS quadratic in $x$ is less than $0$ as the LHS has no roots)
$$(a^2+b^2-c^2)^2y^2-4a^2b^2y^2<0$$
$\implies$
$$(a^2+b^2-c^2)^2-4a^2b^2<0$$
