<s>[INST] 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.
 [/INST] Multiply the inequality by $4a^2$ then add & subtract $(a^2+b^2-c^2)y^2$
\begin{eqnarray*}
\underbrace{4a^4 x^2 +4a^2(a^2+b^2-c^2)xy +\color{red}{(a^2+b^2-c^2)^2y^2}}_{} -\color{red}{(a^2+b^2-c^2)^2y^2}+4a^2b^2 y^2>0
\end{eqnarray*}
Now complete the square 
\begin{eqnarray*}
\left(2a^2x+(a^2+b^2-c^2)y\right)^2+(4a^2b^2-(a^2+b^2-c^2)^2)y^2>0
\end{eqnarray*}
The first term is zero when $2a^2x+(a^2+b^2-c^2)y=0$ . So 
\begin{eqnarray*}
(4a^2b^2-(a^2+b^2-c^2)^2)y^2>0
\end{eqnarray*}
or
\begin{eqnarray*}
(a^2+b^2-c^2)^2-4a^2b^2<0
\end{eqnarray*}
</s>