Let $x, y \in B_1(0)\subset \mathbb{C}$. I am trying to show that $|x|^2 + |y|^2 < 1 + |xy|^2$. This is an inequality that I should be able to prove, but apparently I am unable to do so in a rigorous way. My work so far is the following:
We can write the expression as
$$1 - |x|^2\left(1 - |y|^2\right) - |y|^2 > 0$$
which represents a parabola after the variables have been converted so real ones, as $|x| \in [0, 1)$, we do not really care about the complex properties of the numbers. We can see that the polynomial has two roots, namely at $x = \pm 1$ (in terms of x) and these are all the roots of the polynomial. Since the highest order monomial has a negative coefficient, the parabola opens down and hence it has to be positive on $|x| \in [-1, 1]$.
You might see what I meant with unable to do so in a rigorous way, since this argument relies on the intuitive properties of parabolas from HS level mathematics.
(Question:) So, how could one massage the expression to prove the strict inequality (or even inequality in the case that $x, y \in \overline{B_1(0)}$)?
$|x|^2 + |y|^2 < 1 + |xy|^2$ is equivalent to $$ 0 < (1-|x|^2)(1-|y|^2) $$ and that is true for all $x, y \in B_1(0)$ since both factors on the right are strictly positive.
For $x, y \in \overline{B_1(0)}$ the same argument shows that $|x|^2 + |y|^2 \le 1 + |xy|^2$.