Prove $\lim_{n\to\infty}(|x_n + y_n| - |x_n - y_n|) = +\infty \iff \lim_{n\to\infty}|x_n| =\lim_{n\to\infty}|y_n| =\lim_{n\to\infty}x_ny_n =+\infty$ 
Let $x_n$ and $y_n$ denote sequences such that:
  $$
\lim_{n\to\infty}(|x_n + y_n| - |x_n - y_n|) = +\infty
$$
  Prove:
  $$
\lim_{n\to\infty}(|x_n + y_n| - |x_n - y_n|) = +\infty \iff \lim_{n\to\infty}|x_n| =\lim_{n\to\infty}|y_n| =\lim_{n\to\infty}x_ny_n =+\infty
$$


I've started with the first case ($\implies$):
$$
\lim_{n\to\infty}(|x_n + y_n| - |x_n - y_n|) = +\infty \\ 
\stackrel{\text{def}}{\iff} \forall \epsilon > 0\ \exists N\in\Bbb N: \forall n>N \implies ||x_n + y_n| - |x_n - y_n|| \ge \epsilon
$$
We want to show that $|x_n| \ge \epsilon$ and $|y_n|\ge \epsilon$. By triangular inequality:
$$
|x_n + y_n| + |x_n - y_n| \ge ||x_n + y_n| - |x_n - y_n|| \ge \epsilon
$$
At the same time:
$$
|x_n + y_n| + |x_n - y_n| \ge |(x_n + y_n) + (x_n - y_n)| = 2|x_n|
$$
Or:
$$
|x_n + y_n| + |y_n - x_n| \ge |(x_n + y_n) + (y_n - x_n)| = 2|y_n|
$$
Here is where the first questionable case comes in. It seems that:
$$
|x_n + y_n| + |x_n - y_n| \ge 2|x_n| \ge ||x_n + y_n| - |x_n - y_n|| \ge \epsilon\\
\text{and}\\
|x_n + y_n| + |y_n - x_n| \ge 2|y_n| \ge ||x_n + y_n| - |x_n - y_n|| \ge \epsilon \tag1
$$
But I'm not sure how to justify that. I've checked various cases like $x<0 \land y<0$ and 3 others, for all it seems to hold. So based on this:
$$
\forall \epsilon >0 \ \exists N\in\Bbb N: \forall n> N \implies 2|x_n| \ge \epsilon \\
\forall \epsilon >0 \ \exists N\in\Bbb N: \forall n> N \implies 2|y_n| \ge \epsilon
$$
Which shows:
$$
\lim_{n\to\infty}|x_n| = \lim_{n\to\infty}|y_n| = +\infty
$$

In this part I'm interested in how to justify $(1)$ and where from it follows that:
  $$
\lim_{n\to\infty}x_ny_n = +\infty
$$


Second case $(\impliedby)$. This case I've no idea where to start from. We basically given three things:
$$
\forall \epsilon > 0\ \exists N\in\Bbb N: \forall n>N \implies |x_n| \ge \epsilon\\
\forall \epsilon > 0\ \exists N\in\Bbb N: \forall n>N \implies |y_n| \ge \epsilon \\
\forall \epsilon > 0\ \exists N\in\Bbb N: \forall n>N \implies |x_ny_n| \ge \epsilon
$$
The problem is I don't see where to go from this.

Could you please verify the overall reasoning and help with ($\impliedby$) and question from ($\implies$), I still often have troubles with constructing those proofs since i'm a self-learner and have no one to refer to. Thank you!
 A: $(\Longrightarrow)$ Note that by triangle inequality, we have
$$
2|x_n|=|(x_n+y_n)+(x_n-y_n)|\ge|x_n+y_n|-|x_n-y_n|
$$ and
$$
2|y_n|=|(x_n+y_n)-(x_n-y_n)|\ge|x_n+y_n|-|x_n-y_n|.
$$ By taking $n\to\infty$, we get
$$
\lim_{n\to\infty}|x_n|=\lim_{n\to\infty}|y_n|=\infty.
$$ Also note that
$
|x_n+y_n|\ge |x_n-y_n|
$ holds eventually. Hence, by squaring both sides, we have
$x_ny_n\ge 0$. It follows that $\lim_{n\to\infty}x_n y_n=\lim_{n\to\infty}|x_n| |y_n|=\infty.$

$(\Longleftarrow)$ To show the converse, note that $\lim_{n\to\infty}x_n y_n=\infty$ implies that $$
|x_n+y_n|-|x_n-y_n|\ge 0
$$ eventually. And since it is a positive sequence, we have
$$
\lim_{n\to\infty}|x_n+y_n|-|x_n-y_n|=\infty \Longleftrightarrow \lim_{n\to\infty}\left[|x_n+y_n|-|x_n-y_n|\right]^2=\infty.
$$ Now, we have
$$
\left[|x_n+y_n|-|x_n-y_n|\right]^2= 2(|x_n|^2+|y_n|^2) -2|x_n^2-y_n^2|=:L.
$$ Observe that if $x_n^2\ge y_n^2$, then it holds $L=4|y_n|^2$ and otherwise, $L=4|x_n|^2$. This gives $L=4 \min\{|x_n|^2,|y_n|^2\}$ and
$$
\lim_{n\to\infty}\left[|x_n+y_n|-|x_n-y_n|\right]^2= 4\lim_{n\to\infty}\min\{|x_n|^2,|y_n|^2\}=\infty
$$ by the assumption that $\lim_{n\to\infty}|x_n|=\lim_{n\to\infty}|y_n|=\infty$. This proves the converse claim.
