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When $\lvert x + y\rvert > \lvert x - y\rvert$, I am aware that we can square both sides to find that $xy > 0$.

$x^2 + 2xy + y^2 > x^2 - 2xy + y^2$

$4xy > 0$

$xy > 0$

However, I'm wondering if there are other ways to arrive at $xy > 0$, because I am afraid that if I see a problem similar to this, then I won't always know that it's possible to square both sides to arrive at a simplified solution. Is there a more formulaic or step-by-step process that I can follow to arrive at $xy > 0$?

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    $\begingroup$ I won't write it out because it's painful, one always has the recourse of casework where you prove things such as "If $x+y\geq 0$ and $x-y\geq 0$ and $x+y>x-y$, then $xy>0$" - basically make the absolute values go away. This isn't so fun, but it does work. $\endgroup$ – Milo Brandt Feb 16 at 1:35
  • $\begingroup$ @MiloBrandt Ok cool, I'm familiar with this. I'll give this a try. Thanks for the response! $\endgroup$ – johnnyodonnell Feb 16 at 1:36
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    $\begingroup$ Proceed by contradiction, assuming $xy\leq0$, and remember that $|x-y|$ is the distance between $x$ and $y$ on the real line. If $xy\leq0$, then $x$ and $y$ are on opposite sides of $0$ (or one of them equals $0$) and the distance between them is the sum of their distances from $0$, that is, $|x|+|y|$. On the other hand, $|x+y|$ will be smaller than (or equal to) $|x|+|y|$ because of partial cancellation between $x$ and $y$. $\endgroup$ – Andreas Blass Feb 16 at 1:54
  • $\begingroup$ @AndreasBlass Interesting! This is cool, thanks for the response! $\endgroup$ – johnnyodonnell Feb 16 at 2:30
  • $\begingroup$ To explore further the special relationship $xy =\dfrac{|x+y|^2-|x-y|^2}4$ have a look at math.stackexchange.com/questions/21792/… $\endgroup$ – zwim Feb 16 at 2:49
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You can examine all the possible cases for $x$ and $y$:

1) $x>0$, $y>0$, and $x>y$. Then, $|x+y|=x+y$ and $|x-y|=x-y$, and inequality $|x+y|>|x-y|$ will be reduced to $y>0$ which is always true (since we assume that y is positive). Therefore, since $x>0$ and $y>0$ (based on our assumption), then $xy>0$.

2) $x>0$, $y<0$, and $|x|>|y|$ or $|x|<|y|$, following the above approach, finding the sign of $x+y$ and $x-y$, substituting in the inequality, you will reach one of the $x<0$, $y>0$ cases which is a contradiction. Since we assume that $x>0$, $y<0$, then $xy<0$ doesn't satisfy the inequality.

You can examine other cases where ($x<0$ and $y<0$) as well as ($x<0$ and $y>0$) in a similar way.

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We could conduct the same thinking directly with $\pm x$ but let's instead set $y=ax$. We want to prove that $x$ and $y$ are the same sign (i.e. $xy>0\iff a>0$).

First let put aside $x=0$ since the inequality in that case is not verified.

We get $|x(1+a)|>|x(1-a)|$ and since $|x|\neq 0$ let's divide to get $|1+a|>|1-a|$.

Now think of it like : $|a-(-1)|>|a-(+1)|$

So the distance of $a$ to $-1$ is greater than the distance of $a$ to $+1$, if we think geometrically, then $a$ is further away than the middle point $0$ (in the direction toward $+1$), this means $a>0$.

This is often easier to visualize absolute value properties if you think geometrically in term of distances. $|x-y|$ is the distance from $x$ to $y$ and $|x+y|$ is the distance from $x$ to $-y$.

For instance $|x+3|<|x-5|$ then $x$ is closer to $-3$ than to $+5$ so we are on the left of the middle $\frac{5-3}2=1$ so $x<1$.

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let's suppose that $xy≤0$. given the problem is symmetric we can suppose that $x≤0$ and $y≥0$, in this case, we have: $x≤-x \Rightarrow x+y≤y-x$ and $-y≤y \Rightarrow x-y≤x+y$ $\Rightarrow x-y≤x+y≤y-x \Rightarrow |x+y|≤y-x$ $\Rightarrow |x+y|≤|x-y|$ which proves the inequality by contraposition.

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