# If |x + y| > |x - y|, then how to arrive at $xy > 0$?

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 $$xy > 0$$?

• 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. – Milo Brandt Feb 16 at 1:35
• @MiloBrandt Ok cool, I'm familiar with this. I'll give this a try. Thanks for the response! – johnnyodonnell Feb 16 at 1:36
• 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 $y$. – Andreas Blass Feb 16 at 1:54
• @AndreasBlass Interesting! This is cool, thanks for the response! – johnnyodonnell Feb 16 at 2:30
• To explore further the special relationship $xy =\dfrac{|x+y|^2-|x-y|^2}4$ have a look at math.stackexchange.com/questions/21792/… – zwim Feb 16 at 2:49

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.

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 $$x<1$$.

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.