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I have to demonstrate that $|x+y| < \max(|x|, |y|) \Leftrightarrow xy < 0$. I'm bit lost as how to proceed on this. I know I have to separate in two cases and that the right side is $x$, when $x > 0$ and $y$ when $y > 0$ (only one of the two can be positive).

I can separate the cases $|x+y|>0$ ($x > 0$ and $x > |y|$, though the cases should be just $x > 0$ and $y > 0$, but this doesn't help much) and $|x+y|<0$ but I don't know what to do from there. So for the first case I have $x+y < x$, which is pretty obvious I guess, as $y$ is negative? Do I have to write anything else to prove this?

Thanks in advance!

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  • $\begingroup$ $\max(|x|,|y|)$ isn’t necessarily $x$ when $y<0<x$: $\max(|2|,|-3|)=3$, not $2$. Have you looked at some examples? You should consider some with both $x$ and $y$ positive, some with both negative, and some with one positive and one negative. $\endgroup$
    – Brian M. Scott
    Nov 3, 2011 at 22:14
  • $\begingroup$ thanks @BrianM.Scott! @edit will try now and write here what I got $\endgroup$
    – Clash
    Nov 3, 2011 at 22:16
  • $\begingroup$ Both and x and y positive? But I thought $x*y<0$ ment one of them must be negative? $\endgroup$
    – Clash
    Nov 3, 2011 at 22:18
  • $\begingroup$ $|x+y|$ is always $\ge 0$, and it’s $0$ only when $x+y=0$, so it’s pretty pointless to consider a case $|x+y|>0$. $\endgroup$
    – Brian M. Scott
    Nov 3, 2011 at 22:18
  • $\begingroup$ $xy<0$ if and only if one of $x$ and $y$ is positive and one is negative, but you’re trying to prove a logical equivalence: you not only have to show that $|x+y|<\max(|x|,|y|)$ when $xy<0$, you also have to show that $|x+y|\ge\max(|x|,|y|)$ when $xy\ge 0$. $\endgroup$
    – Brian M. Scott
    Nov 3, 2011 at 22:21

3 Answers 3

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There are clever ways to prove it, but a straightforward argument by cases is certainly possible. If you take that route, I’d break it into three cases: $xy<0$, $xy=0$, and $xy>0$. Specifically, you’d need to show that

  • $|x+y|<\max(|x|,|y|)$ when $xy<0$;
  • $|x+y|\ge\max(|x|,|y|)$ when $xy=0$; and
  • $|x+y|\ge\max(|x|,|y|)$ when $xy>0$.

It might be easiest to look at the last case first. If $xy>0$, either $x$ and $y$ are both positive, or $x$ and $y$ are both negative. If $x,y>0$, then $x+y>0$, so $|x+y|=x+y$, and $\max(|x|,|y|) =$ $\max(x,y)$. Is it true that $x+y\ge\max(x,y)$ when $x$ and $y$ are both positive? Sure: if $0<x\le y$, then $\max(x,y)=y>x+y$, and the if $0<y\le x$, then $\max(x,y)=x>x+y$. Now what happens if $x,y<0$? Then $x+y<0$ as well, so $|x|=-x,|y|=-y$, and $|x+y|=$ $-(x+y)=(-x)+(-y)$, and $|x+y|=(-x)+(-y)>\max(-x,-y)=\max(|x|,|y|)$ by the same reasoning that we just used (since $-x$ and $-y$ are positive).

It should be pretty easy to deal with the middle case. To deal with the first one, you could begin by observing that everything is symmetric with respect to $x$ and $y$: if you interchange $x$ and $y$ everywhere, nothing really changes. Thus, you might as well assume that $x<0<y$, because the case $y<0<x$ is going to be exactly the same with $x$ and $y$ interchanged. You might then want to break it into two subcases: $|x|<|y|$, and $|x|\ge|y|$. Then peel off the absolute values much as I did above, using the fact that $|z|=-z$ when $z<0$, and see whether it really is true that $|x+y|<\max(|x|,|y|)$ in both subcases of this case.

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  • $\begingroup$ Thanks for the great detailed answer! $\endgroup$
    – Clash
    Nov 3, 2011 at 22:53
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Hint: Intuitively, if $|x+y| < \max(|x|, |y|)$ then they have to be of opposite signs because if they were of the same sign the addition would take the sum further from zero.

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  • $\begingroup$ Yes, I agree, that's intuitive. If x and y are positive, then $|x+y|$ becomes $x+y$, which is obviously bigger or equal to max(|x|, |y|). But how can I prove this? $\endgroup$
    – Clash
    Nov 3, 2011 at 22:25
  • $\begingroup$ Should I suppose it's right and then come to something like y<0, which is impossible? $\endgroup$
    – Clash
    Nov 3, 2011 at 22:26
  • $\begingroup$ Right, I have now proven that $x*y < 0$, thanks for the hint! $\endgroup$
    – Clash
    Nov 3, 2011 at 22:33
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Hint: $|x+y|^2 -|x|^2=2x\cdot y+|y|^2$

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  • $\begingroup$ Sorry, I don't follow your hint $\endgroup$
    – Clash
    Nov 3, 2011 at 22:35
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    $\begingroup$ Suppose that $|x+y|<|x|$, then $|x+y|^2<|x|^2$, then $2x\cdot y+|y|^2<0$, and that says that $x\cdot y<0$. $\endgroup$
    – robjohn ♦
    Nov 3, 2011 at 22:43

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