$|x-z| \leq |x-y|+|y-z|$
We know that both LHS and RHS are non negatives. So, I thought of proving this by comparing the squares of both sides but can't advance beyond that step. Any help would be appreciated.
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1Do you know the triangle inequality? – Feb 14 '14 at 05:14
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@T.Bongers: I suspect the point is to prove the triangle inequality for the reals, so you can't use that. – Ross Millikan Feb 14 '14 at 05:16
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1In any case, http://math.stackexchange.com/q/457513/11994 is related, which has proofs in various styles. – MarnixKlooster ReinstateMonica Feb 14 '14 at 05:32
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See also: http://math.stackexchange.com/questions/708431/proving-the-inequality-a-b-leq-a-c-c-b-for-real-a-b-c – Martin Sleziak Mar 12 '14 at 13:05
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If two of the variables are equal, it is easy. If they are not, there are six orderings. Each order gives you a way to resolve the absolute value signs. You can check them all.
Ross Millikan
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@gary: my argument supports this. What is your point? One order is $x \lt y \lt z$ Then you want to prove $z-x \le (y-x) + (z-y)$ which is $z-x \le z-x$ There are five more. – Ross Millikan Feb 14 '14 at 05:35
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If you know that $|a+b|\le|a|+|b|$ holds for any real numbers $a$, $b$, $c$ (this is triangle inequality), then you can simply plug in $b=x-y$ and $c=y-z$.
You get $a+b=x-z$ and $$|x-z|\le|x-y|+y-z|.$$
Martin Sleziak
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Another way, as we are concerned only with distances between points on the real line WLOG we can shift the origin, so set $y=0$, to simplify the inequality. Now easy to check cases or use your squaring argument.
Macavity
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@saturatedexpo This can be done every time the inequality depends only on distances between points. – Macavity Nov 12 '16 at 04:08