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I've just started learning about the triangle inequality. I've got the following two statements which at first glance look very similar.

(1) $x < y$ and $|x − y| ≤ | x − z| + | z − y|$
(2) $x < y$ and $|x − y| < | x − z| + | z − y|$

I've seen the triangle inequality written as (1) without the condition x < y but I can't see any reason that this condition would actually affect the inequality so is it just the same thing?

Secondly, I've been told that for (2) one of the following will be true:
a) holds for all $z$ in the reals
b) holds if and only iff $z$ is an element of $(x, y)$
c) holds if and only iff $z$ is an element of $(-∞, x)\cup(y, ∞)$
d) none of the above

I'm inclined to believe that d is the best answer since I know the triangle inequality to be true if $z$ AND $x$, $y$ are in the reals (which eliminates a as a choice) As far as I know there doesn't need to be any relation between $z$ and $x$, $y$ beyond that (which eliminates $b$ and $c$) thus leaving me with $d$. What are the flaws in my logic?

Kamal Saleh
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user131875
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1 Answers1

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(1): yes, the triangle inequality as stated holds for all $x,y,z$ so no condition is needed.

(2): your explanations are not very clear as justifications. Let's see.

a) is not true because you can take $z=x$ and the inequality is not strict.

b) If $x<y$ and $x<z<y$, then $$ |x-z|+|z-y|=z-x+y-z=y-x=|y-x|, $$ so there is no strict inequality and b) is false.

c) If $z<x<y$, we have $$ |x-z|+|z-y|=x-z+y-z=y+x-2z=y-x+2(x-z)=|y-x|+2|x-z|. $$ Since $|x-z|>0$, we get $$ |y-x|<|x-z|+|z-y|. $$ Similarly, if $x<y<z$, now $$ |x-z|+|y-z|=z-x+z-y=2(z-y)+y-x=2|z-y|+|y-x|, $$ and again we get $$ |y-x|<|x-z|+|z-y|. $$ Conversely, if $|x-y|<|x-z|+|z-y|$ we cannot have $z=x$ nor $z=y$ (we would have equality), and by part b) we would also have equality if $z\in(x,y)$. Thus $z\in (-\infty,x)\cup(y,\infty)$.

So the answer is c).

Martin Argerami
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