So I am reading a derivation and I came to a point where they reach this point: $$ \text{Something} = |y-z| - |x-z|.$$

Then they continue, and say, that from triangle inequality $|y-z| - |x-z| \le |x-y|$.

I found that triangle inequality is defined like this: $|x-y|+|x-z| \le |y-z|$. However when I solve this for $|x-y|$, (i.e subtracting $|x-z|$ from both sides) I get this: $|x-y|\le |y-z| - |x-z|$.

What I am missing here?

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    $\begingroup$ You have your "triangle inequality" the wrong way round. $\endgroup$ – Lord Shark the Unknown Jan 26 at 16:55
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    $\begingroup$ You should learn the reverse triangle inequality which states $$||a|-|b||\leq |a-b|$$ It's a straighforward consequence of the usual triangle inequality. $\endgroup$ – Gabriel Romon Jan 26 at 16:57

You have the $\le$ in your equation backwards.

The triangle inequality usually introduced as

$|a| + |b| \ge |a + b|$.

Which if we replace $a = x-y$ and $b = z-x$ we get

$|x-y| + |z-x| \ge |(x-y) + (z-x)| = |z-y|= |y-z|$

Which is what you should have written down.


Now to your case:

$|y−z|−|x−z|\le|x−y|\iff $

$|y-z| \le |x-y| + |x-z|$.

And here we can either note that $|x-y| + |x-z| = |y-x| + |x-z| \ge |(y-x) + (x-z)| = |y-z|$ and we are done.

or note.

$|y -z| = |(y-x) + (x-z)| \le |y-x| + |x-z|=|x-y| + |x-z|$

So we are done.

I supposed it sometimes gets confusing when you see

$|a| + |b| ?? |a+b|$

$|a+b| ?? |a|+|b|$

$|(x -y) + (y-z)| ?? |x-y| + |y-z|$

to remember whether the correct symbol is "$\le$" or "$\ge$".

The thing to remember is that a sum inside an absolute sign is smaller than that sum outside the absolute value signs.


By the triangle inequality $$|x-y|+|x-z|=|x-y|+|z-x|\geq|x-y+z-x|=|z-y|=|y-z|$$


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