1
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ You have your "triangle inequality" the wrong way round. $\endgroup$ – Lord Shark the Unknown Jan 26 at 16:55
  • 2
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.