# How to derive that $|y-z| - |x-z| \le |x-y|$

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?

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

....

$$|y−z|−|x−z|\le|x−y|\iff$$

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

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 $$|x-y|+|x-z|=|x-y|+|z-x|\geq|x-y+z-x|=|z-y|=|y-z|$$