# Prove that ||x|−|y||≤|x−y|||x|-|y||\le |x-y|

I’ve seen the full proof of the Triangle Inequality

However, I haven’t seen the proof of the reverse triangle inequality:

Would you please prove this using only the Triangle Inequality above?

Thank you very much.

$$|x|+|y−x|≥|x+y−x|=|y||x| + |y -x| \ge |x + y -x| = |y|$$

$$|y|+|x−y|≥|y+x−y|=|x||y| + |x -y| \ge |y + x -y| = |x|$$

Move $$|x||x|$$ to the right hand side in the first inequality and $$|y||y|$$ to the right hand side in the second inequality. We get

$$|y−x|≥|y|−|x||y -x| \ge |y| - |x|$$

$$|x−y|≥|x|−|y|.|x -y| \ge |x| -|y|.$$

From absolute value properties, we know that $$|y−x|=|x−y|,|y-x| = |x-y|,$$ and if $$t≥at \ge a$$ and $$t≥−at \ge −a$$ then $$t≥|a|t \ge |a|$$.

Combining these two facts together, we get the reverse triangle inequality:

$$|x−y|≥||x|−|y||.|x-y| \ge \bigl||x|-|y|\bigr|.$$