Prove that $||x|-|y||\le |x-y|$

Question

I've seen the full proof of the Triangle Inequality \begin{equation*} |x+y|\le|x|+|y|. \end{equation*} However, I haven't seen the proof of the reverse triangle inequality: \begin{equation*} ||x|-|y||\le|x-y|. \end{equation*} Would you please prove this using only the Triangle Inequality above?

Thank you very much.

Answer 1

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

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

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

$$|y -x| \ge |y| - |x|$$

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

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

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

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

Answer 2

$\left| |x|-|y| \right|^2 - |x-y|^2 = \left( |x| - |y| \right)^2 - (x-y)^2 = |x|^2 - 2|x| \cdot |y| +y^2 - x^2 + 2xy-y^2 = 2 (xy-|xy|) \le 0 \Rightarrow \left| |x|-|y| \right| \le |x-y|.$

$"=" \iff xy\ge 0.$ Q.E.D.

The Triangle Inequality can be proved similarly.