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

I've seen the complete evidence of the Triangle Inequality \begin{equation*} |x+y|\le|x|+|y|. \end{equation*} However, I have not seen the evidence of the reverse triangular inequality: \begin{equation*} ||x|-|y||\le|x-y|. \end{equation*} Would you please confirm this making use of just the Triangle Inequality over?

Thanks significantly.

2022-07-16 17:10:41
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Answers: 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:

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

2022-07-16 19:02:18