Backwards Form of Triangle Inequality

Theorem

Whether $x$ and $y$ are in $\R$ or $\C$, the following hold:

• $\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$

• $\left\vert{x - y}\right\vert \ge \left\vert{\left\vert{x}\right\vert - \left\vert{y}\right\vert}\right\vert$

Proof

• First we show that $\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$:

By the triangle inequality, $\left\vert{x + y}\right\vert - \left\vert{y}\right\vert \le \left\vert{x}\right\vert$.

Substitute $z = x + y \implies x = z - y$ and so:

$\left\vert{z}\right\vert - \left\vert{y}\right\vert \le \left\vert{z - y}\right\vert$

Renaming variables as appropriate gives:

$\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$

$\blacksquare$

• Next we show that $\left\vert{x - y}\right\vert \ge \left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert$:

From $\left\vert{x - y}\right\vert \ge \left\vert{x}\right\vert - \left\vert{y}\right\vert$ (proved above), we have:

If $\left\vert{x}\right\vert \ge \left\vert{y}\right\vert$:

$\left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert = \left\vert{x}\right\vert - \left\vert{y}\right\vert$, and we're done.

If $\left\vert{y}\right\vert \ge \left\vert{x}\right\vert$:

$\left\vert{y - x}\right\vert \ge \left\vert{y}\right\vert - \left\vert{x}\right\vert = \left\vert{ \left\vert{y}\right\vert - \left\vert{x}\right\vert }\right\vert$.

But $\left\vert{y - x}\right\vert = \left\vert{x - y}\right\vert$ and $\left\vert{ \left\vert{y}\right\vert - \left\vert{x}\right\vert }\right\vert = \left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert$.

Note from this we have:

$-\left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert \ge -\left\vert{x - y}\right\vert$

Since, by Negative of Absolute Value, we have that $\left\vert{x}\right\vert - \left\vert{y}\right\vert \ge -\left\vert{ \left\vert{x}\right\vert - \left\vert{y}\right\vert }\right\vert$, it follows that:

$-\left\vert{x - y}\right\vert \le \left\vert{x}\right\vert - \left\vert{y}\right\vert \le \left\vert{x - y}\right\vert$

The result follows.

$\blacksquare$

Also see

• Reverse Triangle Inequality (a generalization of the theorem presented on this page)

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.18$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.20 \ (2)$