Triangle Inequality

Real Numbers
Let $$x, y \in \R$$ be real numbers.

Let $$\left|{x}\right|$$ be the absolute value of $$x$$.

Then:
 * $$\left|{x + y}\right| \le \left|{x}\right| + \left|{y}\right|$$;

Complex Numbers
Let $$z_1, z_2 \in \C$$ be complex numbers.

Let $$\left|{z}\right|$$ be the modulus of $$z$$.

Then:
 * $$\left|{z_1 + z_2}\right| \le \left|{z_1}\right| + \left|{z_2}\right|$$;
 * $$\left|{z_1 - z_2}\right| \ge \left|{z_1}\right| - \left|{z_2}\right|$$;
 * $$\left|{z_1 - z_2}\right| \ge \left|{\left|{z_1}\right| - \left|{z_2}\right|}\right|$$.

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


 * $$\left|{x - y}\right| \ge \left|{x}\right| - \left|{y}\right|$$;
 * $$\left|{x - y}\right| \ge \left|{\left|{x}\right| - \left|{y}\right|}\right|$$.

Proof for Real Numbers
$$ $$ $$ $$ $$ $$

Then by Order of Squares in Totally Ordered Ring, $$\left|{x + y}\right| \le \left|{x}\right| + \left|{y}\right|$$.

Note it can be seen that this is a special case of Minkowski's Inequality, with $$n = 1$$.

Proof for Complex Numbers
Let $$z_1 = a_1 + \imath a_2, z_2 = b_1 + \imath b_2$$.

Then from the definition of the modulus, the above equation translates into $$\left({\left({a_1 + b_1}\right)^2 + \left({a_2 + b_2}\right)^2}\right)^{\frac 1 2} \le \left({a_1^2 + a_2^2}\right)^{\frac 1 2} + \left({b_1^2 + b_2^2}\right)^{\frac 1 2}$$.

This is a special case of Minkowski's Inequality, with $$n = 2$$.

Proof of Backwards Form

 * First we show that $$\left|{x - y}\right| \ge \left|{x}\right| - \left|{y}\right|$$:

From the above we see $$\left|{x + y}\right| - \left|{y}\right| \le \left|{x}\right|$$.

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

$$\left|{z}\right| - \left|{y}\right| \le \left|{z - y}\right|$$.

Renaming variables as appropriate gives $$\left|{x - y}\right| \ge \left|{x}\right| - \left|{y}\right|$$.


 * Next we show that $$\left|{x - y}\right| \ge \left|{\left|{x}\right| - \left|{y}\right|}\right|$$:

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

Also, $$\left|{x - y}\right| = \left|{y - x}\right| \ge \left|{y}\right| - \left|{x}\right| = - \left({\left|{x}\right| - \left|{y}\right|}\right)$$.

Thus $$- \left({\left|{x}\right| - \left|{y}\right|}\right) \le \left|{x - y}\right| \le \left({\left|{x}\right| - \left|{y}\right|}\right)$$ from the corollary to Negative of Absolute Value.

Note
There is also a geometric interpretation of the triangle inequality. It is in fact a special case of this algebraic triangle equality in the Euclidean metric space.