Triangle Inequality

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

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

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

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

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

Then:
 * $\left\vert{z_1 + z_2}\right\vert \le \left\vert{z_1}\right\vert + \left\vert{z_2}\right\vert$

Backwards Form
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 1 for Real Numbers
Then by Order of Squares in Totally Ordered Ring:
 * $\left\vert{x + y}\right\vert \le \left\vert{x}\right\vert + \left\vert{y}\right\vert$

Proof 2 for Real Numbers
This can be seen to be a special case of Minkowski's Inequality, with $n = 1$.

Proof 3 for Real Numbers
From Real Numbers form Ordered Integral Domain, we can directly apply Sum of Absolute Values, which is applicable on all ordered integral domains, of which $\R$ is one.

Note that this result can not directly be used for the complex numbers $\C$ as they do not form an ordered integral domain.

Proof Using Vectors
Let $a$ and $b$ be vectors, and $||a||$ be the magnitude of $a$.

$(||a||+||b||)^2=||a||^2+||b||^2+2||a||*||b|| \implies\sqrt{||a||^2+||b||^2+2||a||*||b||} = ||a||+||b||$.

$||a+b||^2=(a+b)\cdot (a+b)=||a||^2+||b||^2+2a\cdot b\le||a||^2+||b||^2+2|a\cdot b|$.

$||a+b||= \sqrt{||a||^2+||b||^2+2a\cdot b}\le\sqrt{||a||^2+||b||^2+2|a\cdot b|}$.

Then by the Cauchy Schwarz Inequality, $|a\cdot b|\le||a||*||b|| \implies \sqrt{||a||^2+||b||^2+2|a\cdot b|}\le\sqrt{||a||^2+||b||^2+2||a||*||b||}$.

$||a+b||\le \sqrt{||a||^2+||b||^2+2|a\cdot b|}\le \sqrt{||a||^2+||b||^2+2||a||*||b||}= ||a||+||b||$.

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

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

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

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

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

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


 * 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:

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

Thus $- \left({\left\vert{x}\right\vert - \left\vert{y}\right\vert}\right) \le \left\vert{x - y}\right\vert \le \left({\left\vert{x}\right\vert - \left\vert{y}\right\vert}\right)$ 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.