Triangle Inequality

Theorem

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

Proof 1 for Real Numbers

	\(\displaystyle \)	\(\displaystyle \left\vert{x + y}\right\vert^2\)	\(=\)	\(\displaystyle \left({x + y}\right)^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x^2 + 2xy + y^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert{x}\right\vert^2 + 2xy + \left\vert{y}\right\vert^2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle \left\vert{x}\right\vert^2 + 2\left\vert{xy}\right\vert + \left\vert{y}\right\vert^2\)	\(\displaystyle \)	Negative of Absolute Value‎
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left\vert{x}\right\vert^2 + 2\left\vert{x}\right\vert\cdot \left\vert{y}\right\vert + \left\vert{y}\right\vert^2\)	\(\displaystyle \)	Product of Absolute Values
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\left\vert{x}\right\vert + \left\vert{y}\right\vert}\right)^2\)	\(\displaystyle \)

Then by Order of Squares in Totally Ordered Ring:

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

$\blacksquare$

Proof 2 for Real Numbers

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

$\blacksquare$

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.

$\blacksquare$

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 $\left\Vert{a}\right\Vert$ be the magnitude of $a$.

$\left({\left\Vert{a}\right\Vert + \left\Vert{b}\right\Vert}\right)^2 = \left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 \left\Vert{a}\right\Vert * \left\Vert{b}\right\Vert \implies \sqrt {\left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 \left\Vert{a}\right\Vert * \left\Vert{b}\right\Vert} = \left\Vert{a}\right\Vert + \left\Vert{b}\right\Vert$

$\left\Vert{a + b}\right\Vert^2 = (a+b) \cdot (a+b) = \left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 a \cdot b \le \left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 \left\vert{a \cdot b}\right\vert$

$\left\Vert{a + b}\right\Vert = \sqrt {\left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 a \cdot b} \le \sqrt {\left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 \left\vert{a \cdot b}\right\vert}$

Then by the Cauchy Schwarz Inequality:

$\left\vert{a \cdot b}\right\vert \le \left\Vert{a}\right\Vert * \left\Vert{b}\right\Vert \implies \sqrt {\left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 \left\vert{a \cdot b}\right\vert} \le \sqrt {\left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 \left\Vert{a}\right\Vert * \left\Vert{b}\right\Vert}$

$\left\Vert{a + b}\right\Vert \le \sqrt {\left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 \left\vert{a \cdot b}\right\vert} \le \sqrt {\left\Vert{a}\right\Vert^2 + \left\Vert{b}\right\Vert^2 + 2 \left\Vert{a}\right\Vert * \left\Vert{b}\right\Vert} = \left\Vert{a}\right\Vert + \left\Vert{b}\right\Vert$

$\blacksquare$

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 $n = 2$.

$\blacksquare$

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.

Sources

• James M. Hyslop: Infinite Series (1942): $\S 4$: Footnote
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): Chapter $\text{III}$
• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Proposition $1.1.9$, Corollary $1.1.10$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 1.17$