The Triangle Inequality

# The Triangle Inequality

The triangle inequality is a very important geometric and algebraic property that we will use frequently in the future.

Theorem 1 (Triangle Inequality): Let $a$ and $b$ be real numbers. Then $\mid a + b \mid ≤ \mid a \mid + \mid b \mid$. |

**Proof of Theorem:**For $a$ and $b$ as real numbers we have that $-\mid a \mid ≤ a ≤ \mid a \mid$ and $-\mid b \mid ≤ b ≤ \mid b \mid$. If we add these inequalities together we get that $-\mid a \mid - \mid b \mid ≤ a + b ≤ \mid a \mid + \mid b \mid$ or rather $-\left ( \mid a \mid + \mid b \mid \right ) ≤ a + b ≤ \left ( \mid a \mid + \mid b \mid \right )$ which is equivalent to saying that $\mid a + b \mid ≤ \mid a \mid + \mid b \mid$. $\blacksquare$

There are also some other important results similar to the triangle inequality that are important to mention.

Corollary 1: If $a$ and $b$ are real numbers then $\mid \mid a \mid - \mid b \mid \mid ≤ \mid a - b \mid$. |

**Proof of Corollary 1:**We first write $a = a - b + b$ and therefore applying the triangle inequality we get that $\mid a \mid = \mid (a - b) + b \mid ≤ \mid a - b \mid + \mid b \mid$ and therefore $\mid a \mid ≤ \mid a - b \mid + \mid b \mid$. Subtracting $\mid b \mid$ from both sides we get that $\mid a \mid - \mid b \mid ≤ \mid a - b \mid$.

- Now we write $b = b - a + a$ and therefore applying the triangle inequality we get that $\mid b \mid = \mid (b - a) + a \mid ≤ \mid b - a \mid + \mid a \mid$ and therefore $\mid b \mid ≤ \mid b - a \mid + \mid a \mid$ and subtracting $\mid a \mid$ from both sides we get that $\mid b \mid - \mid a \mid ≤ \mid b - a \mid$ which is equivalent to $\mid a \mid - \mid b \mid ≥ - \mid b - a \mid$.

- Therefore $\mid \mid a \mid - \mid b \mid \mid ≤ \mid a + b \mid$. $\blacksquare$

Corollary 2: If $a$ and $b$ are real numbers then $\mid a - b \mid ≤ \mid a \mid + \mid b \mid$. |

**Proof of Corollary 2:**By the triangle inequality we get that $\mid a + b \mid ≤ \mid a \mid + \mid b \mid$ and so then $\mid a + (-b) \mid ≤ \mid a \mid + \mid -b \mid = \mid a \mid + \mid b \mid$. Therefore $\mid a - b \mid ≤ \mid a \mid + \mid b \mid$. $\blacksquare$

Corollary 3: If $a_1, a_2, ..., a_n \in \mathbb{R}$ then $\mid a_1 + a_2 + ... + a_n \mid ≤ \mid a_1 \mid + \mid a_2 \mid + ... + \mid a_n \mid$. |

**Proof of Corollary 3:**We note that $\mid a_1 + a_2 + ... + a_n \mid = \mid a_1 + (a_2 + ... + a_n) \mid ≤ \mid a_1 \mid + \mid a_2 + ... + a_{n} \mid$ by the triangle inequality. Applying the triangle inequality multiple times we eventually get that $\mid a_1 + a_2 + ... + a_n \mid ≤ \mid a_1 \mid + \mid a_2 \mid + ... + \mid a_n \mid$. $\blacksquare$

*A more formal proof of Corollary 3 can be carried out by Mathematical Induction.*