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 $\mid a+b\mid \leq \mid a\mid +\mid b\mid$.

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

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 $\mid \mid a\mid -\mid b\mid \mid \leq \mid a-b\mid$.

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

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

• Therefore $\mid \mid a\mid -\mid b\mid \mid \leq \mid a+b\mid$. $\blacksquare$

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

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

Corollary 3: If $a_{1},a_{2},...,a_{n}\in \mathbb {R}$ then $\mid a_{1}+a_{2}+...+a_{n}\mid \leq \mid a_{1}\mid +\mid a_{2}\mid +...+\mid a_{n}\mid$.

• Proof of Corollary 3: We note that $\mid a_{1}+a_{2}+...+a_{n}\mid =\mid a_{1}+(a_{2}+...+a_{n})\mid \leq \mid a_{1}\mid +\mid a_{2}+...+a_{n}\mid$ by the triangle inequality. Applying the triangle inequality multiple times we eventually get that $\mid a_{1}+a_{2}+...+a_{n}\mid \leq \mid a_{1}\mid +\mid a_{2}\mid +...+\mid a_{n}\mid$. $\blacksquare$

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

Licensing

Content obtained and/or adapted from:

• The Triangle Inequality, mathonline.wikidot.com under a CC BY-SA license