The Triangle Inequality – KindiakMath

The Triangle Inequality

Problem 1. Consider the triangle below with $a < b$ .

Show that $c \leq a + b$ . This result is known as the triangle inequality.

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Solution. Draw the altitude $h$ of the triangle with base $c$ and write $c = r + s$ .

Using Pythagoras’ theorem,

$\begin{aligned} c = r + s & \leq \sqrt{a^2 - h^2} + \sqrt{b^2 - h^2} \\ &\leq \sqrt{a^2} + \sqrt{b^2} \\ &= a + b. \end{aligned}$

Problem 2. Show that $a-b \leq c$ . This result is known as the reverse triangle inequality.

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Solution. Apply Problem 1 to obtain the inequality

$a \leq b + c.$

Subtracting $b$ on both sides, $a-b \leq c$ .

Problem 3. For any real number $x$ , define the absolute value of $x$ by

$|x| := \begin{cases} x, & x \geq 0, \\ -x, & x < 0. \end{cases}$

Show that $|x| = \sqrt{x^2}$ . Deduce that $|x| = |{-x}|$ and $-|x| \leq x \leq |x|$ .

(Click for Solution)

Solution. For $x \geq 0$ ,

$\sqrt{x^2} = x = |x|.$

For $x < 0$ , $|x| = -x > 0$ . Since

$(-x)^2 = (-x) \cdot (-x) = x\cdot x = x^2,$

we have

$\sqrt{x^2} = \sqrt{(-x)^2} = -x = |x|.$

Therefore, $|x| = \sqrt{x^2}$ . In particular,

$|{-x}| = \sqrt{(-x)^2} = \sqrt{x^2} = |x|.$

For the last inequality, the case $x \geq 0$ yields

$-|x| \leq -0 = 0 \leq x = |x| \leq |x|.$

The case $x < 0$ yields $-x > 0$ , so applying the previous result gives

$-|{-x}| \leq -x \leq |{-x}|.$

By negating all sides of the inequality,

$-|x| = -|{-x}| \leq x \leq |{-x}| = |x|.$

Problem 4. Using the definition in Problem 3, show that for any real number $a, b$ ,

$|a \pm b| \leq |a| + |b|.$

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Solution. Using Problem 3, we have the inequalities

$-|a| \leq a \leq |a|,\quad -|b| \leq b \leq |b|.$

Adding them together,

$-(|a| + |b|) \leq a + b \leq |a| + |b|.$

Negating all sides,

$-(|a| + |b|) \leq -(a+b) \leq -(-(|a| + |b|)) = |a| + |b|.$

There are two cases to consider:

If $a+b \geq 0$ , then $|a+b| = a+b \leq |a| + |b|$ .
If $a + b < 0$ , then $|a+b| = -(a+b) \leq |a|+|b|$ .

Therefore, we always have $|a+b| \leq |a| + |b|$ . Finally,

$|a-b| = |a+(-b)| \leq |a| + |{-b}| = |a| + |b|.$

—Joel Kindiak, 17 Jan 26, 1303H

Exercises, O-Level Math

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