Inequalities and absolute value

The absolute value of the real number $a$

is given by

$\displaystyle \vert a\vert = \left\{\begin{array}{ll} a & (a \geq 0)\\ -a & ... ...end{array}\right. \vert a\vert = \max\{a, -a\}, \vert a\vert = \sqrt{a^{2}}$

We first consider the inequality

$\displaystyle \vert x\vert < \delta$

Here $\delta$ is some positive number. Now $\vert x\vert < \delta$ can be thought of the points on the number line whose distance from the origin is less than $\delta$ . Thus,

\colorbox{calc-color}{ \begin{minipage}{0.9\textwidth} \begin{equation} |x| < \delta \Leftrightarrow -\delta < x < \delta \label{eq:ineq1-1} \end{equation} \end{minipage} }

Next $\vert x - c\vert < \delta$ can be thought of the points whose distance from the point $c$ is less than $\delta$ . Thus,

$\displaystyle \vert x - c\vert < \delta \Leftrightarrow -\delta < x-c < \delta \Leftrightarrow c -\delta < x < c + \delta$

(0.1)

Lastly $0 < \vert x - c\vert < \delta$ can be thought of $0 < \vert x - c\vert$ and $\vert x - c\vert < \delta$ The first inequality is $x \neq c$ . Thus

$\displaystyle 0 < \vert x-c\vert < \delta \Leftrightarrow c-\delta < x < c {\rm or} c < x < c + \delta$

(0.2)

Example 0..4 Solve the inequality

$\displaystyle \vert x + 2\vert < 3$

SOLUTION By the inequality (1)

$\displaystyle \vert x+2\vert < 3 \Leftrightarrow -3 < x+2 < 3 \Leftrightarrow -5 < x < 1$

Thus the solution is $(-5, 1)$

$\blacksquare$

Let $\varepsilon > 0$ . Then $\vert a\vert > \varepsilon$ can be think of the distance from the origin to $a$ is larger than $\varepsilon$ Thus

$\displaystyle \vert a\vert > \varepsilon \Leftrightarrow a > \varepsilon {\rm or} a < -\varepsilon$

(0.3)

Example 0..5 Solve the inequality

$\displaystyle \vert 2x + 3\vert > 5$

SOLUTION By the inequality (3)

$\displaystyle \vert 2x+3\vert > 5 \Leftrightarrow 2x+3 > 5 {\rm or} 2x+3 < -5$

By the first inequality, we have $2x > 2$

. Thus

. By the second inequalitywe have $2x < -8$

. Thus

From this, the solution is $(-\infty,-4)\cup(1, \infty)$ $\blacksquare$

One of the popular inequalities of calculus is the triangle inequality

Theorem 0..1 For all real numbers $a,b$

$\displaystyle \vert a+b\vert \leq \vert a\vert + \vert b\vert$

Traiangle Inequality $\vert a+b\vert$ can be thought of the length of hypotenuse of triangle and $\vert a\vert + \vert b\vert$ can be thought of the sum of the length of oppsite and adjacent

PROOF If you think of $\vert x\vert$ as $\sqrt{x^{2}}$ , then the proof is easy. Note that

$\displaystyle (a+b)^{2} = a^{2} + 2ab + b^{2} \leq \vert a\vert^{2} + 2\vert a\vert\vert b\vert + \vert b\vert^{2} = (\vert a\vert + \vert b\vert)^{2}.$

Now taking square roots we have

$\displaystyle \sqrt{(a+b)^{2}} \leq \vert a\vert + \vert b\vert$

Now note that

$\displaystyle \sqrt{(a+b)^{2}} = \vert a+b\vert$

Then the result follows $\blacksquare$

Here is another inequality used in calculus.

$\displaystyle \vert\vert a\vert - \vert b\vert\vert \leq \vert a - b\vert$

Exercise A

1.

Solve the following inequalities

(a) $\displaystyle{2+3x \leq 5}$ (b) $\displaystyle{\frac{1}{2}(1+x) < \frac{1}{3}(1 - x)}$ (c) $\displaystyle{4(x^{2} - 3x + 2) > 0}$

(d) $\displaystyle{\frac{1}{x} < x}$ (e) $\displaystyle{\frac{x^{2} - 9}{x+1} > 0}$ (f) $\displaystyle{\frac{1}{x-1} + \frac{4}{x-6} > 0}$

renshu0-2-2

Which is larger?

$\displaystyle \sqrt{\frac{x}{x+1}}, \sqrt{\frac{x+1}{x+2}}$

3.

Solve the following inequalities

(a) $\displaystyle{\vert x\vert < 2}$ (b) $\displaystyle{\vert x + 2\vert < \frac{1}{4}}$ (c) $\displaystyle{0 < \vert x-3\vert < 8}$ (d) $\displaystyle{\vert 2x + 5\vert > 3}$

renshu0-2-4

Show that for all real numbers $a,b$

$\displaystyle \vert a - b\vert \leq \vert a\vert + \vert b\vert$

is true