Inequalities and absolute value

The absolute value of the real number $a$ is given by

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

We first consider the inequality

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

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,

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

$$0 < \vert x-c\vert < \delta \Leftrightarrow c-\delta < x < c {\rm or} c < x < c + \delta$$ (0.2)

Example 0..4   Solve the inequality

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

SOLUTION By the inequality (1)

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

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

$$\vert a\vert > \varepsilon \Leftrightarrow a > \varepsilon {\rm or} a < -\varepsilon$$ (0.3)

Example 0..5   Solve the inequality

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

SOLUTION By the inequality (3)

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

By the first inequality, we have $2x > 2$. Thus $x > 1$. By the second inequalitywe have $2x < -8$. Thus $x < -4$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 $a,b$,

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

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

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

Now taking square roots we have

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

Now note that

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

Then the result follows $\blacksquare$

Here is another inequality used in calculus.

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

Exercise A

1.

Solve the following inequalities

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

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

renshu0-2-2

Which is larger?

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

3.

Solve the following inequalities

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

renshu0-2-4

Show that for all real numbers $a,b$

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

is true