#### Chapter 3 Inequalities in one Variable

Section 3.3 Absolute Value Inequalities and Quadratic Inequalities

# 3.3.3 Further Types of Inequalities

Many other types of inequalities can be transformed into quadratic inequalities. Sometimes, case analyses have to be done or excluded values in the domain have to be taken into account:
##### Info 3.3.9

An inequality containing fractions, where the variable $x$ occurs in the denominator of composite terms, can be transformed into a form without fractions by multiplying the inequality by the least common denominator. However, in doing so, the roots of the denominators have to be excluded from the domain of the new inequality.
Additionally, if the inequality is multiplied by a term, different cases have to be distinguished depending on the sign of the term.

##### Example 3.3.10
The inequality $2-\frac{1}{x}\le x$ can be transformed by multiplying the inequality by $x$. Here, three cases have to be distinguished:
• For $x>0$, the comparator in the inequality is unchanged. The new inequality reads $2x-1\le {x}^{2}$ and is equivalent to ${x}^{2}-2x+1\ge 0$ or $\left(x-1{\right)}^{2}\ge 0$, respectively. This inequality is always satisfied. Because of the case condition one obtains the solution set $L{}_{1}=\text{}\right]0;\infty \left[\text{}$.
• For $x<0$, the comparator in the inequality is inverted. The new inequality reads $2x-1\ge {x}^{2}$ and is equivalent to ${x}^{2}-2x+1\le 0$ or $\left(x-1{\right)}^{2}\le 0$, respectively. This inequality is only satisfied for $x=1$. But this value is excluded by the case condition, i.e. $L{}_{2}=\left\{\right\}$.
• The single value $x=0$ is not in the domain of the initial inequality and hence it is no solution.

So, altogether one obtains the union set $L=\text{}\right]0;\infty \left[\text{}$ as solution set of the initial inequality.

Inequalities involving composite fraction and root terms often do not have solution sets of the types described in info box 3.3.7:
##### Example 3.3.11
Find the solution set of the inequality $\sqrt{x}+\frac{1}{\sqrt{x}}>2$. The domain of the inequality is $\text{}\right]0;\infty \left[\text{}$. Multiplying by $\sqrt{x}$ results in the inequality $x+1>2\sqrt{x}$. Here, no case analysis is required since $\sqrt{x}>0$ is in the domain. Transformation results in $x-2\sqrt{x}+1>0$ or $\left(\sqrt{x}-1{\right)}^{2}>0$, respectively, which is satisfied for all $x\ne 1$ in the domain. Hence, the solution set of the initial inequality is $L=\text{}\right]0;\infty \left[\text{}\setminus \left\{1\right\}$: 