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Exercises on Inequalities

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Prove that if and are positive real numbers then their product is positive.We assume that and . Since and it follows from the order axiom for the real number system that

and the fact that allows us to deduce that .
Prove that if and are negative real numbers then their product is positive.
If and are negative then and are positive and since

the fact that is positive follows from Exercise 1.
Given real numbers and prove that

Since

it follows from the triangle inequality that

and therefore
Given real numbers and prove that

From Exercise 3 we know that

and

Since is one of the numbers and the inequality

follows at once.
In each of the following cases, find the numbers for which the given inequality is true. Compare your answers with the answers given by Scientific Notebook
1. Method 1: We separate the problem into three cases as illustrated in the figure:
  
  Case 1: When $x\leq \frac{3}{2}$ the inequality says that
  
  which tells us that $x\geq -3$ .
  Case 2: When the inequality says that
  
  which tells us that $x\leq 3$ .
  Case 3: When the inequality says that
  
  which tells us that $x\leq -3$ (which is impossible). The set of numbers for which the inequality holds is therefore
  
  Method 2: We look first at the equation
  
  which says that either or . The latter condition says that either or and so we separate the problem into the cases indicated by the next figure
  
  This method leads us even more quickly to the solutions set . We omit the details.
2. We provide one solution here. Another approach is suggested by the solution provided below for part c. We begin by looking at the equation
  
  which says that either or . When $x\geq 0$ these equations say that either (which is impossible) or , which tells us that $x=\frac{11}{2}$ .
  When the equations say that either or which are both impossible. Therefore the only value of at which the inequality can switch from true to false or from false to true is $\frac{11}{2}$ and by looking at a specimen value of less than $\frac{11}{2}$ and a specimen value greater then $\frac{11}{2}$ we see that the inequality holds if and only if $x<\frac{11}{2}$ .
3. Solution: The inequality
  
  can be written as
  
  which says that
  
  In order to express this inequality without any absolute value signes we shall look separately at the cases and $x\geq 0$ and also the cases and $x\geq 1$ .
  
  When , the inequality
  
  becomes
  
  which says that $x\leq -2$ and $-8/3\leq x$ . In other words, when , we must have
  
  When $0\leq x\leq 1$ , the inequality
  
  becomes
  
  which says that $x\geq 2/3$ and $x\leq 8$ . Therefore the required inequality holds when
  
  When $x\geq 1$ , the inequality
  
  becomes
  
  which says that $x\geq 0$ and $x\leq 10/3$ . In other words
  
  Thus the solution of the required inequality is the set
Prove that if and are any real numbers then

One way to produce this inequality is to oberve that

There are several other possible approaches.
Given that and are positive numbers and that prove that

We observe that

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