In the systems of equations chapter, we looked at a company that produces a basic and premium version of its product, and we determined how many of each item they should produce to fully utilize all staffed hours. For that, we need linear inequalities

A linear equation in two variables is an equation like (f(x)=2 x+1), which is sometimes written without function notation as (y=2 x+1 .) Recall that the graph of this equation is a line, formed by all the points ((x, y)) that satisfy the equation. A linear inequality in two variables is similar, but involves an inequality. Some examples:

[y<2 x+1 quad y>4 x-1 quad y leq frac{2}{3} x+4 quad y geq 5-2 x onumber]

Linear inequalities can also be written with both variables on the same side of the equation, like (2 x-3 y<4)

The solution set to a linear inequality will be all the points ((x, y)) that satisfy the inequality. Notice that the line (y=2 x+1) divides the coordinate plane into two halves:

on one half (y<2 x+1), and on the other (y>2 x+1 .) The solution set to a linear inequality will be a half plane, and to show the solution set we shade the part of the coordinate plane where the points lie in the solution set.

To graph the solution to a linear inequality

1. If necessary, rewrite the linear inequality into a form convenient for graphing, like slope-intercept form

2. Graph the corresponding linear equation.

a. For a strict inequality (< or (>) ), draw a dashed line to show that the points on the line are not part of the solution

b. For an inequality that includes the equal sign ((leq) or (geq)), draw a solid line to show that the points on the line are part of the solution.

3. Choose a test point, not on the line.

a. Substitute the test point into the inequality

b. If the inequality is true at the test point, shade the half plane on the side including the test point

c. If the inequality is not true at the test point, shade the half plane on the side that doesn't include the test point

Example (PageIndex{1})

Graph the solution to (y<2 x+1)

**Solution**

Since this is a strict inequality, we will graph (y = 2x + 1 onumber ) as a dashed line.

Now we choose a test point not on the line. Usually picking a point where one coordinate is zero will make things easier. Let's use ((3,0)).

Substituting (3,0) into the inequality, we get

[egin{align*} 0 &< 2(3) + 1 0 &< 7 end{align*} onumber ]

This is a true statement, so we will shade the side of plane that includes ((3,0)).

Example (PageIndex{2})

Graph the solution to (x geq 2 y+4)

**Solution**

This is not written in a form we are used to use for graphing, so we might solve it for (y) first.

(x geq 2 y+4 quad )

Subtract (2 v) from both sides

(x-2 y geq 4 quad )

Subtract (x) from both sides

(-2 y geq 4-x)

Divide by (-2), reversing the order of the inequality

(y leq-2+frac{1}{2} x)

Since this inequality includes the equals sign, we will graph (y=-2+frac{1}{2} x) as a solid line

Now we choose a test point not on the line. (0,0) is a convenient choice. Substituting (0,0) into the inequality, we get

[egin{align*} 0 &leq - 2 + frac{1}{2}(0) 0 &leq - 2 end{align*} onumber ]

This is a false statement, so we shade the half of the plane that does not include ((0,0)).

Exercise (PageIndex{1})

Graph the solution to [y geq - frac{1}{2}x + 1 onumber ]

**Answer**Maximize (P = 14x + 9y ) subject to the constraints:

[egin{align*} x + y &leq 9 3x + y &leq 15 x geq 0, &; y geq 0 end{align*} onumber ]

Example (PageIndex{3})

A store sells peanuts for ($ 4 /) pound and cashews for ($ 8 /) pound, and plans to sell a ney mixed-nut blend in a jar. What combinations of peanuts and cashews are possible, if they want the mix to cost ($ 6) or less?

**Solution**

We start by defining our variables:

(p): The number of pounds of peanuts in 1 pound of mix

(c): The number of pounds of cashews in 1 pound of mix

The cost of one pound of mix will be (4 p+8 c), so all the mixes costing ($ 6) or less will satisfy the inequality (4 p+8 c leq 6)

We can graph the equation (4 p+8 c=6) fairly easily by finding the intercepts:

When (p=0)

[egin{align*} 4(0)+8 c &= 6 8 c &= 6 c &= frac{6}{8}=frac{3}{4}. end{align*}]

So the point (left(0, frac{3}{4} ight)) on the line.

When (c=0,4 p=6), so (p=frac{6}{4}=frac{3}{2}), giving the point (left(frac{3}{2}, 0
ight)).

Notice the test point ((0,0)) will satisfy the inequality, so we will shade the side of the line including the origin. Because of the context, only values in the first quadrant are reasonable. The graph shows all the possible combinations the store could use, including 1 pound of peanuts with (1 / 4) pound of cashews, or (1 / 2) a pound of each.

## Systems of Linear Inequalities

In the systems of equations chapter, we looked for solutions to a system of linear equations - a point that would satisfy all the equations in the system. Likewise, we can consider a system of linear inequalities. The solution to a system of linear inequalities is

the set of points that satisfy all the inequalities in the system.

With a single linear inequality, we can show the solution set graphically. Likewise, with a system of linear inequalities we show the solution set graphically. We find it by looking for where the regions indicated by the individual linear inequalities overlap.

Example (PageIndex{4})

Graph the solution to the system of linear inequalities

[egin{align*} y &leq x + 2 y &geq 1 - x end{align*} onumber ]

**Solution**

If we graph the solution set to each inequality individually, we get the two solutions sets shown here.

Graphing these solution sets on the same axes reveals the solution to the system of inequalities as the region where the two overlap.

The solution set, where the regions overlap.

The solution set, drawn alone.

Exercise (PageIndex{2})

Graph the solution to the system of linear inequalities

[egin{align*} y &leq 3 - 2x y &geq frac{1}{2}x + 1 end{align*} onumber ]

**Answer**

The following question is similar to a problem we solved using systems in the systems chapter.

Example (PageIndex{5})

A company produces a basic and premium version of its product. The basic version requires 20 minutes of assembly and 15 minutes of painting. The premium version requires 30 minutes of assembly and 30 minutes of painting. If the company has staffing for 3,900 minutes of assembly and 3,300 minutes of painting each week. How many items can then produce within the limits of their staffing?

**Solution**

Notice this problem is different than the question we asked in the first section, since we are no longer concerned about fully utilizing staffing, we are only interested in what is possible. As before, we'll define

(b): The number of basic products made, (p): The number of premium products made.

Just as we created equations in the first section, we can now create inequalities, since we know the hours used in production needs to be less than or equal to the hours available. This leads to two inequalities:

[egin{array}{*{20}{c}} 20b + 30p leq 3900 15b + 30p leq 3300 end{array} onumber ]

Graphing these inequalities gives us the solution set.

Since it isn't reasonable to consider negative numbers of items, we can further restrict the solutions to the first quadrant. This could also be represented by adding the inequalities

[b geq 0,quad p geq 0. onumber ]

In the solution set we can see the solution to the system of equations we solved in the previous section: 120 basic products and 50 premium products. The solution set shows that if the company is willing to not fully utilize the staffing, there are many other possible combinations of products they could produce.

The techniques we used above are key to a branch of mathematics called linear programming, which is used extensively in business. We will explore linear programming further in the next section.

Important Topics of this Section

Graphing linear inequalities

Dashed line for strict inequalities, solid for ≤ or ≥

Graphing systems of linear inequalities

### LINEAR INEQUALITIES

**Definition of Inequality:** “ Two real numbers or two algebraic expressions related by any one of the following symbols:

form an **“ INEQUALITY ”**

__Types of Inequalities:__

- Numerical Inequality
- Linear Inequality of one variable
- Linear Inequality of two variable
- Literal Inequality
- Double Inequality
- Quadratic Inequality
- Strict Inequality
- Slack Inequality

- 20 < 100 or 30 > 400
- 30x > 300 or 4y < 8 or 2x – 3 <_ 9 or 5z + 2 – _> 12
- 3x + y > 30 or 4y < 2x – 8 or x + 7 <_ y or 5z + 2y _> -9
- x >_ 7 or y <_ -4 or x < 5 or x > 8
- 3 < 5 < 7 , 2 < y <_ 5
- 4x Square + 3x 6 > 0
- Inequalities with ‘<’ or ‘>’ sign
- Inequalities with ‘≤’ or ‘≥’ sign

__Solving an inequality:__

It is the process of obtaining all possible solutions of an inequality. All values of the variable satisfying the inequality are the solutions of inequality.

__Solution Set :__

The set of all possible solutions of an inequality is called its solution set.

__Replacement Set :__

A set given to us from which the values of variable are replaced in inequality in is called replacement set.

## Which system of linear inequalities has the point (3, –2) in its solution set?

The points were the systems are true in the white areas, the colored areas are the excluded ones.

Now we need to put our point in both systems and see if the point is a solution or not,

In the first one, you can see that y needs to be less than -2, and in out point y is equal to 2, then the point (3,2) cant is a solution of the first system.

let's see the second system:

valuate it in the point (3,2)

Then the point (3,2) is not a solution for neither system, and you can see it in the graphs, in the first graph the point (3,2) is in the black area, and in the second one is in the red area.

and

Which system of linear inequalities has the point (3,-2) in its solution set?

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequalities (makes true both inequalities)

----> inequality A

----> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

----> is not true

The ordered pair is not a solution of the system A

----> inequality A

----> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

----> is true

----> is true

The ordered pair is a solution of the system B

----> inequality A

----> inequality B

Substitute the value of x and y of the point (3,-2) in both inequalities and then compare the results

----> is not true

The ordered pair is not a solution of the system C

----> inequality B

----> is not true

## Operations on Inequalities

When we multiply or divide the same **positive number** from both sides of an inequality, the inequality sign remains **unchanged** .

Find the new inequality when:

a) 3 < 5 is multiplied both sides by 2

b) 18 > 9 is divided both sides by 3

When we multiply or divide the same **negative number** from both sides of an inequality, the inequality sign must be **reversed** . (change < to > and > to <).

Find the new inequality when:

a) 4 < 11 is multiplied both sides by &ndash 2

b) 30 > &ndash9 is divided both sides by &ndash3

a) 4 < 11

4 × (&ndash 2) < 11 × (&ndash 2)

&ndash8 > &ndash22 ( **reverse the inequality sign** )

b) 30 > &ndash9

30 ÷ (&ndash3) > (&ndash9) ÷ (&ndash3)

&ndash10 < 3 ( **reverse the inequality sign** )

The following videos show more examples of solving inequalities:

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.

## 3.2: Linear Inequalities - Mathematics

**Solving Inequalities: An Overview****(page 1 of 3)**

Solving linear inequalities is very similar to solving linear equations , except for one small but important detail: you flip the inequality sign whenever you multiply or divide the inequality by a negative. The easiest way to show this is with some examples:

Graphically, the solution is:

The only difference between the linear equation " *x* + 3 = 2 " and this linear inequality is that I have a "less than" sign, instead of an "equals" sign. The solution method is exactly the same: subtract 3 from either side.

Note that the solution to a "less than, but not equal to" inequality is graphed with a parentheses (or else an open dot) at the endpoint, indicating that the endpoint is not included within the solution.

Graphically, the solution is:

Note that " *x* " in the solution does not "have" to be on the left. However, it is often easier to picture what the solution means with the variable on the left. Don't be afraid to rearrange things to suit your taste.

Graphically, the solution is:

The only difference between the linear equation " 4 *x* + 6 = 3*x* &ndash 5 " and this inequality is the "less than or equal to" sign in place of a plain "equals" sign. The solution method is exactly the same.

Note that the solution to a "less than or equal to" inequality is graphed with a square bracket (or else a closed dot) at the endpoint, indicating that the endpoint is included within the solution.

Graphically, the solution is:

Copyright © Elizabeth Stapel 1999-2011 All Rights Reserved

Graphically, the solution is:

The rule for example 5 above often seems unreasonable to students the first time they see it. But think about inequalities with numbers in there, instead of variables. You know that the number four is larger than the number two: 4 > 2 . Multiplying through this inequality by &ndash1 , we get &ndash4 < &ndash2 , which the number line shows is true:

If we hadn't flipped the inequality, we would have ended up with " &ndash4 > &ndash2 ", which clearly isn't true.

## 4.7 Graphs of Linear Inequalities

We have learned how to solve inequalities in one variable. Now, we will look at inequalities in two variables. Inequalities in two variables have many applications. If you ran a business, for example, you would want your revenue to be greater than your costs—so that your business would make a profit.

### Linear Inequality

A linear inequality is an inequality that can be written in one of the following forms:

### Solution of a Linear Inequality

### Example 4.69

Determine whether each ordered pair is a solution to the inequality y > x + 4 y > x + 4 :

#### Solution

Determine whether each ordered pair is a solution to the inequality y > x − 3 y > x − 3 :

Determine whether each ordered pair is a solution to the inequality y < x + 1 y < x + 1 :

### Recognize the Relation Between the Solutions of an Inequality and its Graph

Now, we will look at how the solutions of an inequality relate to its graph.

### Boundary Line

For an inequality in one variable, the endpoint is shown with a parenthesis or a bracket depending on whether or not a a is included in the solution:

Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to indicate whether or not it the line is included in the solution. This is summarized in Table 4.48

Now, let’s take a look at what we found in Example 4.69. We’ll start by graphing the line y = x + 4 y = x + 4 , and then we’ll plot the five points we tested. See Figure 4.32.

In Example 4.69 we found that some of the points were solutions to the inequality y > x + 4 y > x + 4 and some were not.

Let’s take another point on the left side of the boundary line and test whether or not it is a solution to the inequality y > x + 4 y > x + 4 . The point ( 0 , 10 ) ( 0 , 10 ) clearly looks to be to the left of the boundary line, doesn’t it? Is it a solution to the inequality?

Any point you choose on the left side of the boundary line is a solution to the inequality y > x + 4 y > x + 4 . All points on the left are solutions.

Similarly, all points on the right side of the boundary line, the side with ( 0 , 0 ) ( 0 , 0 ) and ( −5 , −15 ) ( −5 , −15 ) , are not solutions to y > x + 4 y > x + 4 . See Figure 4.33.

### Example 4.70

#### Solution

Since the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality y ≥ 2 x − 1 y ≥ 2 x − 1 .

Write the inequality shown by the graph with the boundary line y = −2 x + 3 y = −2 x + 3 .

Write the inequality shown by the graph with the boundary line y = 1 2 x − 4 y = 1 2 x − 4 .

### Example 4.71

#### Solution

(You may want to pick a point on the other side of the boundary line and check that 2 x + 3 y > 6 2 x + 3 y > 6 .)

Since the boundary line is graphed as a dashed line, the inequality does not include an equal sign.

The graph shows the solution to the inequality 2 x + 3 y < 6 2 x + 3 y < 6 .

Write the inequality shown by the shaded region in the graph with the boundary line x − 4 y = 8 x − 4 y = 8 .

Write the inequality shown by the shaded region in the graph with the boundary line 3 x − y = 6 3 x − y = 6 .

### Graph Linear Inequalities

Now, we’re ready to put all this together to graph linear inequalities.

### Example 4.72

#### How to Graph Linear Inequalities

Graph the linear inequality y ≥ 3 4 x − 2 y ≥ 3 4 x − 2 .

#### Solution

Graph the linear inequality y ≥ 5 2 x − 4 y ≥ 5 2 x − 4 .

Graph the linear inequality y ≤ 2 3 x − 5 y ≤ 2 3 x − 5 .

The steps we take to graph a linear inequality are summarized here.

### How To

#### Graph a linear inequality.

### Example 4.73

Graph the linear inequality x − 2 y < 5 x − 2 y < 5 .

#### Solution

Graph the linear inequality 2 x − 3 y ≤ 6 2 x − 3 y ≤ 6 .

Graph the linear inequality 2 x − y > 3 2 x − y > 3 .

What if the boundary line goes through the origin? Then we won’t be able to use ( 0 , 0 ) ( 0 , 0 ) as a test point. No problem—we’ll just choose some other point that is not on the boundary line.

### Example 4.74

Graph the linear inequality y ≤ −4 x y ≤ −4 x .

#### Solution

Now, we need a test point. We can see that the point ( 1 , 0 ) ( 1 , 0 ) is not on the boundary line.

Graph the linear inequality y > −3 x y > −3 x .

Graph the linear inequality y ≥ −2 x y ≥ −2 x .

Some linear inequalities have only one variable. They may have an *x* but no *y*, or a *y* but no *x*. In these cases, the boundary line will be either a vertical or a horizontal line. Do you remember?

### Example 4.75

Graph the linear inequality y > 3 y > 3 .

#### Solution

So we shade the side that does not include (0, 0).

Graph the linear inequality y < 5 y < 5 .

Graph the linear inequality y ≤ −1 y ≤ −1 .

### Section 4.7 Exercises

#### Practice Makes Perfect

**Verify Solutions to an Inequality in Two Variables**

In the following exercises, determine whether each ordered pair is a solution to the given inequality.

Determine whether each ordered pair is a solution to the inequality y > x − 1 y > x − 1 :

Determine whether each ordered pair is a solution to the inequality y > x − 3 y > x − 3 :

Determine whether each ordered pair is a solution to the inequality y < x + 2 y < x + 2 :

Determine whether each ordered pair is a solution to the inequality y < x + 5 y < x + 5 :

Determine whether each ordered pair is a solution to the inequality x + y > 4 x + y > 4 :

Determine whether each ordered pair is a solution to the inequality x + y > 2 x + y > 2 :

**Recognize the Relation Between the Solutions of an Inequality and its Graph**

In the following exercises, write the inequality shown by the shaded region.

Write the inequality shown by the graph with the boundary line y = 3 x − 4 . y = 3 x − 4 .

Write the inequality shown by the graph with the boundary line y = 2 x − 4 . y = 2 x − 4 .

Write the inequality shown by the graph with the boundary line y = 1 2 x + 1 . y = 1 2 x + 1 .

Write the inequality shown by the graph with the boundary line y = − 1 3 x − 2 . y = − 1 3 x − 2 .

Write the inequality shown by the shaded region in the graph with the boundary line x + y = 5 . x + y = 5 .

Write the inequality shown by the shaded region in the graph with the boundary line x + y = 3 . x + y = 3 .

Write the inequality shown by the shaded region in the graph with the boundary line 2 x + y = −4 . 2 x + y = −4 .

Write the inequality shown by the shaded region in the graph with the boundary line x + 2 y = −2 . x + 2 y = −2 .

Write the inequality shown by the shaded region in the graph with the boundary line 3 x − y = 6 . 3 x − y = 6 .

Write the inequality shown by the shaded region in the graph with the boundary line 2 x − y = 4 . 2 x − y = 4 .

Write the inequality shown by the shaded region in the graph with the boundary line 2 x − 5 y = 10 . 2 x − 5 y = 10 .

Write the inequality shown by the shaded region in the graph with the boundary line 4 x − 3 y = 12 . 4 x − 3 y = 12 .

**Graph Linear Inequalities**

In the following exercises, graph each linear inequality.

Graph the linear inequality y > 2 3 x − 1 y > 2 3 x − 1 .

Graph the linear inequality y < 3 5 x + 2 y < 3 5 x + 2 .

Graph the linear inequality y ≤ − 1 2 x + 4 y ≤ − 1 2 x + 4 .

Graph the linear inequality y ≥ − 1 3 x − 2 y ≥ − 1 3 x − 2 .

Graph the linear inequality x − y ≤ 3 x − y ≤ 3 .

Graph the linear inequality x − y ≥ −2 x − y ≥ −2 .

Graph the linear inequality 4 x + y > −4 4 x + y > −4 .

Graph the linear inequality x + 5 y < −5 x + 5 y < −5 .

Graph the linear inequality 3 x + 2 y ≥ −6 3 x + 2 y ≥ −6 .

Graph the linear inequality 4 x + 2 y ≥ −8 4 x + 2 y ≥ −8 .

Graph the linear inequality y > 4 x y > 4 x .

Graph the linear inequality y > x y > x .

Graph the linear inequality y ≤ − x y ≤ − x .

Graph the linear inequality y ≤ −3 x y ≤ −3 x .

Graph the linear inequality y ≥ −2 y ≥ −2 .

Graph the linear inequality y < −1 y < −1 .

Graph the linear inequality y < 4 y < 4 .

Graph the linear inequality y ≥ 2 y ≥ 2 .

Graph the linear inequality x ≤ 5 x ≤ 5 .

Graph the linear inequality x > −2 x > −2 .

Graph the linear inequality x > −3 x > −3 .

Graph the linear inequality x ≤ 4 x ≤ 4 .

Graph the linear inequality x − y < 4 x − y < 4 .

Graph the linear inequality x − y < −3 x − y < −3 .

Graph the linear inequality y ≥ 3 2 x y ≥ 3 2 x .

Graph the linear inequality y ≤ 5 4 x y ≤ 5 4 x .

Graph the linear inequality y > −2 x + 1 y > −2 x + 1 .

Graph the linear inequality y < −3 x − 4 y < −3 x − 4 .

Graph the linear inequality x ≤ −1 x ≤ −1 .

Graph the linear inequality x ≥ 0 x ≥ 0 .

#### Everyday Math

**Money.** Gerry wants to have a maximum of $100 cash at the ticket booth when his church carnival opens. He will have $1 bills and $5 bills. If *x* is the number of $1 bills and *y* is the number of $5 bills, the inequality x + 5 y ≤ 100 x + 5 y ≤ 100 models the situation.

**Shopping.** Tula has $20 to spend at the used book sale. Hardcover books cost $2 each and paperback books cost .50 each. If *x* is the number of hardcover books Tula can buy and *y* is the number of paperback books she can buy, the inequality 2 x + 1 2 y ≤ 20 2 x + 1 2 y ≤ 20 models the situation.

#### Writing Exercises

Lester thinks that the solution of any inequality with a > sign is the region above the line and the solution of any inequality with a < sign is the region below the line. Is Lester correct? Explain why or why not.

Explain why in some graphs of linear inequalities the boundary line is solid but in other graphs it is dashed.

#### Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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## Solving a Three-part Linear Inequality - Concept

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

In mathematics, it can be useful limit the solution or even have multiple solutions for an inequality. For this we use a

**compound inequality**, inequalities with multiple inequality signs. When solving compound inequalities, we use some of the same methods used in solving multi-step inequalities. The solutions to compound inequalities can be graphed on a number line, and can be expressed as intervals.Solving a three part inequality, basically it's the same as solving a two part inequality but instead of now dealing with just one side we're dealing with another side as well. So first step is always just to get our x's by itself, subtract one over we're trying to solve for this x so just subtract 1 and instead of doing it from both sides we now have to do it from all three. So 11 minus 1, 10 less than 3x and then lastly it equal to 6 okay. Solving for x, we need to divide by negative 3, remember when we are in an inequality form if we divide by a negative we have to actually flip that sign. So 10 divided by a negative 3 we can't simplify that so we're just left with negative 10 thirds our sign switches x this sign has to switch as well 6 divided by negative 3 is negative 2 okay.

So what we actually have is x has to be between negative 10 thirds and -2, so this answer right now is in inequality form okay. We want to throw it into another form we could do our brackets and remember this is called "interval noation" we are not including negative ten thirds so we have a soft bracket here we are including -2 so we have this hard bracket here okay. We could also plot this on a number line we are going from negative ten thirds to negative 2, including negative 2 so that gets filled in not including ten thirds so that doesn't, filling an outline okay. Set bill notation again just taking the same exact thing here with a x such that bracket, x such that negative ten thirds is less than x is less than equal to -2. So solved it out and the four different ways of representing the same exact answer.

## 3.2: Linear Inequalities - Mathematics

Exercise 3.2.1. a) Consider the vectors and where and are arbitrary positive real numbers. Use the Schwarz inequality involving and to derive a relationship between the arithmetic mean and the geometric mean .

b) Consider a vector from the origin to point , a second vector of length from to the point and the third vector from the origin to . Using the triangle inequality

derive the Schwarz inequality. (Hint: Square both sides of the inequality and expand the expression .)

Answer: a) From the Schwarz inequality we have

From the definitions of and , on the left side of the inequality we have

assuming we always choose the positive square root.

From the definitions of and we also have

so that the right side of the inequality is

again assuming we choose the positive square root. (We know is positive since both and are.)

or (dividing both sides by 2)

We thus see that for any positive real numbers and the geometric mean is less than the arithmetic mean .

b) From the triangle inequality we have

for the vectors and . Squaring the term on the left side of the inequality and using the commutative and distributive properties of the inner product we obtain

Squaring the term on the right side of the inequality we have

is thus equivalent to the inequality

Subtracting and from both sides of the inequality gives us

and dividing both sides of the inequality by 2 produces

Note that this is almost but not quite the Schwarz inequality: Since the Schwarz inequality involves the absolute value we must also prove that

(After all, the inner product might be negative, in which case the inequality would be trivially true, given that the term on the right side of the inequality is guaranteed to be positive.)

We have . Since the triangle inequality holds for any two vectors we can restate it in terms of and as follows:

Since squaring the term on the right side of the inequality produces

as it did previously. However squaring the term on the left side of the inequality produces

The original triangle inequality

Since we have both and we therefore have

which is the Schwarz inequality.

So the triangle inequality implies the Schwarz inequality.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition , Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

## Bounded Intervals

reads “−1 one is less than or equal to

*x*and*x*is less than three.” This is a compound inequality because it can be decomposed as follows:The logical “and” requires that both conditions must be true. Both inequalities are satisfied by all the elements in the intersection The set formed by the shared values of the individual solution sets that is indicated by the logical use of the word “and,” denoted with the symbol ∩ . , denoted ∩ , of the solution sets of each.

**Example 5:**Graph and give the interval notation equivalent: x < 3 and x ≥ − 1 .**Solution:**Determine the intersection, or overlap, of the two solution sets. The solutions to each inequality are sketched above the number line as a means to determine the intersection, which is graphed on the number line below.Here x = 3 is not a solution because it solves only one of the inequalities.

Answer: Interval notation: [ − 1 , 3 )

Alternatively, we may interpret − 1 ≤ x < 3 as all possible values for

*x***between**or bounded by −1 and 3 on a number line. For example, one such solution is x = 1 . Notice that 1 is between −1 and 3 on a number line, or that −1 < 1 < 3. Similarly, we can see that other possible solutions are −1, −0.99, 0, 0.0056, 1.8, and 2.99. Since there are infinitely many real numbers between −1 and 3, we must express the solution graphically and/or with interval notation, in this case [ − 1 , 3 ) .**Example 6:**Graph and give the interval notation equivalent: − 3 2 < x < 2 .**Solution:**Shade all real numbers bounded by, or strictly between, − 3 2 = − 1 1 2 and 2.Answer: Interval notation: ( − 3 2 , 2 )

**Example 7:**Graph and give the interval notation equivalent: − 5 < x ≤ 15 .**Solution:**Shade all real numbers between −5 and 15, and indicate that the upper bound, 15, is included in the solution set by using a closed dot.Answer: Interval notation: ( − 5 , 15 ]

In the previous two examples, we did not decompose the inequalities instead we chose to think of all real numbers between the two given bounds.

## 3.2.4 Sequences

Additional foundation content

generate terms of a sequence from either a term-to-term or a position-to-term rule

**Notes**: including from patterns and diagrams.Additional foundation content

recognise and use sequences of triangular, square and cube numbers and simple arithmetic progressions

including Fibonacci-type sequences, quadratic sequences, and simple geometric progressions where is an integer and is a rational number > 0)

including other sequences

including where is a surd

**Notes**: other recursive sequences will be defined in the question.

## Watch the video: #5 Linear Inequalities in two variables Business Mathematics Exercise (October 2021).