**Harder Inequalities**

## Algebraic Inequalities

Contents

In the extension 1 course, we have looked at solving simple inequalities with the unknown on the denominator and proving basic inequalities using Induction. This topic is an extension on the extension 1 course where we prove inequalities using algebra, calculus and Mathematical Induction. However, we will cover inequalities involving Mathematical Induction later on in the Induction topic. Finally, we will look at solving inequalities using algebraic and graphical methods.

### Inequalities Involving

Using the fact that the square of a real number is always greater than or equal to zero allows us to solve more complicated inequalities. As a simple exercise using , let’s prove the following relationships where :

(i)

(ii)

**Proof**

i)

ii)

From i)

The methods of obtaining many inequality relationships involve:

- Expansion and simplification
- Substitution of an expression into an existing relationship
- Rearrangement
- Addition and multiplication
- Work forward
- Work backward

The following examples demonstrate how these concepts can be used in establishing inequality relationships.

**Example 1**

Prove that , where and .

### Solution 1

**Example 2**

Use the expression , show that

(i)

(ii)

### Solution 2

i)

Similarly,

and

Summing up the LHS and RHS, we have

ii)

Given

Let

Now, we can substitute for , and similarly for without loss of generality.

**Example 3**

Prove that , where .

### Solution 3

Working backward,

**<–>**

**<–>**

**<–>**

**<–>**

**<–>**

Which is clearly true,

**Example 4**

Prove that , where, where

### Solution 4

Recall that

Therefore, substituting for

We have

**Example 5**

[HSC 2011, Question 5]

If and are positive real numbers and , prove that

### Solution 5

Consider the numerator:

Numerator is greater than 0.

Since the denominator is also greater than 0, therefore .

**Example 6**

[HSC 2012, Question 15]

(i) Prove that , where and .

(ii) If , show that .

(iii) Let and be positive integers with . Prove that

(iv) For integers, prove that

### Solution 6

i)

ii)

Noting that

iii)

Consider

Let

Finally, consider

To maximise this, we solve for

Test: Therefore, it is a maximum.

Thus, maximum value for But

Thus,

iv)

From iii), we have for

Substituting we have

Multiplying both sides, we get

## Further Inequalities

One of the inequalities that we have looked at earlier is a famous inequality which states the geometric mean of two positive numbers is always less than or equal to the arithmetic mean. The inequality can also be generalised for terms and gives:

Note: Students are not allowed to assume the above inequality in examinations.

In fact, we can extend the above inequality further into the following inequality.

The Harmonic Mean, Geometric Mean, Arithmetic Mean and Quadratic Mean Inequalities:

Another famous inequality is the Cauchy-Schwartz Inequality which states:

where,

Note: these two above inequalities are important to know and understand but they are not mentioned in the syllabus thus students are not required to learn these.

## Application of Calculus in Inequalities

### Differential Calculus

Differential calculus can be applied to prove inequalities that require us to show whether a function is increasing or decreasing on a given interval.

If the question requires us to show for , then we can follow the steps below:

- prove that , which means is increasing for .
- hence, .
- show that, if , then for .

and vice versa for decreasing functions.

Note: The above method can also be applied for second and higher derivatives if the first derivative is insufficient to prove for .

### Integral Calculus

Integral calculus is the other branch of calculus that can be applied to prove inequalities and of course, on the bases that we are able to integrate the inequalities. Now consider the following diagram,

We can see that, on the interval and therefore,

Note: In many inequality questions, integral calculus is often used in conjunction with differential calculus.

Let’s now look at some examples!

**Example 3**

[HSC 2007, Question 7]

(i) Show that for .

(ii) Let . Show that the graph of is concave up for .

(iii) By considering the first two derivatives of , show that for .

### Solution 3

i)

Consider

Also,

Thus, is always decreasing and is equal to 0 for .

Therefore, for , .

ii)

From i), we know that for .

, and is concave up.

iii)

Consider

Since , that means is always increasing.

Furthermore,

Thus, for since it starts at 0 and is increasing.

Therefore, since and for , that means .

for.

**Example 4**

[HSC 2009, Question 8]

Let be a positive integer greater than 1.

The area of the region under the curve from to is between the areas of two rectangles, as shown in the diagram.

Show that

### Solution 4

Note that the exact area under the graph from n-1 to n is greater than the area of the smaller rectangle, and smaller than the area of the large rectangle.

Thus,

**Example 5**

[HSC 2006, Question 8]

Suppose .

(i) Show that .

(ii) Hence show that .

(iii) By integrating the expressions in the inequality in part (ii) with respect to from and (where ), show that

(iv) Hence show that for

### Solution 5

i)

Now,

(noting that )

Also,

.

ii)

iii)

iv)

Raise iii) to the power of

## Solving Inequalities

We have studied solving inequations in the Maths Extension 1 course such as . In the Extension 2 course, the problems will be a lot more difficult and so to solve them more efficiently, we will be using both the algebraic and graphical methods.

The following examples demonstrate how we can use both the algebraic and graphical methods to solve harder inequations.

**Example 6**

Solve the following inequality for ,

### Solution 6

Since , we can multiply both sides without changing sign (noting also that our final answer is now restricted to ).

Below is the plot the cubic on the LHS:

Thus, the inequality holds when or .

However, recall that

Thus, solutions is .

**Example 7**

Solve the following inequality for ,

### Solution 7

Consider two cases:

Case 1:

When this is true, the numerator will be positive. Thus, for the LHS to be positive, we only require that the denominator also be positive.

Solving , we have:

But, is only true when or

Thus, combining these, we have or

Case 2:

When this is true, the numerator will be negative. Thus for the LHS to be positive, we only require the denominator to be negative.

Solving , we have:

or

But, is only true when

Thus, combining these, we have no solution for this case.

Therefore, the overall solutions is or .