# 4.12 Inequalities

In a similar way to how equations express that two mathematical objects must be the same, an inequality expresses that two mathematical objects must be different in a specific way. There are a number of different types of inequalities, for example

 $\displaystyle a>b$ Read \say$a$ is greater than $b$ (4.152) $\displaystyle a Read \say$a$ is less than $b$ (4.153) $\displaystyle a\geqq b$ Read \say$a$ is greater than or equal to $b$ (4.154) $\displaystyle a\geqq b$ Read \say$a$ is less than or equal to $b$ (4.155)
2828 28 The less than symbol can be written as $\leq$ or $\leqq$ - both symbols mean the same thing.

As with equations, inequalities can be manipulated according to a set of rules while maintaining the properties of the inequality.

## 4.12.1 Problems in applying functions to inequalities

One has to apply functions to inequalities. For example, where it is fine to add the same numbers to both sides of an inequality, multiplication can often cause problems! For example, while

2929 29 Read: \say$2$ is less than $3$
$2<3$ (4.157)

It is not the case that the inequality will be true if we multiply both sides by a negative number.

$-2\nless-3$ (4.158)

This same problem occurs when we take the reciprocal3030 30 The reciprocal function is defined as $f(x)=\frac{1}{x}$. of both sides of an inequality. For example, it is true that

$8>4$ (4.159)

However, if we take the reciprocal of both sides, this inequality ceases to hold:

$\frac{1}{8}\ngtr\frac{1}{4}$ (4.160)

Why is this the case? It helps to look at the graph of the reciprocal function here

On the diagram, I have marked two different inputs3131 31 Yes, these happen to be two specific points on the curve, but the principle extends to the entire domain of the reciprocal function - including the negative numbers to the reciprocal function (these are the vertical lines running from the $x$-axis to the curve) and the corresponding outputs (these are the horizontal inputs running from the curve to the $y$-axis). The main thing to notice is that although the orange input was smaller when it was inputted, the actual output is larger than that of the larger red input (but smaller output).

These kinds of informal arguments are very useful when solving problems, but it is also possible to prove this in a more formal manner (see Section 10.7.1 for specifics), but this is not the place.

## 4.12.2 Solutions for problems in applying functions to inequalities

If we have a function $f(x)$ which is decreasing (i.e. as the inputs become larger, the outputs decrease) then to apply it to both sides of an inequality, it is necessary to \sayflip the inequality. For example, if we pick the reciprocal function3232 32 Lucky reciprocal function to be $f(x)$, (i.e. we set $f(x)=\frac{1}{x}$), we might want to apply $f(x)$ to both sides of

$a>b$ (4.161)

To ensure that we still have a valid inequality, we end up with

$\frac{1}{a}<\frac{1}{b}$ (4.162)

## 4.12.3 Adding or removing terms

It seems \sayobvious3333 33 But be careful; so does the pigeonhole principle (Section 6.5) and applying that is far from obvious to say that if $a\leqq b$ then the validity of the inequality is unchanged by any operation which either makes $a$ smaller or $b$ bigger, for example, if we have $a\leqq b$, then both of these inequalities must also be true

 $\displaystyle a\leqq b+1$ (4.163) $\displaystyle a-1\leqq b$ (4.164)

Or, more generally, for any $x>0$, $a\leqq b$ implies that both

 $\displaystyle a\leqq b+x$ (4.165) $\displaystyle a-x\leqq b$ (4.166)

This is a fundamental (read: very useful) principle for manipulating inequalities.

###### Example 4.12.1

Show that $n!\geqq\left(\frac{n}{2}\right)^{\frac{n}{2}}$

As when proving identities (see Section 6.2), we can prove this by starting with one side of the inequality and showing that this is smaller/larger than the other half. Starting with $n!$ we can apply the definition, which gives the equality that

 $\displaystyle n!$ $\displaystyle=n\times(n-1)\times(n-2)\times...\times 3\times 2\times 1$ (4.167)

We now want to make the inequality smaller, and not just in any old way; we specifically want to make it smaller and look like $\left(\frac{n}{2}\right)^{\frac{n}{2}}$. It really helps to consider the structure of $\left(\frac{n}{2}\right)^{\frac{n}{2}}$ here - i.e. it looks something a bit like $\frac{n}{2}\times\frac{n}{2}\times...\times\frac{n}{2}$ (or it would if we could be sure that $\frac{n}{2}$ was an integer, which we can by rounding $\frac{n}{2}$ up - it’s fine to show that $n!$ is bigger than (or equal to) something which is in turn bigger than (or equal to) $\frac{n}{2}$, as this still proves the fact that we are after - which we can write as $\left\lceil\frac{n}{2}\right\rceil$). Let us try to remove some terms to \sayfit $n!$ to $\left(\frac{n}{2}\right)^{\frac{n}{2}}$

 $\displaystyle n!$ $\displaystyle\geqq n\times(n-1)\times...\times\left(\left\lceil\frac{n}{2}% \right\rceil\right)$ (4.169)

We can then make all the terms smaller, like this

 $\displaystyle n!$ $\displaystyle\geqq\left(\left\lceil\frac{n}{2}\right\rceil\right)\times...% \times\left(\left\lceil\frac{n}{2}\right\rceil\right)$ (4.171) $\displaystyle=\left(\left\lceil\frac{n}{2}\right\rceil\right)^{\left\lceil% \frac{n}{2}\right\rceil}$ (4.172) $\displaystyle\geqq\left(\frac{n}{2}\right)^{\frac{n}{2}}.$ (4.173)

## 4.12.4 Some interesting inequality proofs

###### Example 4.12.2

Let $a,b,c$ be positive reals such that

$a^{2}-b^{2}>c^{2}-a^{2}.$ (4.174)

Prove that

$a-b>c-a$ (4.175)

We want to prove the statement

 $\displaystyle a^{2}-b^{2}>c^{2}-a^{2}\implies a-b>c-a.$ (4.176)

We can prove the equivalent statement,

$a-b\leqq c-a\implies a^{2}-b^{2}\leqq c^{2}-a^{2}.$ (4.177)

Which in turn is the same as

$2a\leqq b+c\implies 2a^{2}\leqq c^{2}+b^{2}.$ (4.178)

Let us fix some $a,b,c\in\mathbb{R}$ such that $a,b,c\geqq 0$. Then let us assume further that $2a\leqq b+c$ in which case

 $\displaystyle 2a^{2}$ $\displaystyle\leqq 2\left(\frac{b+c}{2}\right)^{2}$ (4.179) $\displaystyle\leqq\frac{1}{2}c^{2}+cb+\frac{b^{2}}{2}$ (4.180)

This is pretty close to what we want - in fact if we could prove that $cb\leqq\frac{1}{2}c^{2}+\frac{1}{2}b^{2}$ we would be done.

 $\displaystyle cb\leqq\frac{1}{2}c^{2}+\frac{1}{2}b^{2}$ (4.181) $\displaystyle\iff 0\leqq\frac{1}{2}c^{2}-cb+\frac{1}{2}b^{2}$ (4.182) $\displaystyle\iff 0\leqq c^{2}-2cb+b^{2}$ (4.183) $\displaystyle\iff 0\leqq(c-b)^{2}$ (4.184)

and the last statement is always true.

Therefore,

 $\displaystyle 2a^{2}$ $\displaystyle\leqq\frac{1}{2}c^{2}+cb+\frac{b^{2}}{2}$ $\displaystyle\leqq\frac{1}{2}c^{2}+\frac{1}{2}c^{2}+\frac{1}{2}b^{2}+\frac{b^{% 2}}{2}$ (4.185) $\displaystyle\leqq c^{2}+b^{2}$ (4.186)

from which it follows that

 $\displaystyle a^{2}-b^{2}\leqq c^{2}-a^{2}$ (4.187)

and then we can forget that we fixed $a,b,c$ and we have a proof for any $a,b,c$, so we are done.

$\Box$

## 4.12.5 The Cauchy-Schwarz inequality

This is a really significant inequality which is useful for proving lots of things. The Cauchy-Schwarz inequality states that for any sequence of real numbers $a_{1},a_{2},...,a_{n}$ and $b_{1},b_{2},...,b_{n}$, the pairwise products are less than products of the square roots of the sum of the squares, that is, that

$a_{1}b_{1}+a_{2}b_{2}+...+a_{n}b_{n}\leqq\sqrt{(a_{1})^{2}+(a_{2})^{2}+...+(a_% {n})^{2}}\sqrt{(b_{1})^{2}+(b_{2})^{2}+...+(b_{n})^{2}}$ (4.188)

This theorem says a lot, so it helps to consider what it says in some specific cases, for example if we have the sequences $1,-1,5,12,-64$ and $-12,7,3,90,4$ then the Cauchy-Schwarz inequality tells us that

 $\displaystyle(1)(-12)+(-1)(7)+(5)(3)+(12)(90)+(-64)(4)$ (4.189) $\displaystyle\quad\leqq\sqrt{(1)^{2}+(-1)^{2}+(5)^{2}+(12)^{2}+(-64)^{2}}$ (4.190) $\displaystyle\quad\quad\sqrt{(-12)^{2}+(7)^{2}+(3)^{2}+(90)^{2}+(4)^{2}}$ (4.191)

This is true as (at least when I plugged the numbers into my calculator) the left-hand side is equal to $820$ (what a nice, whole number) whereas the right-hand side is approximately equivalent to $5957.59229891$ (which is not a nice, whole number).

Of course, this isn’t a proof - it’s just an example! To prove this in the general case (i.e. for every $n\in\mathbb{N}$) there’s a non-zero chance that we should use induction3434 34 As introduced in Section 6.3 (because whenever we want to prove that some property holds for all natural numbers, it’s likely that induction will help).

How do we prove this in the general case? Well, let’s apply the principle of mathematical induction.

• If $n=0$ the statement is trivially true (as $0\leqq 0$ - note: let’s agree that the sum over $0$ terms is $0$). In case you’re unconvinced that either $0$ is a natural number (don’t worry, there’s no shame in being wrong) or that the inequality is really defined for $n=0$, we can also show that it’s true for $n=1$. As a value is less than or equal to its modulus (as proven in Section 4.9.5)

 $\displaystyle a_{1}b_{1}$ $\displaystyle\leqq|a_{1}b_{1}|$ (4.192) $\displaystyle=\sqrt{(a_{1}b_{1})^{2}}$ def. modulus function (see Section 4.74) (4.193) $\displaystyle=\sqrt{(a_{1})^{2}}\sqrt{(b_{1})^{2}}$ (4.194)
• Now onto the inductive step. We assume the statement for $k$ is true (see Equation 4.195) and then our goal (which it is important not to lose sight of, in life as well as mathematics problems) is to show that the statement is also true for $n=k+1$ (this is an example of going from $k$ to $k+1$, rather than trying to spot the case of $k$ lurking inside the case of $k+1$ - see Section 6.3.2)

$a_{1}b_{1}+a_{2}b_{2}+...+a_{k}b_{k}\leqq\sqrt{(a_{1})^{2}+(a_{2})^{2}+...+(a_% {k})^{2}}\sqrt{(b_{1})^{2}+(b_{2})^{2}+...+(b_{k})^{2}}$ (4.195)

We would like to make this inequality true for the case of $k+1$, so we should do something to manipulate its structure which gives us this case. There are two options, trying to add an $(a_{k+1})^{2}$ (and also the same for $b$) inside the square roots, or perhaps we can try adding $a_{k+1}b_{k+1}$ to both sides. I know which one I prefer (the latter).

 $\displaystyle a_{1}b_{1}+...+a_{k}b_{k}+a_{k+1}b_{k+1}$ $\displaystyle\leqq\sqrt{(a_{1})^{2}+...+(a_{k})^{2}}\sqrt{(b_{1})^{2}+...+(b_{% k})^{2}}+a_{k+1}b_{k+1}$ (4.196)

Now we need to try to combine $a_{k+1}b_{k+1}$ by using operations which either leave the value of the right-hand side the same, or make it bigger. In the end, we would like a square root, and we do know one way to make a bigger expression out of the numbers $\sqrt{(a_{1})^{2}+...+(a_{k})^{2}}\sqrt{(b_{1})^{2}+...+(b_{k})^{2}}$ and $a_{k+1}b_{k+1}$; the inequality we are trying to prove, but for the case $n=2$ (which is fantastic because proving the statement for $n=2$ is a great deal easier than proving it for all $n$). Therefore, applying the Cauchy-Schwarz inequality in the case where $n=2$, accepting that we have sacrificed some generality (which we will need to regain by proving that the statement is true for $n-2$) this is less than or equal to

 $\displaystyle a_{1}b_{1}+...+a_{k}b_{k}+a_{k+1}b_{k+1}$ $\displaystyle\leqq\sqrt{\sqrt{(a_{1})^{2}+...+(a_{k})^{2}}^{2}+(a_{k+1})^{2}}% \sqrt{\sqrt{(b_{1})^{2}+...+(b_{k})^{2}}+(b_{k+1})^{2}}$ (4.197) $\displaystyle=\sqrt{(a_{1})^{2}+...+(a_{k})^{2}+(a_{k+1})^{2}}\sqrt{(b_{1})^{2% }+...+(b_{k})^{2}+(b_{k+1})^{2}}$ (4.198)

TODO: proof for case $n=2$