10.1 Foundations and Motivation

In this chapter, we consider inequalities.

Definition 10.1 (Inequalities).

Given $x,y\in\mathbb{R}$ , if there exists a positive $z\in\mathbb{R}$ such that

x+z=y

then we say “ $x$ is less than $y$ ” and we write $x<y$ .

Equivalently, we may say “ $y$ is greater than $x$ ” and write $y>x$ .

If $z\in\mathbb{R}$ may have the value 0, we say “ $x$ is at most $y$ ” or “ $x$ is less than or equal to $y$ ”, and we write $x\leqslant y$ . Equivalently, we may say “ $y$ is at least $x$ ” and write $y\geqslant x$ .

Informally, we say $x<y$ exactly when “ $x$ is strictly to the left of $y$ on the number line” and $y>x$ when “ $y$ is strictly to the right of $x$ ”.

Figure 10.1: A real number line with

x<y

and

y>x

When we write $x\leq y$ we mean that either $x<y$ or $x=y$ . Hence it is true that $2\leq 3$ .

Inequalities are particularly useful to allow us to place a constraint on what values a variable can take.

Definition 10.2 (Bounds).

If $x\leqslant y$ or $x<y$ , we say that $y$ is an upper bound for $x$ and $x$ is a lower bound for $y$ .

If the bound we use is the best possible bound (so $x<y-a$ is not valid for any $a>0$ ), then we call the bound sharp.

10.1.1 Axioms of Inequalities

There are some foundational rules¹¹ 1 An axiom is a foundational statement that is accepted as true without any proof. You will see more on axioms next semester in IMA but can read more about them now. that the inequality obeys. Note that the rules below only refer to the strict inequality $<$ but the corresponding rules for $>$ , $\leqslant$ and $\geqslant$ can be derived from these rules as a result of their definitions.

Definition 10.3 (Axioms of Inequalities on the Reals).

For every $x,y,z\in\mathbb{R}$ , the inequality $<$ on $\mathbb{R}$ obeys the following rules.

(R1)

Exactly one of the following is true: $0<x$ , $x=0$ , or $x<0$ .
(R2)

If $x<y$ and $c\in\mathbb{R}$ , then $x+c<y+c$ .
(R3)

If $0<x$ and $0<y$ , then $0<xy$ .
(R4)

If $x<y$ and $y<z$ , then $x<z$ . [Transitivity of $<$ ]

Again, you may be able to think of some more rules for inequalities that aren’t listed in Definition 10.3 but we can derive them from these rules for inequalities and our core rules for addition and multiplication over $\mathbb{R}$ .

Proposition 10.4 (More rules for inequalities).

For every $x,y,z\in\mathbb{R}$ ,

(a)

if $x>0$ , then $-x<0$ .
(b)

if $y<x$ , then $-x<-y$ .
(c)

if $x\neq 0$ , then $x^{2}>0$ .
(d)

if $x>0$ then $\frac{1}{x}>0$ .
(e)

if $x>0$ , then $y>z$ if and only if $xy>xz$ .
(f)

if $0<x$ and $0<y$ , then $x^{2}>y^{2}$ if and only if $x>y$ .

Proof.

(a)

Suppose $x>0$ . Then taking $c=-x$ in (R2), we have $x+(-x)>0+(-x)$ and thus $0>-x$ .

Exercise 10.5.

Prove the remaining claims in Proposition 10.4 using the axioms in Definition 10.3.

Solution (please try for yourself before looking)

For every $x,y,z\in\mathbb{R}$ ,

(b)

Suppose $y<x$ . Adding $c=-x-y$ to both sides of the inequality using (R2), we have $-x<-y$ .
(c)

Suppose $x\neq 0$ .

If $x>0$ , then by (R3) we have $x^{2}>0$ .

If $x<0$ , then $0<-x$ by part (b), and

$(-x)(-x)=x^{2}>0$

by (R3).
(d)

Suppose $x>0$ and define $y=\frac{1}{x}\neq 0$ . By (R1) either $y>0$ or $y<0$ .

If $y<0$ then $0<-y$ (by part (b)) and so

$0<x\cdot(-y)=-x\frac{1}{x}=-1.$

This is a contradiction. So we cannot have $y<0$ .

Hence $y=\frac{1}{x}>0$ .
(e)

Suppose $x>0$ .

If $y>z$ , then $y-z>0$ by (R2). Next $xy-xz>0$ by (R3), and so $xy>xz$ by (R2).

Now suppose $xy>xz$ . Then $x(y-z)>0$ by (R2). By part (d) we have $\frac{1}{x}>0$ , so multiplying both sides of our inequality by $\frac{1}{x}$ gives us $y-z>0$ by (R3). Finally $y>z$ by (R2).
(f)

Suppose $0<x$ and $0<y$ .

If $x>y$ then (i) $x^{2}>xy$ by part (e), and (ii) $xy>y^{2}$ by part (e). Combining these with (R4) gives $x^{2}>y^{2}$ .

If $x^{2}>y^{2}$ then $(x+y)(x-y)=x^{2}-y^{2}>0$ by (R2). We know $x+y>0$ by (R2) so $\frac{1}{x+y}>0$ by part (d). Hence $x-y=\frac{1}{x+y}(x+y)(x-y)>0$ by (R3). Therefore, $x>y$ by (R2).

Many inequalities are derived from the fact that if two factors have the same algebraic sign then their product is positive.

Exercise 10.6.

If $a,A,b,B$ are all positive, and $a>b$ and $A>B$ then prove that $Aa>Bb$ .

Solution (please try for yourself before looking)

Since $A>0$ and $a>b$ , then by Proposition 10.4 (e) $Aa>Ab$ .

Since $b>0$ and $A>B$ , then by Proposition 10.4 (e) $Ab>Bb$ .

Since $Aa>Ab$ and $Ab>Bb$ it follows that $Aa>Bb$ by transitivity (Definition 10.3 (R4)).

Proposition 10.7.

For all $a,b\in\mathbb{R}$ , $ab>0$ if and only if $a,b>0$ , or $a,b<0$ .

Proof.

Suppose $a,b>0$ or $a,b<0$ . If $a,b>0$ , then $ab>0$ follows from Definition 10.3 (R3). If $a,b<0$ , then Proposition 10.4 (a) tells us $-a,-b>0$ . Thus, by Definition 10.3 (R3), we get $(-a)\cdot(-b)>0$ and hence, $ab>0$ .

For the other direction, suppose $ab>0$ . We have two cases.

Case 1: $a>0$ . Then, since $ab>0=a\cdot 0$ , by Proposition 10.4 (e) we have $b>0$ .

Case 2: $a<0$ . Then, by Proposition 10.4 (b), $-a>0$ and thus, $\frac{1}{-a}>0$ by part Proposition 10.4 (d). Hence, by Definition 10.3 (R3), we have $\frac{1}{-a}\cdot ab>0\implies-b>0$ and hence, $b<0$ by Proposition 10.4 (b).

NOTE: For the basic rules of inequalities stated above, we usually use them in proofs without explicit reference to these results every time, just as we would typically take for granted that the order of multiplication and addition in the reals doesn’t matter (commutativity), or the product of two integers produces an integer.

We can continue building on our rules for inequalities to prove more complicated statements.

Exercise 10.8.

Prove that for any $a,b\in\mathbb{R}$ ,

ab\leq\frac{a^{2}+b^{2}}{2}

(10.1)

Solution (please try for yourself before looking)

For any $a,b\in\mathbb{R}$ we have $(a-b)^{2}\geq 0$ . Expanding out the brackets gives

a^{2}-2ab+b^{2}\geq 0.

Adding $2ab$ to both sides, and dividing both sides by $2$ (which of course is positive, so we don’t change the inequality) gives

ab\leq\frac{a^{2}}{2}+\frac{b^{2}}{2}.

Exercise 10.9.

Prove that

a^{2}+b^{2}+c^{2}>bc+ca+ab,

holds for all $a,b,c\in\mathbb{R}$ , except when $a=b=c$ .

Solution (please try for yourself before looking)

Solution 1: First note that

2(a^{2}+b^{2}+c^{2}-bc-ca-ab)=(b-c)^{2}+(c-a)^{2}+(a-b)^{2}.

Unless $a,b,c$ are equal, the right hand side is positive. This gives us our desired inequality.

Solution 2: Use the result (10.1) three times, namely,

\frac{b^{2}+c^{2}}{2}>bc

\frac{c^{2}+a^{2}}{2}>ca

\frac{a^{2}+b^{2}}{2}>ab

and add these together.