# Basic Techniques for Proving Algebraic Inequalities

## The Need for Proving Algebraic Inequalities

A central part of the methodology underlying all applications of algebraic inequalities to reduce uncertainty and risk consists of techniques for proving algebraic inequalities. Despite the existence of extensive literature on algebraic inequalities and numerous solved examples involving algebraic inequalities, there is a clear absence of systematic exposition of the techniques through which an algebraic inequality can be proved. A good awareness of the available tools by which an inequality can be proved significantly increases the benefit from using the algebraic inequalities. The systematic exposition of such techniques reduces the dependence on inspiration and clever tricks, w'hich enhances the power of researchers and practitioners in using inequalities to reduce uncertainty and risk.

A number of techniques for proving inequalities have already been covered in the literature related to algebraic inequalities. However, there are also many powerful methods and techniques for proving algebraic inequalities which have not received adequate coverage in the available literature. Consequently, an important task of this chapter is compiling well-known methods and introducing little-known methods for proving algebraic inequalities.

## Proving Inequalities by a Direct Algebraic Manipulation and Analysis

This technique is based on direct algebraic manipulation and analysis demonstrating that the inequality indeed holds. This technique will be illustrated by proving the arithmetic mean - harmonic mean (AM-HM) inequality

where a, are positive real numbers. To prove this inequality, it suffices to prove that
Clearly, for two real numbers only, the inequality
is true. Indeed, expanding the left-hand side of (3.3) gives
For any positive numbers a, and *a _{2}*

is fulfilled. This follows from the inequality

which follows directly from the inequality *(yfai - л/сь) ^{2}* > 0.

Indeed, squaring both sides of inequality (3.6) (which are positive numbers) and dividing by *a _{{}a_{2}* yields inequality (3.5), directly.

Expanding the left-hand side of inequality (3.2) gives *n ^{2}* terms of the form f-, 1 <

*ij < n.*From these

*n*terms,

^{1}*n*terms have the same index

*i = j*and for these terms “■ — 1. The rest of the

*re -n*terms can be paired in the sums

*%- + %-■*According to what has been proved earlier, for each of the paired sums

*^- + ^->2*holds. The number of these paired sums is («' -

*n)/2*and the left-hand side of (3.2) becomes

which proves inequalities (3.2) and (3.1). Equality is attained for *а _{л}=а*

_{2}=... =

*a„.*

**Example 3.1**

Let a,,a_{2},...,a„ be real numbers and *b _{b}b_{2},...,b_{n}* be positive real numbers. Prove the

*Bergstrom inequality*(Pop, 2009) by using a direct algebraic manipulation.

Proof. Consider only two sequences *a, a _{2}* and b,,

*b*containing only two real numbers each. Proving that

_{2}

is equivalent to proving that

Simplifying the left side gives

(ba; a_{2}) ^ q _{w}[_{1}j_{c}[_{1} j_{s a}|_{wa}y_{S} non-negative. Equality is present only if *Ь]02{Ь]* + *O**2**) *bi/a, *= b _{2}/a_{2}.*

Now, if the value is added to both sides of inequality (3.9), the same reasoning can be applied to the two terms *—г~Г~ 'T~* ^{anc}^ the inequality

D| + p2 O}

is obtained. Continuing this reasoning proves the inequality for a,,a_{2},...,a„ and bi,b_{2},...,b„.

Equality is attained only if a, / b| = a_{2} / *b _{2} *=... = a„ / b„.

The next example demonstrates proving an inequality by a direct analysis.

**Example 3.2**

For the four positive numbers *x >* 0, *у >* 0 and *z* > 0, f > 0 prove that
Inequality (3.12) is equivalent to the inequality

Note that *2yfxt because *

*(y[x-yfi)*0 and

^{2}>*l^jyz^y + z*because

*(Jy-fz)*0. The left-hand part of inequality (3.13) is a difference of two non-negative numbers 2^IL and each of which is smaller than 1. Therefore their difference is smaller than 1, inequality (3.13) is true, and therefore inequality (3.12) is also true.

^{2}>

## Proving Inequalities by Presenting Them as a Sum of Non-Negative Terms

This technique is also based on a direct algebraic manipulation following a powerful strategy. The inequality *f(x _{]},...,x„)>0* is proved by transforming it into a sum of terms gy(x,,...,

*x„) >*0,

*j*= 1 each of which is non-negative:

To illustrate this important technique, consider the inequality

where *x,y,z* are real numbers. The inequality is equivalent to the inequality

Starting with the identity *(a* - *b*)^{2} = *a ^{2}* -

*lab*

*+ b*a presentation for the left side of inequality (3.16) as a sum of non-negative terms is sought. The left-hand side of inequality (3.16) can be presented as a sum of squares which are guaranteed to be non-negative:

^{1},

and is therefore non-negative which completes the proof of inequalities (3.16) and (3.15). Equality in (3.15) is attained for *x = y = z-*

A similar technique can also be used to prove the Cauchy-Schwarz inequality,

which holds for any two sequences of real numbers *a _{u}a_{2},...,a„* and

*b*

_{x},b_{2},...,b„.Since the difference between the left and right side of inequality (3.18) can be presented as a sum of non-negative terms:

inequality (3.18) and the conditions for attaining equality can be deduced directly from the equivalent inequality:

As can be seen, equality holds for proportional *a _{i}* and

*b*if and only if for any

_{h}*i Ф j, a,*/ /;, =

*а*/

_{}}*bj*are fulfilled.

Presenting an inequality as a sum of non-negative terms is not necessarily confined to presenting it as a sum of squares. Non-negative terms can be formed by relentless factoring of the separate parts of the inequality. This powerful technique is illustrated by the next example.

Example 3.3

If *a,b* are real numbers such that 0 < a < 1 and 0 < *b <* 1, prove the inequality

Proof. First note that the first term *a ^{2}b*

^{2}in (3.20) can be presented as (1-a

^{2})(1-b

^{2})-1 + a

^{2}+b

^{2}- As a result, the left side of inequality (3.18) can be presented as

Factoring the expression -ab + a + b-1 in the right side of (3.21) results in *-ab* + a + b -1 = —(1 — b)(1 - a) and the right side of equality (3.21) becomes

Taking out the factor (1 — a) in the right side of equality (3.22) gives

and taking out another factor (1-b) in the right side of (3.23) finally results in

As a result, the left side of inequality (3.20) has been presented as a sum of two non-negative terms:

because (1 — a) > 0 and (1-b) >0. This completes the proof of the original inequality (3.20).

Often, by exploiting the symmetry in the inequality, the variables in the inequality can be ordered, which permits presenting the inequality as a sum of non-negative terms. Section 3.6 demonstrates this approach.