# Weighted Arithmetic Mean–Geometric Mean (AM-GM) Inequality

Table of Contents:

The weighted arithmetic mean-geometric mean (AM-GM) inequality is an important application of the Jensen inequality.

For a set of positive real numbers x,,x2,...,x„, and a set tht2,...,t„ of positive weights t{ + t2 + ... + /„ = 1, the AM-GM inequality can be generalised with the weighted AM-GM inequality

Introduce the values a„ i = 1 ,.,.,/r as follows: a, = lnx,,...,a„ = lnx„. Because ex is a convex function, from the Jensen inequality, the next inequality is obtained

Since e'm +-+,''“n = x['x'i...x'„" and tea> + ... + tnean =tX + ... + tnx,„ the weighted AM- GM inequality (2.23) follows directly.

An important corollary of the weighted AM-GM inequality is the Young inequality. If x,y are positive numbers and a,b are positive numbers such that Ma + /b = 1, the Young inequality states that

The Young inequality follows directly from the weighted AM-GM inequality by noticing that xy = (x‘')Ua x(yh)uh. According to the weighted AM-GM inequality (2.23)

# Hölder Inequality

Let au,al2,...,al„;a2l,a22,—,a2„',.:;ambam2,...,am be m sequences of positive real numbers and be m positive real numbers adding up to 1 (Z'"|A, = l).

Then the Holder inequality holds (Hardy et al, 1999):

Inequality (2.25) is also known as the elementary form of the Holder inequality.

For two sequences х[,х^,...,хЦ and У,Уг,—,у1, of positive real numbers (m = 2, p>0,q>0,) and weights A, = 1 Ip; A2 = /q, (A, + A2 = l), the Holder inequality yields

Inequality (2.26) is the Holder inequality for sums which can be proved by using the Young inequality.

Many inequalities can be proved by a direct application of the Holder inequality. If a, b, c, x, у and z are positive real numbers, by a direct application of the Holder inequality, for example, it can be shown that the following inequality

holds.

Indeed, note that the inequality

is equivalent to the original inequality because the original inequality is obtained by dividing both sides by 3(x + у + г).

Taking A, = A2 = A3 = 1/3 (A! + A2 + A3 = l) and applying the Holder inequality to the three sequences (1,1,1), (х,у,г) and (o’lx, b*/y, c}/z) gives:

Raising both sides of this inequality to the power of 3 gives which completes the proof of the original inequality.

# Cauchy-Schwarz Inequality

One of the most important algebraic inequalities is the Cauchy-Schwarz inequality (Cauchy, 1821), which states that for the sequences of real numbers aua2,...,a„ and bt,b2,...,b„, the following tight inequality holds:

Equality holds if and only if for any i Ф j, cijbj = aft are fulfilled.

The Cauchy-Schwarz inequality is a powerful inequality. Many algebraic inequalities can be proved by reducing them to the Cauchy-Schwarz inequality by using an appropriate substitution. It has also a simple geometrical interpretation.

The Cauchy-Schwartz inequality (2.27) is equivalent to the inequality

If the sequences of real numbers a,a2,...,a„ and b,b2,...,b„ are interpreted as the components of two vectors in the и-dimensional space, the numerator on the left side of (2.28) is the dot product of the two vectors and the denominator is the product of the magnitudes of the vectors. Consequently, the expression on the left side of inequality (2.28) is the cosine of the angle between these two vectors. In the three- dimensional space (Figure 2.1) the cosine of the angle between two vectors with components , a2,a} and b, b2, b} is given by

and the Cauchy-Schwarz inequality expresses the fact that the cosine of an angle is a number which cannot exceed 1.

FIGURE 2.1 Geometric interpretation of the Cauchy-Schwartz inequality.

The Cauchy-Schwarz inequality is a special case of the Holder inequality (2.26) for p = q = 2.

An alternative way to prove the Cauchy-Schwarz inequality (2.27) is to use the properties of the quadratic trinomial introduced in Section 2.4. Consider the expression

For any sequences of real numbers aua2,...,a„ and b{,b2,...,b„, у is non-negative (y > 0). Expanding the right-hand side of (2.29) and collecting the coefficients in front of t2 and t, gives the quadratic trinomial

with respect to t. In equation (2.30), у is non-negative (y > 0) only if D < 0, where

is the discriminant of the quadratic trinomial. Therefore, the condition

must hold for a non-negative y. The condition (2.31) is identical to the Cauchy- Schwarz inequality (2.27) which completes the proof.

The Cauchy-Schwarz inequality will be illustrated by proving that

where a,b,c are positive real numbers. _ _ _

Proof. Consider the two sequences fa,[b,yfc and Msfa, l/yfb, 1 Д/с. From the Cauchy-Schwarz inequality:

from which the inequality follows directly.

# Rearrangement Inequality

The rearrangement inequality is a powerful yet underused inequality that can be applied to provide bounds for the uncertainty associated with reliability-critical parameters.

Consider the two sequences aua2,...,a„ and .........b„ of real numbers. It can be

shown that: a. The sum S = axbx + a2b2 +... + anbn is maximal if the sequences are sorted in the same way - both monotonically decreasing:

«, > a2 >,...,> «„; 6, > b2 >,...,> 6„ or both monotonically increasing:

«1 S a2 «„; 6, < b2 <,...,< 6„.

b. The sum S = «,6, +a2b2 + ... + «„6„ is minimal if the sequences are sorted in the opposite way: one monotonically increasing and the other monotonically decreasing.

To prove the first statement, the extreme principle will be used. Suppose that there is a sum

where the «-sequence and 6-sequence are not both monotonically increasing or both monotonically decreasing and which is the largest possible sum. If the «-sequence and 6-sequence are not both sorted in ascending order or in a descending order, there will certainly be values ar, br and as, bs (r < л) for which either < as and 6, > bs is true or > as and br < bs is true. If no such pairs can be found, then the «-sequence and 6-sequence are already either both increasing or both decreasing and the hypothesis that the «-sequence and 6-sequence are not both sorted in ascending or in descending order does not hold.

Suppose that ar < «., and br > 6, is true. Without loss of generality, it can be assumed that in equation (2.32) the terms «,6, (/ = 1,...,«) have been sorted in ascending order of «, . This can always be done by a simple permutation of the terms. Now, consider the sum 5,

which has been obtained from the sum S0 by switching the positions of 6, and bs only. Subtracting S, from S0 gives:

Because ars and 6,. > 6V is true, then So -Si = (ar -as)(br -6,) < 0; therefore the sum 5, is larger, which contradicts the assumption that S0 is the largest sum. Consequently, the hypothesis that the largest sum can be attained for sequences that are not both increasing or both decreasing is incorrect. The case ar > as and 6, < bs leads to a contradiction in a similar manner.

In a similar fashion, statement (b) can also be proved.

An important application of the rearrangement inequality is its use as a basis for generating new useful inequalities that can be used to produce bounds for the uncertainty in reliability-critical parameters. For two sequences a,,«2,...,«„ and 6|,62,...,6„ of real numbers, the notation

can be introduced (Engel, 1998). This is similar to the definition of a dot product of two vectors with components specified by the two row's of the matrix.

Tw'o important corollaries of the rearrangement inequality are the following:

i. Given a set of real numbers a,a2,...,a,„ for any permutation als,a2s,—,a„s, the following inequality holds:

Proof. Without loss of generality, it can be assumed that a, 2 a„.

Applying the rearrangement inequality to the pair of sequences (aua2,...,a„), (al,a2,...,a„) and the pair of sequences (a,,a2,...,a„ 2„...,й,и) then yields

because the first pair of sequences are similarly ordered and the second pair of sequences are not. This completes the proof of inequality (2.34).

ii. Given a set of positive real numbers aha2,...,a„ different from zero, for any permutation als,a2s,...,a„s the following inequality holds:

Proof. Without loss of generality, it can be assumed that a, < a2 <,...,< a„. The reciprocals can then be ranked as follow's: 1/a, >l/a2 1 /a„.

Applying the rearrangement inequality to the pair of sequences иа2,...,а„), (l/ai,l/a2,...,l/a„) and the pair of sequences {au,a2s,...,am), (1/а|,1/а2,—,1/яи) then yields

because the first pair of sequences are oppositely sorted (therefore associated with the smallest sum ■(l/al) + a2-(l/a2) + ... + a„-(l/a„) = n) and the second pair of sequences are not oppositely sorted and therefore, the sum «iv/c/i + chjch +... + a,Ja„ is not the smallest sum, which completes the proof of inequality (2.35).