# Proving Inequalities by Using the Properties of Sub-Additive and Super-Additive Functions

The properties of sub-additive and super-additive functions can be used for proving inequalities. A function f(x,y) is sub-additive if for any real numbers x,y, the next inequality holds:

A function f(x,y) is super-additive if for any real numbers x,y, the next inequality holds:

Multiplying the definition inequalities (3.65) and (3.66) by -Г, reverses the sign of the inequality. Consequently, if f(x,y) is sub-additive, -f(x,y) is super-additive and if /(x,y) is super-additive, -f(x,y) is sub-additive.

An example of a super-additive function defined for -oo < x < is the function ex. A key result related to sub-additive and super-additive functions can now be stated. If a function f(x,y) with a domain [0,oo) and range [0,oo) is concave, then the function is sub-additive (/(x + у) < f(x) + f(y)). If the function is convex, then it is super-additive (Alsina and Nelsen, 2010).

Proof. Consider all x, у such that x + y*0. Because f(x,y) is a concave function:

Adding inequalities (3.67) and (3.68) yields

Since /(0) > 0 (the range of the function f(x,y) is [0,oo)),

Note that equality can only be attained if /(0) = 0.

Indeed, for >• = 0. f(x + 0) = f(x) + f(0) = f(x). Similarly, for x = 0, f(0+ y) =

m + f(y) = f(y).

However, if /(0) > 0 the inequality is not sharp:

This proves inequality (3.66). Inequality (3.65) can be proved by using similar reasoning. Inequalities (3.65) and (3.66) can be generalised easily by induction for n variables X|,x2,....,xn:

Example 3.11

Here is an example, illustrating the use of sub-additive functions.

For any real x > 0; у > 0, prove the inequality

The function f(u) = '-e~u is a concave function because the second derivative is negative.

Since f(0) = 1-e° =0, the inequality follows from the key result stated earlier. Equality is attained for x = 0 or у = 0.

Inequalities (3.65) and (3.66) have very important potential applications. If the function/(z) measures the effect of a particular factor z, inequalities (3.65) and (3.66) provide the unique oppotunity to increase the effect of the factor by segmenting the factor (z = z, + z2;/(z,) +/(z2) >= /(z)) or by aggregating the factor (z = z, + z2); /(z) >=/(z,) +/(z2)), depending on whether the function/(z) is concave or convex.

# Proving Inequalities by Transforming Them to Known Inequalities

Inequalities do not normally appear in the form of the standard inequalities discussed in Chapter 2. It is therefore very important to recognise what transformation of variables leads to standard inequalities or to inequalities which have already been proved.

## Proving Inequalities by Transforming Them to an Already Proved Inequality

The inequality is transformed until an inequality is reached which has already been proved. This technique will be illustrated by the next example. Let щ,а2,...,а„ be a set of positive real numbers and a, +a2 + ... + a„ = .v. Then, it can be shown that the inequality

holds.

Multiplying the left-hand side by the sum of the reciprocals of the separate terms gives

which is the well-known inequality (x, + x2 +... + x„)(l /x, +1 /x2 +... +1 /x„) > n2, valid for any set of n positive real numbers x,.

Since the second factor on the left side can be presented as

inequality (3.70) follows directly.

Example 3.12

The inequality can be easily transformed into the already proved Bergstrom inequality (3.8):

Many of the standard inequalities can be proved by reducing them to already proved inequalities. If, for example, the variables c, in the Sedrakian inequality

(3.49) are selected such that Cj = n; i = 1,2then £,"=1c, = n2 and inequality

which is the Chebyshev inequality.

## Proving Inequalities by Transforming Them to the Holder Inequality

Transforming an inequality to the Holder inequality introduced in Chapter 2 is a powerful technique for proving inequalities. The method is often appropriate for cyclic inequalities and will be illustrated by the next two examples. Transformation to the Holder inequality also works in the presence of constraints.

Example 3.13

Let a,b,c and x,y,z be real positive numbers such that a + b + c = x + y + z. Then, the next inequality holds:

Proof. Consider the three sequences (x,y,z),(x,y,z), and (a323 / y2,c3 / z2). Applying the Holder inequality with weights A] = A2 = A, =1/3 (ХыЛ =l) gives

Considering the constraint a + b + c = x + y + z, raising both sides into a power of 3 and dividing both sides by the positive value (x + y + z)2 yields

Example 3.14

Let a,b,c be real numbers such that a + b + c = 1. Then, the next inequality holds.

Proof. Consider the four sequences (a,b,c), ib + c,a + c,a + b), (b + c,a + c,a + b) and | (' ,, -——y-j. Applying the Holder inequality with weights

A] = A, = Aj = A4 =1/4 = l) gives

Since a + b + c = 1 and b + c + a + c + a + 6 = 2, dividing both sides of the last inequality by 4 gives

which completes the proof.

## An Alternative Proof of the Cauchy-Schwarz Inequality by Reducing It to a Standard Inequality

This alternative way of proving the Cauchy-Schwarz inequality has been given in Alsina and Nelsen (2010). Proving the Cauchy-Schwarz inequality (3.18) is equivalent to proving the next inequality.

Consider the numbers x = !a>' 1 and у = / ^ ^ . According to a standard

J5 JP

inequality (proved in Chapter 2), xy < (x1 + y2)/2. Consequently,

## An Alternative Proof of the GM-HM Inequality by Reducing It to the AM-GM Inequality

Let xux2,...,x„ be a set of positive real numbers. The geometric mean - harmonic mean inequality states that

with equality attained only if x, =x2 = ... = x„. Proving inequality (3.75) is equivalent to proving

the left-hand side of which can be transformed into

The right side of the last equality is the average of the numbers — ^Х|Х2...х„, i = 1According to the AM-GM inequality x

This completes the proof of the original inequality (3.75).

## Proving Inequalities by Transforming Them to the Cauchy-Schwarz Inequality

This is a very important technique that can be applied to prove a vast range of inequalities.

Let al,a2,...,a„ be real numbers and bub2,...,b„ be positive real numbers. Then, the inequality (Bergstrom inequality)

holds (see Section 3.2). An alternative proof of inequality (3.76) can be obtained by transforming the inequality to the Cauchy-Schwarz inequality

valid for any two sequences of real numbers xi,x2,...,x„ and y,y2,-,yn-

Note that the substitutions x, = -J- (z = 1,...,/;) and y, = *Jb, (z = l,...,zz) of the original variables applied with the Cauchy-Schwarz inequality (3.77) leads to inequality (3.76):

which shows that inequality (3.76) is a special case of the Cauchy-Schwarz inequality (3.77).

The inequality

where xh..., x„ is a set of real numbers is known also as arithmetic mean-root-mean square (AM-RMS) inequality. It can also be proved by showing that it is a special case of the Cauchy-Schwarz inequality.

Indeed, consider the set of n real numbers xhx2,...,x„ and the set of n numbers 1,1,...,1. According to the Cauchy-Schwarz inequality:

Dividing both sides of inequality (3.79) by the positive value n yields inequality (3.78). Example 3.15

For any positive real numbers x,y and z such that x2 + y2 + z2 =1, prove the inequality

Proof. Factoring xyz from the left hand side of the inequality results in x2yz + y2zx + z2xy = xyz(x + у + z).

From the AM-GM inequality

2 2 2

X + у + Z Inn—T ')')')

-^-> цх у z~ and from the constraint x + у + z = 1 the next

inequality is obtained:

From the Cauchy-Schwarz inequality

Therefore, x2yz + y2zx + z2xy = xyz(x + y + z)< —^ x yjb = which completes the proof. 3/3 3