Proving Inequalities by Using the Properties of Sub-Additive and Super-Additive Functions
- Proving Inequalities by Transforming Them to Known Inequalities
- Proving Inequalities by Transforming Them to an Already Proved Inequality
- Proving Inequalities by Transforming Them to the Holder Inequality
- An Alternative Proof of the Cauchy-Schwarz Inequality by Reducing It to a Standard Inequality
- An Alternative Proof of the GM-HM Inequality by Reducing It to the AM-GM Inequality
- Proving Inequalities by Transforming Them to the Cauchy-Schwarz Inequality
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 <
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:
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
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.
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.
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 (a3 /х2,Ь3 / 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
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
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).
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 I—n—n—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