# Proving Inequalities by Segmentation

This is a powerful technique whose essence is to segment (split) the original inequality into simpler inequalities by using some of the standard inequalities (frequently the AM-GM inequality) and to sum the segmented inequalities in order to assemble the original inequality.

Here is an example of this technique demonstrated on the inequality

which is valid for any arbitrary real x,y,z. The left side of this inequality can be segmented into three inequalities by using the standard AM-GM inequality:

Adding the three inequalities gives the original inequality (3.80) and completes the proof. Another example of this technique is proving the inequality

valid for non-negative real numbers x, у and z-

Since the terms xjyz, yjzx, z-Jxy in the right-hand side can be obtained easily by applying the AM-GM inequality to xy + yz, yz + zx and zx + xy, correspondingly, the segmentation yields the following inequalities:

To make connection with the term (x + y+z)2 on the left-hand side, note that the terms Xyfyz, yjzx, Zyfxy on the right-hand side can also be obtained by applying the AM-GM inequality to x2 + x2 + y2 + z2, y2 + y2 + x2 + z2 and z2 + z2 + x2+y2. Indeed, applying the AM-GM inequality yields:

Finally, adding inequalities (3.89) and (3.85) yields the inequality

which is equivalent to inequality (3.81).

## Determining Bounds by Segmentation

Segmentation can also be used for determining bounds of various symmetric expressions. This technique will be demonstrated by the next example determining the lower bound of the expression

where x,y,z are positive real numbers that satisfy the constraint x2 + y2 + z2 = 1.

The lower bound will be found by segmenting expression (3.90) through the AM-GM inequality

which, after the substitution of the constraint x2 +y2 + z2 = 1 results in for the lower bound. Equality is attained for .v = у = z = 1 / V3 •

# Proving Algebraic Inequalities by Combining Several Techniques

Often, a rigorous proof of an algebraic inequality may require combining several techniques. Consider the next example which involves substitution, the standard Cauchy-Schwarz inequality and finally, the AM-GM inequality.

The example is related to proving the inequality (problem 2 from the International Mathematical Olympiad 1995)

where a,b,c are positive real numbers for which abc = 1.

Initially, the inequality is transformed by applying the substitution technique'. Consider the substitution x = 1 / а;у = 1 / b;z = 1 / c. Then xyz = 1 and substituting a = 1 / x = 1 / y;c = 1 / z in the original inequality and by using yz = 1 / x, ух = 1 / z and xz = 1 / y, the original inequality (3.95) is transformed into the inequality

Consider now the two sequences of real numbers

Using the Cauchy-Schwarz inequality to these two sequences gives

or

Using the AM-GM inequality for the right part of inequality (3.97) gives (jt + y + г) > 3ifxyz = 3. Hence, inequality (3.96) is obtained from inequality (3.97), which completes the proof of inequality (3.95).

# Using Derivatives to Prove Inequalities

Consider proving an inequality of the type f(x) > g(x) defined on an interval [a,b, where f(x) and ,g(x) are differentiable functions.

Given that f(a) > g(a) and f(x) > g'(x), it can be shown that /(x) > gQr) holds for any x e [a,b].

To show why this is true, consider the function t(x) which is the difference t(x)= f(x)-g(x). This function is non-decreasing because t'(x) = f'(x)-g'(x)> 0.

Therefore, t(x) > t(a) for any x e [a,b]. Since t(a) = f(a)-g(a) > 0, this means that t(x) > 0 or f(x) > g(x) for any x e [a,b].

This technique will be demonstrated by proving the Bernoulli inequality,

for a non-negative x and any positive integer n> 1.

Proof./(х)=(1 + х)'',£(х) = 1 + их.Clearly,/(0)=g(0) = 1 and /'(x)=n(l + x)'1'1 > g'(x) = n, because (1 + x)"~' > 1. As a result, t(x) = (1 + х)" -(1 + nx) is an increasing function over the interval [0, +oo). Since f(0) = /(0) - g(0) = 0, and t(x) is an increasing function, it follows that t(x) > 0, which means that f(x) > g(x) or (1 + х)я > 1 + nx for any x e [0, +oo].

Example 3.16

Another example is the proof of the inequality

for xe [0,1).

Proof.

g(x) = 2x2. Clearly, f(0) = g(0) = 0 and f'(x)= ^ >g'(x)=4x because jA- -4x= is non-negative for x e [0,1).

As a result, f(x)=ln(T^r) —2x2 is an increasing function on the interval [0,1). Since f(0) = f(0)-g(0) = 0 and t(x) is an increasing function, it follows that f(x) > 0, which means that f(x) > g(x) or Ц-pj-) >2x2 for any x e [0,1).

Another technique employing derivatives for proving inequalities is the mean value theorem. The mean value theorem holds for a continuous function f(x) on a closed interval [a, b] which is differentiable on the open interval (a, b). The theorem states that under these conditions, there exists a value c for which

where /'(I) is the derivative of the function at a point | from the open interval (a, b). The mean value theorem can be used to prove the inequality

for any x, 2.

Applying the mean value theorem yields

where £ e(x],x1)-

From (3.102) it follows

Inequality (3.101) follows directly from equation (3.103) because I cos(^) l< 1.

Another useful inequality can be derived from inequality (3.101) by substituting

Jr, = 0 and x2 = x: