# Proving Inequalities by Exploiting Homogeneity

This is a powerful yet underestimated method for proving inequalities. A function /(is homogeneous of order *к* if /(A.ri,A.i'_{2},...,A.v„) = *?^f(x _{[},x_{2},...,x_{n}) *for every A > 0.

Consider the inequality /(х_{ь}х_{2},...,л:_{л})<0. The homogeneity of the inequality permits assuming without loss of generality additional constraints, for example, jci *+x _{2}+... + x_{n} =* 1, j:J" +

*x'{‘*+ ... +

*x!"*= 1,

*m >*1;

*x*= 1. Indeed, the positive constant A can always be selected to scale the variables

_{x}x_{2}...x_{n}*x,x*so that, for example, they add up to 1: Ax,

_{2},...,x„*+Лх*=1. If instead of the original variables

_{2}+... + kx_{n}*x,x*the scaled variables

_{2},...,x„,*a*= Ax„ are substituted, the inequality will not be altered. As a result, the equivalent inequality

_{t}=Xx,a_{2}= Xx_{2},...,a„*f(a*0 is obtained, with a constraint

_{b}a_{2},...,a„)<*a*1.

_{t}+ a_{2}+... + a„ =This technique will be illustrated with proving the special case of Holder’s inequality:

where *a*, > 0, *b, >* 0, c, > 0; *i* = 1

Constants Я, ju and *v* can always be chosen such that

Substituting in (3.41) the original variables a,,fe,,c, with the scaled variables Я*a _{h}jibi,vcj* gives

which, after the cancellations, yields the original inequality (3.41).

By setting *x, =* Aa,, y_{f} = *kb,* and г, = Я/у, the equivalent inequality

is obtained, with constraints

After substituting the constraints (3.43) in (3.42), the proof reduces to demonstrating that

According to the AM-GM inequality, the following inequalities hold:

* ^{x}i* +

*yf*+

*zf . .*

-*-*-^ *x _{t}y,zi, i =* 1

Adding the *n* inequalities and considering (3.43), yields *xyz+ ■■■ + x„y _{n}z_{n}* < 1, which proves inequality (3.42) and the original inequality (3.41).

# Proving Inequalities by a Mathematical Induction

Commonly, the method of mathematical induction for proving an inequality, stated for an arbitrary number *n* of variables, consists of the following steps.

- 1. Proof of a base case (for example, that the inequality holds for
*n =*2 variables). - 2. Induction hypothesis: assuming that the inequality holds for
*n*=*к*variables - (*>2).
- 3. Inductive step: Proof that whenever the inequality holds for
*n=k*variables, it also holds for*n*=*к*+1 variables. - 4. Conclusion: If steps 1, 2 and 3 have been proved, the inequality holds for
*n*variables.

The method of mathematical induction will be illustrated by proving the inequality

where a, are non-negative real numbers.

1. Proof that the inequality holds for *n =* 2 variables. For *n =* 2, the inequality

becomes

Expanding the left-hand side gives

*a a f* 1 1 ^{>}

Since — + — > 2, *(a, + a _{2})* — + — > 4 =

*2*and the base case has been

^{2}*a*' ^ a,

_{2}a,*a*

_{2},

proved.

2. Assume that the inequality holds for *n = k* (induction hypothesis).

3. It will be proven that if inequality (3.45) holds, the inequality also holds for *n =k +*1, (inductive step), which means proving the inequality

Transforming the left-hand side of inequality (3.46) gives

Because of the induction hypothesis and because Л!_ + > 2;

* ^{a}t** i "I

_5_+ >2*;+ Ht±> 2,* for equality (3.47), which is the left-hand

^{fl}t+1 “2 *«Ш “k* ^{1} ^

side of inequality (3.46), we have

The inductive step has been proved, therefore inequality (3.44) holds for all of integer *n.*

**Example 3.6**

This example is from Sedrakyan and Sedrakyan (2010).

Suppose that for the real numbers a,,fo,,c_{i(} *i* = 1 the following conditions hold:

Then, it can be shown that the following general inequality *(Sedrakyan inequality) *holds:

Inequality (3.49) will be proved by induction. The proof starts with the basic case corresponding to *n = 2.* For, *n = 2,* inequality (3.49) becomes

To prove (3.50), it suffices to prove

The left-hand side of (3.51) can be transformed to:

From the conditions (3.48), a,/ci>a_{2}/c_{2} and b|/ci>6_{2}/c_{2}; therefore, (aic_{2}-a_{2}cd>0 and (b|C_{2}-feed>0. As a result, the right side of (3.52) is nonnegative, —^{a}2^{c}i)(b|C_{2}—_{w}|_{1}j_{c}[_{1} p_{roves} inequalities (3.51) and (3.50).

c,c_{2}(c, +c_{2})

Now assume that inequality (3.49) is true for n = *к >* 2 (induction hypothesis):

It will be shown that inequality (3.53) is also true for *n = k *+1 (induction step).

Adding the term ^{t0} both sides of (3.53), *(a _{k}/c_{k}>=* a

_{k+1}/c

_{k+l};

*b*

_{k}/c_{k}>= Ь_{м}/с_{ы}),results in the inequality

Denoting x = Xf=i^{a}i< *У* = X,*L|b;, c,, the right-hand side of (3.54) becomes

*xy* _{|} a_{M}b_{M}

* ^{z}* c<.+i . It can be demonstrated that x/z>=a

_{k+1}/c

_{k+}, and

*y/z>=b*(the

_{M}/c_{M}proof is very similar to the one given in Section 9.4 and will not be duplicated here).

Because inequality (3.49) holds for the basic case *n =* 2, the next inequality holds:
Finally,

This proves the induction step and inequality (3.49) for arbitrary *n.*

The inequality is sharp because equality is attained if a, / c, =a, / *c _{jr} i Ф j* and

*b;* / C; = *bj* / *Cj, i Ф j.*

# Proving Inequalities by Using the Properties of Convex/Concave Functions

The concept of convex/concave functions has already been introduced in formulating the Jensen inequality in Chapter 2. The concept will be discussed in greater detail here because it is a very powerful tool for proving inequalities. A function /(x) with a domain *[a,b]* is said to be convex if for all values *x*, and *x _{2}* in its domain (x,,x

_{2}

*e [a,b*]), the next inequality holds

where 0< *w<* 1.

The function *f(x)* is said to be strictly convex if equality in (3.56) holds only when *w* = 0; *w* = 1 or x, = *x _{2}.* The function

*f(x)*is strictly convex if for 0 <

*w <*1, x,

*Ф x*the inequality

_{2},

holds.

If /(x) is twice differentiable for all *xe(a,b*), and if the second derivative is non-negative (/"(x) > 0), the function /(x) is convex on *(a,b).* If /(x) is twice differentiable for all x e (*a,b*), and if the second derivative is positive (/"(x) > 0), the function /(x) is strictly convex on *(a,b).* For a strictly convex function/(x) of a single argument, the graph between any two points x, < x_{2} remains below the line segment joining the values /(x,), /(x_{2}) corresponding to these points (Figure 3.1a). An example of a convex function is /(x) = x'" for *m* > 1 and x > 0.

For concave functions, the inequalities are reversed.

If /(x) is twice differentiable for all *xe(a,b),* and if for the second derivative /"(x) < 0 is fulfilled, the function /(x) is concave on *(a,b).* If /(x) is twice differentiable for all x e (*a,b),* and if the second derivative is negative, *f"(x) <* 0, the function /(x) is strictly concave on *(a,b).* For a strictly concave function/(x) of a single argument, the graph between any two points x, < x_{2} remains above the line segment joining

FIGURE 3.1 (a) A strictly convex function and (b) a strictly concave function of a single

argument.

the values /(x,), /(x_{2}) corresponding to these points (Figure 3.1b). An example of a concave function is *f(x) =* In .v. If a function *f(x)* is convex, the function *-f(x)* is concave and *vice versa.* Thus, *f(x)* = In x and *g(x) = [x* are both concave functions but -ln(x) and -л/х are both convex functions.

A sum of convex functions is a convex function and a sum of concave functions is a concave function.

*A convex function does not mean that the function is increasing or decreasing.* The absolute maximum of a convex function is never found in the interior of the domain where the arguments vary. It occurs at the boundary of the domain. Similarly, the absolute minimum of a concave function is never found in the interior of the domain where the arguments vary. It occurs at the boundary of the domain.

The concepts of convex/concave functions are very powerful and can be used to prove various inequalities.

**Example 3.7**

Consider the quadratic trinomial *f(x) = ax ^{2}+bx+c* where xeh,u

_{2}|, If a>the second derivative of the quadratic trinomial is positive, the function

*f(x) a ax*is convex and its global maximum is attained either at x = Uj or x =

^{2}+bx + c*u*The maximum is found by comparing f(uj) and

_{2}.*f(u*whichever is greater.

_{2}),If a < 0, the second derivative of the quadratic trinomial is negative, the function *f(x)* = *ax ^{2} + bx + c* is concave and its global minimum is attained either at x = Ui or x =

*u*The global minimum is found by comparing

_{2}.*f(U])*and

*f(u*whichever is smaller. This provides a useful technique for proving inequalities.

_{2}),Here is an example.

For the non-negative *x,y,z* for varying in the intervals 0

holds.

This inequality can be proved by using convex functions.

Because the second partial derivatives of the function *f(x,y,z)= *x^{2}(1-y)+x^{2}(1-z)+y^{2}(1-z)+y^{2}(1-x)+z^{2}(1-x)+z^{2}(1-y) with respect to each of the variables x,yandz are non-negative *(d ^{2}f(x,y,z)* / <5x

^{2}= 2 x(2 - y- z) > 0;

*d*=2x(2-x-z)>0 and

^{2}f(x,y,z)/8y^{2}*d*= 2x(2-x-y)>0), the function

^{2}f(x,y,z)/8z^{2}*f(x,y,z)*is convex and attains its global maximum at the boundary of its domain. Because of the symmetry, the combination of values to check for a maximum is /"(0,0,0) = 0, /"(1,0,0) = 2,

*f(*1,1,0) = 2 and

*f(*1,1,1) = 0. The maximum value of the function

*f(x,y,z)*is equal to 2, which proves the inequality.

**Example 3.8**

Determine the smallest constant C such that where 1 < x < 2.

*f(x) = 2x ^{2} -yfx* -2lnx is a convex function because it is a sum of three convex functions:

*f(x) = lx*=

^{2}, f_{2}(x)*-fx*and

*f*

_{3}(x)*=*-2ln(x).

Consequently, the maximum of *f(x*) is obtained at the ends of the interval [1,2]. There are two values to be compared: f(1) = 1 and *f(*2) = 5.2. The maximum is attained at x = 2 and is equal to 5.2. Therefore, the smallest value of the constant is C = 5.2.

Example 3.9

In the next example, for the non-negative numbers 0 < a < 1, 0 < Ь < 1 and 0 < c < 1, by using the concept of convexity, it can be shown that

The function f(a,b,c) =-^{J}-r + --7-r + 7r-^{!}-r + a^{2}b^{2}+fo^{2}c^{2}+c^{2}a^{2} is continuous on a closed domain 0 < a < 1, 0 < fo < 1 and 0 < c < 1; therefore, according to the *extreme value theorem* from calculus, the function must attain its maximum and minimum in that domain.

The function *f(a,b,c) *=——^- +—V-r *+—-—т + з ^{2}Ь^{2} +b^{2}c^{2} + c^{2}a^{2}* is strictly (1 + a)

^{2}(1 + fo)

^{2}(1 + c)

^{2 1}

convex with respect to each of the variables *a, **b* and c because

Because the function is strictly convex and continuous on the bounded and closed domain 0ies of the domain. The boundary points are:

The maximum is attained at f(1,1,1) = 15/4, hence inequality (3.58) holds.

## Jensen Inequality

For any convex function *f(x),* the Jensen’s inequality states that

where vv, (/ = 1,...,«) are numbers that satisfy 0 < w, < 1 and w, + w, +... + vc„ = 1. For concave functions, the direction of inequality (3.59) is reversed.

The function In x is concave for positive values x > 0 because its second derivative (lnx)" = -l/x^{2} is negative. According to the Jensen inequality for concave functions,

holds. Using the properties of the logarithms

Because both sides of (3.60) are positive and *e ^{x}* is an increasing function, it follows that
or finally

which is the AM-GM inequality.

If xi,x_{2},...,x„ _{are} positive numbers, and *a*-* 0 is a real number, the number

*(**а* ^{1/П}

^{л}Г ^{+}-*2 * ^{+}—^{+x}n* I i

_{s}known as a

*power mean*of the positive numbers.

Applying the AM-GM inequality to the numbers xf ,x“,...,x“ yields

from which is obtained.

If *a >* 1 and x, > 0 then /(x) = x“ is a convex function because /"(x) = *a(a* - l)x“'^{2} > 0- According to the Jensen inequality for convex functions,

If 0 < *a* < 1 and x, > 0 then /(x) = x“ is a concave function because /"(x) = *a(a* - l)x“~^{2} < 0. According to the Jensen inequality for concave functions

The next example illustrates another application of the Jensen inequality.

**Example 3.10**

For the non-negative numbers *x >* 0, *у >* 0 and *z >* 0, prove that

Proof. The function *f(u) = u~ ^{V2}* is a strictly convex function for

*и*

*0 because*

**>***f"(u)=(3/4)u~*

^{5/2}>0. According to the Jensen inequality, for a convex function

*f(x,y,z),*the inequality

holds and for the convex function *f(u)* = и , the inequality

holds, from which inequality (3.64) follows. Because the function is strictly convex, the equality in (3.64) is attained only for *x = у = z.*