Mean Inequalities. Inequalities of Minkowski and Holder

Definition of means

The arithmetic mean of numbers \( a_1 \), \( a_2 \), \( \dots \), \( a_n \) is defined as \[ A(a_1,a_2,\dots, a_n)=\frac{a_1+a_2+\cdots+a_n}n.\] In the case that all of the numbers are positive, the geometric mean is defined as \[ G(a_1, a_2,\dots, a_n)=\sqrt[n]{a_1a_2\cdots a_n}.\] Besides arithmetic and geometric mean, there are other means of sequences of numbers for which we have beautiful and useful inequalities. The most famous and used are quadratic and harmonic means of numbers \( x_1, \dots, x_n \). They are denoted by \( Q(x_1,\dots, x_n) \) and \( H(x_1,\dots, x_n) \) and they are defined as: \[ Q(x_1,\dots, x_n)=\sqrt{\frac{x_1^2+\cdots+x_n^2}n},\;\; H(x_1,\dots, x_n)=\frac{n}{\frac1{x_1}+\cdots+\frac1{x_n}}.\] We have the following inequality \begin{eqnarray*} Q(x_1,\dots, x_n)\geq A(x_1,\dots, x_n)\geq G(x_1,\dots, x_n)\geq H(x_1,\dots, x_n). \end{eqnarray*} All inequalities are strict unless \( x_1=\dots=x_n \) in which case all the inequalities become equalities.

All of these means can be generalized, and in some sense considered all at once. It is possible to define the general mean of order \( r \), denoted by \( M_r(x_1,\dots, x_n) \). Quadratic, arithmetic, geometric, and harmonic means are just the special case of that general mean. More precisely:

Definition 1 (Mean)
 
Given a sequence \( x_1, x_2, \dots, x_n \) of positive real numbers, the mean of order \( r \), denoted by \( M_r(x) \) is defined as \begin{eqnarray*} M_r(x)=\left(\frac{x_1^r+x_2^r+\cdots+x_n^r}n \right)^{\frac1r}. \end{eqnarray*}

Example 1
 
\( M_1(x_1, \dots, x_n) \) is the arithmetic mean, while \( M_2(x_1, \dots, x_n) \) is the quadratic mean of the numbers \( x_1, \dots, x_n \). \( M_{-1} \) is the harmonic mean.

\( M_0 \) can’t be defined using the expression analogous to the expressions for other means, but we will show later that as \( r \) approaches \( 0 \), \( M_r \) will approach the geometric mean. The famous mean inequality can be now stated as \[ M_r(x_1,\dots, x_n)\leq M_s(x_1, \dots, x_n), \;\;\mbox{ for } 0\leq r\leq s.\] However we will treat this in slightly greater generality. We will consider the weighted mean of order \( r \).

Definition 2 (Weighted mean)
 
Let \( m=(m_1, \dots, m_n) \) be a fixed sequence of non-negative real numbers such that \( m_1+m_2+\cdots +m_n=1 \). Then the weighted mean of order \( r \) of the sequence of positive reals \( x=(x_1, \dots, x_n) \) is defined as: \begin{eqnarray*} M^m_r(x)=\left(x_1^rm_1+x_2^rm_2+\cdots + x_n^rm_n \right)^{\frac1r}. \end{eqnarray*}

We will prove later that as \( r \) tends to \( 0 \), the weighted mean \( M_r^m(x) \) will tend to the weighted geometric mean of the sequence \( x \) defined by \( G^m(x)=x_1^{m_1}\cdot x_2^{m_2}\cdots x_n^{m_n} \).

Example 2
 
If \( m_1=m_2=\cdots =\frac1n \) then \( M_r^m(x)=M_r(x) \) where \( M_r(x) \) is previously defined.

Theorem 1 (General Mean Inequality)
 
If \( x=(x_1,\dots, x_n) \) is a sequence of positive real numbers and \( m=(m_1,\dots,m_n) \) another sequence of positive real numbers satisfying \( m_1+\cdots + m_n=1 \), then for \( 0\leq r\leq s \) we have \( M^m_r(x)\leq M^m_s(x) \).

We will postpone the proof for later. It will follow from the Holder’s inequality (or Jensen’s inequality if you like to do it that way). To get accustomed with these new definitions of means, let us solve couple of problems.

Problem 1 (Young’s inequality)
 
If \( a \), \( b \), \( p \), and \( q \) are positive real numbers such that \( \frac1p+\frac1q=1 \), then \[ \frac{a^p}p+\frac{b^q}q\geq ab.\]

Problem 2
 
If \( a \), \( b \), \( c \) are positive real numbers prove that \[ \frac{a^2+b^2+c^2}{a+b+c}\geq \left(a^a b^b c^c\right)^{\frac1{a+b+c}}.\]

Problem 3
 
If \( a,b,c \) are positive real numbers prove that \[ \frac{(b+c)^a(c+a)^b(a+b)^c}{2^{a+b+c}}\leq \left(\frac{a+b+c}3\right)^{a+b+c}\leq a^ab^bc^c.\]

Inequalities of Minkowski, Holder, and Cauchy-Schwarz

Inequalities presented here are sometimes called weighted inequalities of Minkowski, Holder, and Cauchy-Schwartz. The standard inequalities are easily obtained by placing \( m_i=1 \) whenever some \( m \) appears in the text below. Assuming that the sum \( m_1+\cdots + m_n=1 \) one easily get the generalized (weighted) mean inequalities, and additional assumption \( m_i=1/n \) gives the standard mean inequalities.

Lemma 1
 
If \( x,y> 0 \), \( p> 1 \) and \( \alpha\in(0,1) \) are real numbers, then \begin{eqnarray*} (x+y)^p\leq \alpha^{1-p}x^p+ (1-\alpha)^{1-p}y^p. \quad\quad\quad\quad\quad (1)\end{eqnarray*} The equality holds if and only if \( \frac{x}{\alpha}= \frac{y}{1-\alpha} \).

Lemma 2
 
If \( x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n \) and \( m_1, m_2, \dots, m_n \) are three sequences of positive real numbers and \( p> 1 \), \( \alpha\in(0,1) \), then \begin{eqnarray*} \sum_{i=1}^n(x_i+y_i)^pm_i \leq \alpha^{1-p}\sum_{i=1}^nx_i^pm_i+ (1-\alpha)^{1-p}\sum_{i=1}^n y_i^pm_i. \quad\quad\quad\quad\quad (2) \end{eqnarray*} The equality holds if and only if \( \frac{x_i}{y_i}=\frac{\alpha}{1-\alpha} \) for every \( i \), \( 1\leq i\leq n \).

Theorem 2 (Minkowski)
 
If \( x_1 \), \( x_2 \), \( \dots \), \( x_n \), \( y_1 \), \( y_2 \), \( \dots,y_n \), and \( m_1 \), \( m_2 \), \( \dots, m_n \) are three sequences of positive real numbers and \( p> 1 \), then \begin{eqnarray*} \left(\sum_{i=1}^n(x_i+y_i)^pm_i\right)^{1/p} \leq \left(\sum_{i=1}^nx_i^pm_i\right)^{1/p}+ \left(\sum_{i=1}^ny_i^pm_i\right)^{1/p}. \quad\quad\quad (3) \end{eqnarray*} The equality holds if and only if the sequences \( (x_i) \) and \( (y_i) \) are proportional, i.e. if and only if there is a constant \( \lambda \) such that \( x_i=\lambda y_i \) for \( 1\leq i\leq n \).

Problem 4
 
If \( u_1, \dots, u_n, v_1, \dots, v_n \) are real numbers, prove that \[ 1+\sum_{i=1}^n(u_i+v_i)^2\leq \frac43\left(1+\sum_{i=1}^nu_i^2\right)\left(1+\sum_{i=1}^nv_i^2\right).\] When does equality hold?

Problem 5
 
If \( a_1, \dots, a_n \) are real numbers, prove that \[ \sqrt{a_1^2+(1-a_2)^2}+\sqrt{a_2^2+(1-a_3)^2}+\cdots+ \sqrt{a_n^2+(1-a_1)^2}\geq \frac{n\sqrt 2}2.\]

Theorem 3 (Young)
 
If \( a,b> 0 \) and \( p,q> 1 \) satisfy \( \frac1p+\frac1q=1 \), then \begin{eqnarray*} ab\leq \frac{a^p}p+ \frac{b^q}q.\quad\quad\quad\quad(5)\end{eqnarray*} Equality holds if and only if \( a^p=b^q \).

Remark. You might wander why we are proving Young’s theorem twice. It was proved as an application of general mean inequality. But the mean inequality we used was not proved. The proof was postponed, because it uses the Holder’s inequality and the proof of Holder’s inequality will use the Young’s inequality. If there were no proof of it here, we would have a horrible example of circular reasoning in mathematics.

Lemma 3
 
If \( x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n, m_1, m_2, \dots, m_n \) are three sequences of positive real numbers and \( p,q> 1 \) such that \( \frac1p+\frac1q=1 \), and \( \alpha > 0 \), then \begin{eqnarray*} \sum_{i=1}^nx_iy_im_i\leq \frac1p\cdot \alpha^p\cdot \sum_{i=1}^nx_i^pm_i + \frac1q\cdot \frac1{\alpha^q}\cdot \sum_{i=1}^ny_i^qm_i. \quad\quad\quad (6)\end{eqnarray*} The equality holds if and only if \( \frac{\alpha^px_i^p}p= \frac{y_i^q}{q\alpha^q} \) for \( 1\leq i\leq n \).

Theorem 4 (Holder)
 
If \( x_1,x_2,\dots,x_n,y_1,y_2,\dots,y_n, m_1, m_2, \dots, m_n \) are three sequences of positive real numbers and \( p,q> 1 \) such that \( \frac1p+\frac1q=1 \), then \begin{eqnarray*} \sum_{i=1}^nx_iy_im_i\leq \left(\sum_{i=1}^nx_i^pm_i\right)^{1/p}\cdot \left( \sum_{i=1}^ny_i^qm_i\right)^{1/q}. \quad\quad\quad (7) \end{eqnarray*} The equality holds if and only if the sequences \( (x_i^p) \) and \( (y_i^q) \) are proportional.

Problem 6
 
If \( a_1, \dots, a_n \) and \( m_1, \dots, m_n \) are two sequences of positive numbers such that \( a_1m_1+\cdots + a_nm_n=\alpha \) and \( a_1^2m_1+ \cdots + a_n^2m_n = \beta^2 \), prove that \[ \sqrt{a_1}m_1+\cdots + \sqrt{a_n}m_n\geq \frac{\alpha^{3/2}}{\beta}.\]

Problem 7
 
If \( a \), \( b \), \( c \), and \( d \) are positive real numbers such that \( a+b+c+d=1 \), prove that \[ \frac{a^2}{(1+b)(1-c)}+\frac{b^3}{(1+c)(1-d)}+\frac{c^3}{(1+d)(1-a)}+\frac{d^3}{(1+a)(1-b)}\geq\frac1{15}.\]

We are now ready for the proof of the mean inequality.

Theorem 1 (General Mean Inequality)
 
If \( x=(x_1,\dots, x_n) \) is a sequence of positive real numbers and \( m=(m_1,\dots,m_n) \) another sequence of positive real numbers satisfying \( m_1+\cdots + m_n=1 \), then for \( 0\leq r\leq s \) we have \( M^m_r(x)\leq M^m_s(x) \).

Problem 8
 
Let \( x \), \( y \), and \( z \) be positive real numbers such that \( xyz=1 \). Prove that \[ \frac{x^3}{(1+y)(1+z)}+\frac{y^3} {(1+z)(1+x)}+\frac{z^3}{(1+x)(1+y)}\geq\frac{3}{4}.\]

Theorem 5 (Weighted Cauchy-Schwartz Inequality)
 
If \( x_i \), \( y_i \) are real numbers, and \( m_i \) positive real numbers, then \begin{eqnarray*} \sum_{i=1}^nx_iy_im_i&\leq&\sqrt{\sum_{i=1}^nx_i^2m_i} \cdot\sqrt{\sum_{i=1}^ny_i^2m_i}. \quad\quad\quad (8)\end{eqnarray*}

Problem 9
 
If \( a \), \( b \), and \( c \) are positive numbers, prove that \[ \frac ab+\frac bc+\frac ca \geq \frac{(a+b+c)^2}{ab+bc+ca}.\]

Problem 10
 
If \( a \), \( b \), \( c \) are positive real numbers prove that \[ \frac{a^2}b+\frac{b^2}c+\frac{c^2}a\geq \frac{a^2+b^2+c^2}{a+b+c}.\]

Problem 11 (Dušan Djukić)
 
Let \( a,b,c \) be positive real numbers. Prove the inequality \[ \frac{a^2}b+\frac{b^2}c+\frac{c^2}a\geq a+b+c+\frac{4(a-b)^2}{a+b+c}.\]

The following theorem theorem is given here for completeness. It states that as \( r\rightarrow 0 \) the mean of order \( r \) approaches the geometric mean of the sequence. Its proof involves some elementary calculus, and you can skip the proof if you haven’t mastered the calculus yet.

Theorem 6
 
If \( a_1, \dots, a_n \) are positive real numbers, then \[ \lim_{r\rightarrow 0} M_r(a_1,\dots, a_n)= a_1^{m_1}\cdot a_2^{m_2}\cdots a_n^{m_n}.\]


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