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
\[\frac{a^p}p+\frac{b^q}q\geq ab.\]
Let us consider the following two-element sequence of masses \(m=\left(\frac1p,\frac1q\right)\). This is indeed a sequence of masses. If you want to be a sequence of masses (although I don’t know why you would want to be that), all you have to do is to consist of positive real numbers whose sum is one. Now we have that
\(M_1^m\left(a^p,b^q\right)=\frac1p\cdot a^p+\frac1q\cdot b^q\). We also have that
\(M_0^m(a^p,b^q)=G^m(a^p,b^q)=(a^p)^{1/p}\cdot (b^q)^{1/q}=ab\). From general mean inequality we have that \(M_1^m(a^p,b^q)\geq M_0^m(a^p,b^q)\). The equality holds if and only if \(a^p=b^q\).
Problem 2
If \(a\), \(b\), \(c\) are positive real numbers prove that \[\frac{a^2+b^2+c^2}{a+b+c}\geq \left(a^a b^b c^c\right)^{\frac1{a+b+c}}.\]
Consider the sequence of masses \(m=\left(\frac a{a+b+c},\frac b{a+b+c},\frac c{a+b+c}\right)\). Then \(M^m_1(a,b,c)=\frac a{a+b+c}\cdot a+\frac b{a+b+c}\cdot b+\frac{c}{a+b+c}\cdot c=\frac{a^2+b^2+c^2}{a+b+c}\), and \(M^m_0(a,b,c)=a^{\frac a{a+b+c}}\cdot b^{\frac b{a+b+c}}\cdot c^{\frac c{a+b+c}}=\left(a^ab^bc^c\right)^{\frac1{a+b+c}}\). The inequality \(M_1^m(a,b,c)\geq M_0^m(a,b,c)\) follows from the mean inequality.
The equality hold if and only if \(a=b=c\).
Problem 3
If \(a,b,c\) are positive real numbers prove that \[\frac{(b+c)^a(c+a)^b(a+b)^c}{2^{a+b+c}}\leq \left(\frac{a+b+c}3\right)^{a+b+c}\leq a^ab^bc^c.\]
The inequality on the left is equivalent to \[\left(\frac{b+c}2\right)^{\frac a{a+b+c}}
\cdot \left(\frac{c+a}2\right)^{\frac b{a+b+c}}\cdot \left(\frac{a+b}2\right)^{\frac c{a+b+c}}
\leq \frac{a+b+c}3.\]
Consider the sequence of masses \(m=\left(\frac a{a+b+c},\frac b{a+b+c},\frac c{a+b+c}\right)\). According to the mean inequality we have: \(M_0^m\left(\frac{b+c}2,\frac{c+a}2,\frac{a+b}2\right)\leq M_1^m\left(
\frac{b+c}2,\frac{c+a}2,\frac{a+b}2\right)\). Therefore
\begin{eqnarray*}\left(\frac{b+c}2\right)^{\frac a{a+b+c}}
\cdot \left(\frac{c+a}2\right)^{\frac b{a+b+c}}\cdot \left(\frac{a+b}2\right)^{\frac c{a+b+c}}
&\leq& \frac1{2(a+b+c)}(a(b+c)+b(c+a)+c(a+b))\newline&=&\frac{ab+bc+ca}{a+b+c}\newline&=&
\frac{ab+bc+ca+2(ab+bc+ca)}{3(a+b+c)}
\newline&\leq&\frac{a^2+b^2+c^2+2(ab+bc+ca)}{3(a+b+c)}\newline&=&
\frac{a+b+c}3.\end{eqnarray*}
The equality holds if and only if \(a=b=c\).
The inequality on the right-hand side is equivalent to
\[\frac{a+b+c}3\leq a^{\frac a{a+b+c}}b^{\frac b{a+b+c}}c^{\frac c{a+b+c}}.\]
We have that \(M_0^m(a,b,c)=a^{\frac a{a+b+c}}b^{\frac b{a+b+c}}c^{\frac c{a+b+c}}\), while \[M_{-1}^m(a,b,c)=\left(\frac a{a+b+c}\cdot a^{-1}+\frac b{a+b+c}\cdot b^{-1}+
\frac c{a+b+c}\cdot c^{-1}\right)^{-1}=\frac{a+b+c}3.\]
The mean inequality now gives \(M_{-1}^m(a,b,c)\leq M_{0}^m(a,b,c)\). The equality holds if and only if \(a=b=c\).
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 1If \(x,y > 0\), \(p > 1\) 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 \(\frac{x}{\alpha}=
\frac{y}{1-\alpha}\).
For \(p > 1\), the function \(\varphi(x)=x^p\) is
strictly convex
hence \((\alpha a+(1-\alpha)b)^p
\leq \alpha a^p+(1-\alpha)b^p\).
The equality
holds if and only if \(a=b\). Setting \(x=\alpha a\) and
\(y=(1-\alpha) b\) we get (1) immediately.
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
\(i\), \(1\leq i\leq n\).
From (1) we get
\((x_i+y_i)^p\leq \alpha^{1-p} x_i^p+(1-\alpha)^{1-p} y_i^p\).
Multiplying by \(m_i\) and adding as \(1\leq i\leq n\) we
get (2). The equality holds if and only
if \(\frac{x_i}{y_i}=\frac{\alpha}{1-\alpha}\).
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 \(1\leq i\leq n\).
For any \(\alpha\in(0,1)\) we have inequality
(2). Let us write \[A=
\left(\sum_{i=1}^nx_i^pm_i\right)^{1/p},\;\;
B=\left(\sum_{i=1}^ny_i^pm_i\right)^{1/p}.\] In
new terminology (2) reads as
\begin{eqnarray*} \sum_{i=1}^n(x_i+y_i)^pm_i \leq \alpha^{1-p}
A^p+(1-\alpha)^{1-p}B^p. \quad\quad\quad(4)\end{eqnarray*} If we choose
\(\alpha\) such that \(\frac A{\alpha}=\frac{B}{1-\alpha}\),
then (1) implies
\(\alpha^{1-p}A^p+(1-\alpha)^{1-p}B^p=(A+B)^p\) and (4)
now becomes \[\sum_{i=1}^n(x_i+y_i)^pm_i \leq
\left[ \left(\sum_{i=1}^n x_i^pm_i\right)^{1/p}+
\left(\sum_{i=1}^n y_i^pm_i\right)^{1/p}\right]^p\] which
is equivalent to (3).
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?
Let us set \(\displaystyle a=\sqrt{\sum_{i=1}^n u_i^2}\) and \(\displaystyle b=
\sqrt{\sum_{i=1}^n v_i^2}\). By Minkowski’s inequality (for \(p=2\))
we have \(\sum_{i=1}^n(u_i+v_i)^2\leq(a+b)^2\). Hence the LHS of the
desired inequality is not greater than \(1+(a+b)^2\), while the RHS
is equal to \(4(1+a^2)(1+b^2)/3\). Now it is sufficient to prove
that \[3+3(a+b)^2\leq 4(1+a^2)(1+b^2).\] The last inequality can
be reduced to the trivial \(0\leq (a-b)^2+(2ab-1)^2\). The equality
in the initial inequality holds if and only if \(u_i/v_i=c\) for
some \(c\in \mathbb{R}\) and \(a=b=1/\sqrt2\).
Problem 5
If \(a_1, \dots, a_n\) are real numbers, prove that
\[\sqrt{a_1^2+(1-a_2)^2}+\sqrt{a_2^2+(1-a_3)^2}+\cdots+ \sqrt{a_n^2+(1-a_1)^2}\geq \frac{n\sqrt 2}2.\]
We will prove first by induction that \[\sqrt{a_1^2+b_1^2}+\sqrt{a_2^2+b_2^2}+\cdots+\sqrt{a_n^2+b_n^2}\geq \sqrt{(a_1+\cdots+ a_n)^2+(b_1+\cdots + b_n)^2}. \]
The case \(n=2\) is a consequence of the Minkowski’s inequality for \(p=2\) and sequences \(x=(a_1,b_1)\), \(y=(a_2,b_2)\). If the inequality holds for some \(n\), then
\begin{eqnarray*}&&\sqrt{a_1^2+b_1^2}+\cdots+\sqrt{a_n^2+b_n^2}+\sqrt{a_{n+1}^2+b_{n+1}^2}\newline
&\geq &
\sqrt{(a_1+\cdots+ a_n)^2+(b_1+\cdots + b_n)^2}+
\sqrt{a_{n+1}^2+b_{n+1}^2}\newline
&\geq& \sqrt{(a_1+\cdots+ a_n+a_{n+1})^2+(b_1+\cdots + b_n+b_{n+1})^2}.
\end{eqnarray*} The equality holds if and only if \((a_i,b_i)=\lambda_i
\cdot(x,y)\) for some fixed real numbers \((x,y)\) and real numbers \(\lambda_i\).
Applying this to the sequences \((a_1, \dots, a_n)\), \((b_1, \dots, b_n)\) we get
\begin{eqnarray*}&&
\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}\newline&\geq& \sqrt{[a_1+\cdots+a_n]^2+[n-(a_1+\cdots+a_n)]^2}\geq
\sqrt{\frac{\left([a_1+\cdots+a_n]+[n-(a_1+\cdots+a_n)]\right)^2}2}\newline&=&\frac{n\sqrt 2}2.
\end{eqnarray*}
The equality holds if and only if \(a_1=\cdots=a_n=\frac12\).
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
\(a^p=b^q\).
Since \(\varphi(x)=e^x\) is a convex function
we have that \(e^{\frac1p x+\frac1q y}\leq \frac1p e^x+
\frac1q e^y\). The equality holds if and only if \(x=y\), and
the inequality (5) is immediately obtained
by placing \(a=e^{x/p}\) and \(b=e^{y/q}\). The 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 \(1\leq i\leq n\).
From (5) we immediately
get \(x_iy_i= (\alpha x_i)\frac{y_i}{\alpha}\leq
\frac1p \cdot \alpha^px_i^p+\frac1q\cdot\frac1{\alpha^q}
y_i^q\). Multiplying by \(m_i\) and adding as \(i=1,2,\dots, n\)
we get (6). 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.
The idea is very similar to the one
used in the proof of Minkowski’s inequality.
The inequality (6) holds
for any positive constant \(\alpha\). Let \[A=
\left(\alpha^p\sum_{i=1}^nx_i^pm_i\right)^{1/p},\;\;
B=\left(\frac1{\alpha^q}\sum_{i=1}^n y_i^qm_i\right)^{1/q}.\]
By Young’s inequality we have that
\(\frac1pA^p+\frac1qB^q=AB\) if \(A^p=B^q\). Equivalently
\(\alpha^p\sum_{i=1}^nx_i^pm_i=\frac1{\alpha^q}\sum_{i=1}^n
y_i^qm_i\). Choosing such an \(\alpha\) we get
\[ \sum_{i=1}^nx_iy_im_i\leq \frac1pA^p+\frac1qB^q=
AB=\left(\sum_{i=1}^nx_i^pm_i\right)^{1/p}\cdot
\left( \sum_{i=1}^ny_i^qm_i\right)^{1/q}.\]
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 \[\sqrt{a_1}m_1+\cdots + \sqrt{a_n}m_n\geq
\frac{\alpha^{3/2}}{\beta}.\]
We will apply Holder’s inequality
on \(x_i=a_i^{1/3}\), \(y_i=a_i^{2/3}\),
\(p=\frac32\), \(q=3\):
\[\alpha=\sum_{i=1}^n a_im_i\leq
\left(\sum_{i=1}^n a_i^{1/2}m_i\right)^{2/3} \cdot
\left(\sum_{i=1}^n a_i^2m_i\right)^{1/3}=
\left(\sum_{i=1}^n \sqrt{a_i}m_i\right)^{2/3}
\cdot \beta^{2/3}.\]
Hence
\(\sum_{i=1}^n\sqrt{a_i}m_i\geq \frac{\alpha^{3/2}}{\beta}\).
Problem 7
If \(a\), \(b\), \(c\), and \(d\) are positive real numbers such that \(a+b+c+d=1\), prove that
\[\frac{a^2}{(1+b)(1-c)}+\frac{b^3}{(1+c)(1-d)}+\frac{c^3}{(1+d)(1-a)}+\frac{d^3}{(1+a)(1-b)}\geq\frac1{15}.\]
Assume that \((x_i)_{i=1}^n\), \((y_i)_{i=1}^n\), and \((z_i)_{i=1}^n\) are three sequences of positive real numbers. Using Holder’s inequality we obtain:
\[\sum_{i=1}^n x_iy_iz_i=\sum_{i=1}^n (x_iy_i)z_i\leq\left(\sum_{i=1}^n (x_iy_i)^{\frac32}\right)^{\frac23}\cdot\left(\sum_{i=1}^n z_i^3\right)^{\frac13}.\]
Using Holder’s inequality again we get
\[\sum_{i=1}^n x_i^{\frac23}y_i^{\frac23}\leq\left(\sum_{i=1}^n x_i^3\right)^{\frac12}\cdot\left(\sum_{i=1}^ny_i^3\right)^{\frac12}\] which implies
\[\sum_{i=1}^n x_iy_iz_i\leq \left(\sum_{i=1}^n x_i^3\right)^{\frac13}\cdot \left(\sum_{i=1}^n y_i^3\right)^{\frac13}\cdot \left(\sum_{i=1}^n z_i^3\right)^{\frac13}.
\] The equality holds if and only if there are real numbers \(\gamma\) and \(\delta\) such that \(y_i=\gamma x_i\) and \(z_i=\delta x_i\).
We now apply the last inequality to the following three sequences \begin{eqnarray*}&&\left(\frac{a}{\sqrt[3]{(1+b)(1-c)}},\frac{b}{\sqrt[3]{(1+c)(1-d)}}, \frac{c}{\sqrt[3]{(1+d)(1-a)}},
\frac{d}{\sqrt[3]{(1+a)(1-b)}}\right)
\newline
&&\left(\sqrt[3]{1+b}, \sqrt[3]{1+c},\sqrt[3]{1+d},\sqrt[3]{1+a}\right)
\newline
&&\left(\sqrt[3]{1-c},\sqrt[3]{1-d},\sqrt[3]{1-a},\sqrt[3]{1-b}\right)\end{eqnarray*}
and obtain:
\[a+b+c+d\leq \left(\frac{a^2}{(1+b)(1-c)}+\frac{b^3}{(1+c)(1-d)}+\frac{c^3}{(1+d)(1-a)}+\frac{d^3}{(1+a)(1-b)}\right)^{\frac13}\cdot \sqrt[3]{5}\cdot\sqrt[3]{3}\]
which is equivalent to the required inequality. Equality holds if and only if \(a=b=c=d=\frac14\).
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 \(M^m_r(x)\leq M^m_s(x)\).
\(M^m_r=\left(\sum_{i=1}^n x_i^r\cdot m_i
\right)^{1/r}\). We will use the Holders inequality for
\(y_i=1\), \(p=\frac sr\), and \(q=\frac{p}{1-p}\). Then we get
\[M^m_r\leq \left(\sum_{i=1}^nx_i^{rp}\cdot
m_i\right)^{\frac1
{pr}}\cdot\left(\sum_{i=1}^n 1^{q}\cdot
m_i\right)^{p/(1-p)} =M_s. \]
Problem 8
Let \(x\), \(y\), and \(z\) be positive real numbers such that
\(xyz=1\). Prove that \[\frac{x^3}{(1+y)(1+z)}+\frac{y^3}
{(1+z)(1+x)}+\frac{z^3}{(1+x)(1+y)}\geq\frac{3}{4}.\]
The given inequality is equivalent to
\[ x^3(x+1)+y^3(y+1)+
z^3(z+1)\geq\frac34(1+x+y+z+xy+yz+zx+xyz).\] The left-hand
side can be written as \(x^4+y^4+z^4+x^3+y^3+z^3=
3M_4^4+3M_3^3\). Using \(xy+yz+zx\leq x^2+y^2+z^2=3M_2^2\)
we see that the right-hand side is less than or equal
to \(\frac34(2+3M_1+3M_2^2)\). Since \(M_1\geq 3\sqrt[3]{xyz}=1\),
we can further say that the right-hand side of the
required inequality is less than or equal to
\(\frac34(5M_1+3M_2^2)\). Since \(M_4\geq M_3\), and
\(M_1\leq M_2\leq M_3\), the following inequality would
imply the required statement:
\[3M_3^4+3M_3^3\geq \frac34(5M_3+3M_3^2).\]
However the last inequality is equivalent to
\((M_3-1)(4M_3^2+8M_3+5)\geq0\) which is true because
\(M_3\geq 1\). The equality holds if and only if
\(x=y=z=1\).
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*}
After noticing that
\(\sum_{i=1}^nx_iy_im_i\leq
\sum_{i=1}^n|x_i|\cdot |y_i|m_i\), the rest is just a special
case (\(p=q=2\)) of the Holder’s inequality.
Problem 9
If \(a\), \(b\), and \(c\) are positive numbers,
prove that \[\frac ab+\frac bc+\frac ca
\geq \frac{(a+b+c)^2}{ab+bc+ca}.\]
We will apply the Cauchy-Schwartz inequality
with \(x_1=\sqrt{\frac ab}\), \(x_2=\sqrt
{\frac bc}\), \(x_3=\sqrt{\frac ca}\),
\(y_1=\sqrt{ab}\), \(y_2=\sqrt{bc}\), and \(y_3=\sqrt{ca}\).
Then
\begin{eqnarray*}
a+b+c&=&x_1y_1+x_2y_2+x_3y_3\leq \sqrt{x_1^2+x_2^2+x_3^2}
\cdot \sqrt{y_1^2+y_2^2+y_3^2}\newline
&=& \sqrt{\frac ab+\frac bc+ \frac ca}\cdot
\sqrt{ab+bc+ca}.\end{eqnarray*}
The equality holds if and only if \(a=b=c\).
Problem 10
If \(a\), \(b\), \(c\) are positive real numbers prove that
\[\frac{a^2}b+\frac{b^2}c+\frac{c^2}a\geq \frac{a^2+b^2+c^2}{a+b+c}.\]
We apply the Cauchy-Schwarz inequality to the sequences \(\left(\frac a{\sqrt b},\frac b{\sqrt c},\frac c{\sqrt a}\right)\) and \((\sqrt b, \sqrt c,\sqrt a)\). The result follows immediately.
Problem 11 (Dušan Djukić)
Let \(a,b,c\) be positive real numbers. Prove the
inequality \[\frac{a^2}b+\frac{b^2}c+\frac{c^2}a\geq
a+b+c+\frac{4(a-b)^2}{a+b+c}.\]
Starting from \(\frac{(a-b)^2}b=
\frac{a^2}b-2a+b\) and similar equalitites for \((b-c)^2/c\) and
\((c-a)^2/a\) we get that
the required inequality is equivalent to
\begin{eqnarray*}
(a+b+c)\left(\frac{(a-b)^2}b+\frac{(b-c)^2}a+\frac{(c-a)^2}b
\right)\geq 4(a-b)^2. \quad\quad\quad (12)\end{eqnarray*}
By the Cauchy-Schwartz inequality we
have that the left-hand side of (12) is greater than
or equal to \((|a-b|+|b-c|+|c-a|)^2\). The inequality (12) now follows
from \(|b-c|+|c-a|\geq |a-b|\).
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
\[\lim_{r\rightarrow 0} M_r(a_1,\dots, a_n)=
a_1^{m_1}\cdot a_2^{m_2}\cdots
a_n^{m_n}.\]
\[M_r(a_1,\dots, a_n)=e^{\frac1r
\log(a_1^rm_1+\cdots + a_n^rm_n)}.\] Using the
L’Hopital’s theorem we get
\begin{eqnarray*}
\lim_{r\rightarrow 0}\frac1r\log(a_1^rm_1+
\cdots + a_n^rm_n)&=&\lim_{r\rightarrow0}
\frac{m_1a_1^r\log a_1+
\cdots +m_na_n^r\log a_n}{a_1^rm_1+\cdots+a_n^rm_n}\newline
&=& m_1\log a_1+\cdots + m_n\log a_n\newline
&=&\log\left(a_1^{m_1}\cdots a_n^{m_n}\right).\end{eqnarray*}
The result immediately follows.