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REAL ANALYSIS

INEQUALITIES

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\begin{description}

\item[The Inequality of Arithmetic and Geometric Means]

Given $x_1\geq0\ x_2\geq0\ldots x_n\geq0$

$$A=\frac{x_1+x_2+\ldots+x_n}{n}\ G=(x_1x_2\ldots
x_n)^\frac{1}{n}$$

Then $G<A$ unless $x_1=x_2=\ldots=x_n$, when $G=A$.

\item[Proof]
Suppose without loss of generality that $x_1$ is a maximal $x_\nu$
and $x_2$ is a minimal $x_\nu$.

If $x_1=x_2$ the $_r$'s are all equal and there is nothing to
prove.

Suppose then that $x_1>x_2$. We form a  new set of numbers
$x_{11}x_{21}\ldots x_{n1}$ by writing

$$x_{11}=A\ x_{21}=-A+x_1+x_2\ x_{r1}=x_r\ r=3,\ldots,n.$$

Let $A_1,G_1$ be the A.M and G.M of the $x_{r1}$'s.

$A_1=A$ since $x_{11}+x_{21}=x_1+x_2$.

However

\begin{eqnarray*}
x_{11}x_{2x}-x_1x_2&=&A(x_1+x_2-A)-x_1x_2\\ &=&(x_1-A)(A-x_2)>0
\end{eqnarray*}

since $x_1>A>x_2$.

Therefore $G_1>G$

If the $x_{\nu1}$ are not all equal we can again take a largest
$x_{\alpha1}$ and a smallest $x_{\beta1}$ and replace them by $A,\
x_{\alpha1}+x_{\alpha2}-A$.

$A_2=A_1\ G_2>G_1$

After at most $n-1$ steps all the $x$'s are equal.

$G<G_1<\ldots<G_K=A_k=A$ therefore $G<A$.

\item[Cauchy's Inequalities]
Given $x_1\ldots x_n\ y_1\ldots y_n$ real.

Then

$$
\sum_{r=1}^nx_ry_r\leq\left(\sum_{r=1}^na_r^2\right)^\frac{1}{2}\left(\sum_{r=1}^ny_r^2\right)^\frac{1}{2}$$

with equality $\Leftrightarrow$ the two sets are proportionali.e.
$\Leftrightarrow \exists(\lambda\mu)\neq(00)|\lambda x_r+\mu
y_r=0\ (r=1,2,\ldots,n)$

\item[Proof]
Consider the quadratic form $Q(\lambda\mu)$ defined by

\begin{eqnarray*}
Q(\lambda\mu)&=&\sum_{r=1}^n(\lambda x_r+\mu y_r)^2\\
&=&\lambda^2\sum_{r=1}^n
x_r^2+2\lambda\mu\left(\sum){r=1}^nx_ry_r\right)+\mu^2\sum_{r=1}^ny_r^2
\end{eqnarray*}

If $\exists \lambda\mu\neq00|\lambda x_r+\mu y_r=0\
r=1,2,\ldots,n$ then there is nothing to prove.

Suppose $\exists$ no such $(\lambda,\mu)$. Then $Q(\lambda\mu)>0$
for every $(\lambda\mu)\neq(00)$. Hence $Q(\lambda\mu)$ is
positive definite so that

$$\left(\sum_{r=1}^nx_ry_r\right)^2<\sum_{r=1}^nx_r^2\sum_{r=1}^ny_r^2$$

using \lq\lq$b^2<4ac$''.

\item[Weighted Means]
Given a set of non-negative numbers $x_1\ldots x_n$ and a set of
weights $P$, where we attach the weight $P_r$ to $x_r$, each
$P>0$. The weighted means are

$$A_P=\frac{P_1x_1+\ldots+P_nx_n}{P_1+P_2+\ldots +P_n}$$

$$G_P=(x_1^{P_1}x_2^{P_2}\ldots
x_n^{P_n})^\frac{1}{p_1+P_2+\ldots+P_n}$$

\item[Note]
If the weights $p_1\ldots p_n$ are replaced by $tp_1\ldots tp_n$,
then $A_p\ G_P$ are unchanged. In particular if we take
$t=\frac{1}{P_1+\ldots+p_n}$ we get a set of weights $Q:\
q_1\ldots q_n|q_1+\ldots+q_n=1$. Then $G_P\leq A_P$ with equality
$\Leftrightarrow$ all the $x$'s are equal.

\item[Proof]

\begin{description}

\item[(i)]
Result proved when $P_j$ are all integers.

\item[(ii)]
Result follows when $P_j$ are commensurable; i.e. when $\exists
t>0|tP_1\ldots tP_n$ are all integers.

\item[(iii)]
We have to deal with the case where the $P$'s are not
commensurable.

Let $q_1\ldots q_n$ be a set of weights $|\sum q_j=1$.

Let $Q\ (q_1\ldots q_n)$ be a point in $R_n$.

Take a set of rational points

$$P^r=(r_1\ldots r_n)\ \ r_j>0$$

where $P^r\rightarrow Q$ as $r\to\infty$.

$G_{P_r}<A_{P_r}$ unless the $x^\nu$ equal.

Letting $r\to\infty$ $G_Q\leq A_Q$.

We still have to prove strict inequality when the $x$'s are not
all equal. Suppose then that the $x$'s are not all equal. Write

$$q_j=j_j'+q_j''\ j=1,2,\ldots,n$$

where $q_j'>0\ q_j''>0\ q_j'$ is rational.

\begin{eqnarray*}
P'&:&Q_1'\ldots q_n'\ \ P'':q_1''\ldots q_n''\\ r'&=&\sum q_j'\
r''=\sum q_j''\ r'+r''=1
\end{eqnarray*}
$G_{P'}<A_{P'}$ by (ii) $G_{p''}\leq A_{p''}$

$$G_Q=(G_{P'})^{r'}(G_{p''})^{r''}\leq
r'G_{P'}+r''G_{P''}<r'A_{P'}+r''A_{P''}=A_Q$$

using $\sum q_j=1$

\end{description}

\item[H\"{o}lder's Inequality]
We have two sets of numbers

\begin{center}
\begin{tabular}{ll}
$x_1\ldots x_n$&$x_j\geq0$\\ $y_1\ldots y_n$&$y_j\geq0$
\end{tabular}
\end{center}

$\alpha,\beta$ are positive and $\alpha+\beta=1$. Then

$$\sum_{\nu=1}^nx_\nu^\alpha
y_\nu^\beta\leq\left(\sum_{\nu=1}^nx_\nu\right)^\alpha\left(\sum_{\nu=1}^ny_\nu\right)^\beta$$

with equality $\Leftrightarrow $ the sets are proportional.

\item[Alternative Form]
Suppose $\lambda,\mu$ are positive and
$\frac{1}{\lambda}+\frac{1}{\mu}=1$

$$\sum_{\nu=1}^nx_\nu y_\nu\leq\left(\sum_{\nu=1}^n
x_\nu^\lambda\right)^\frac{1}{\lambda}\left(\sum_{\nu=1}^n
y_\nu^\mu\right)^\frac{1}{\mu}$$

[This result above with $\alpha,\beta$ replaced by
$\frac{1}{\lambda},\frac{1}{\mu}$ and $x_\nu^\alpha, x_\nu^\beta$
replaced by new variables $x_\nu,y_\nu$.]

This generalises to $k$ sets and $k$ numbers
$\alpha_1+\ldots+\alpha_k=1$.

Cauchy's inequality follows with $\lambda=\mu=2$.

\item[Proof]
Write $\displaystyle U=\sum_{\nu=1}^n x_\nu\ \ V=\sum_{\nu=1}^n
y_\nu$

Suppose $UV>0$ (nothing to prove otherwise).

\begin{eqnarray*}
U^\alpha V^\beta=\sum_{\nu=1}^nx_\nu^\alpha
y_\nu^\beta&=&\sum_{\nu=1}^n\left(\frac{x_\nu}{U}\right)^\alpha\left(\frac{y_\nu}{V}\right)^\beta\\
&\leq&\sum_{\nu=1}^n\alpha\frac{x_\nu}{\nu}+\beta\frac{y_\nu}{\nu}=\alpha+\beta=1
\end{eqnarray*}

with equality $\Leftrightarrow \frac{x_\nu}{U}=\frac{y_\nu}{V}$
for $\nu=1,2,\ldots,n$.

These inequalities generalise to integrals.

Suppose $f(x)\geq0\ g(x)\geq0$ are continuous on $[a\ b]$

$$\int_a^bf(x)g(x)\,dx\leq\left(\int_a^bf^2\,dx\right)^\frac{1}{2}\left(\int_a^bg^2\,dx\right)^\frac{1}{2}$$

This is known as Schwarz's inequality.

If $\frac{1}{\lambda}+\frac{1}{\mu}=1\ \lambda>0\ \mu>0$ then

$$\int_a^bf(x)g(x)\,dx\leq\left(\int_a^bf^\lambda\,dx\right)^\frac{1}{\lambda}\left(\int_a^bg^{-\mu}\,dx\right)^\frac{1}{\mu}.$$

This is known as H\"{o}lder's inequality.

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