Inequalities - Elementary mean values

Given:
$\mathcal{\bar{A}}_r = \mathcal{\bar{A}}_r(a)=\left(\frac{1}{n}\sum a^r\right)^{1/r} = \left(\frac{1}{n}\sum\limits_{\nu=1}^n a_{\nu}^r\right)^{1/r}$

geometric mean:
$\mathcal{\bar{G}} = \mathcal{\bar{G}}(a) = \sqrt[n]{(a_1a_2...a_n)} = \sqrt[n]{(\Pi a)}$

in paritcular, arithmetic mean:
$\mathcal{A} = \mathcal{A}(a) = \mathcal{\bar{A}}_1$
Thus,

$\mathcal{\bar{A}}_r(a) = \left( \mathcal{A}(a) \right)^{\frac{1}{r}}$
$\mathcal{\bar{G}}(a) = e^{\mathcal{A}(\log a)}$

harmonic mean:
$\mathcal{\bar{H}}(a) = \mathcal{\bar{A}}_{-1}(a)$

Conclusions:

$\mathcal{\bar{A}}_r = \mathcal{\bar{A}}_r(a,p)=\left(\frac{\sum p a^r}{\sum p}\right)^{1/r}$ holds with $p_\nu>0$
$\mathcal{\bar{G}} = \mathcal{\bar{G}}(a,p) = (\Pi a^p)^{1/\sum p}$

Weighted means:

suppose $\sum q = 1$, then we obtain:

$\mathcal{\bar{A}}_r(a) = \left(\sum q a^r\right)^{1/r}$
$\mathcal{\bar{G}}(a) = \Pi a^q$

Suppose $a>0$, then we obtain:

$\mathcal{A}(a+b) = \mathcal{A}(a) + \mathcal{A}(b)$
$\mathcal{\bar{G}}(ab) = \mathcal{\bar{G}}(a)\mathcal{\bar{G}}(b)$
$\mathcal{\bar{A}}_r(b) = k \mathcal{\bar{A}}_r(a)$ if and only if $b=k(a)$
$\mathcal{\bar{G}}(b) = k \mathcal{\bar{G}}(a)$ if and only if $b=k(a)$
$\mathcal{\bar{A}}_r(a) \le \mathcal{\bar{G}}(b)$ if $a_\nu \le b_\nu$, for all $\nu$

Limiting cases of $\mathcal{\bar{A}}_r(a)$. $\min a$ and $\max a$ denote the smallest and the largest value of $(a)$. We obtain:

$\lim\limits_{r\to 0}\mathcal{\bar{A}}_r(a) = \mathcal{\bar{G}}(a)$.

proof:
$\lim\limits_{r\to 0}\mathcal{\bar{A}}_r(a) = \lim\limits_{r\to 0} e^{\frac{1}{r}\log \left(\sum q a^r\right)} = \lim\limits_{r\to 0} e^{\frac{1}{r}(1+r\sum q \log a + O(r^2))} = e^{\sum q \log a} = \Pi a^q = \mathcal{\bar{G}}(a).$

$\lim\limits_{r\to +\infty}\mathcal{\bar{A}}_r(a) = \max a$, and $\lim\limits_{r\to -\infty}\mathcal{\bar{A}}_r(a) = \min a$.

Thus, we have: $\mathcal{\bar{A}}_{-\infty}(a)<\mathcal{\bar{A}}_r(a)<\mathcal{\bar{A}}_{+\infty}(a).$

Cauchy’s inequality:

$\mathcal{\bar{A}}_r(a) < \mathcal{\bar{A}}_{2r}(a).$

$(\sum ab)^2 \le \sum a^2 \sum b^2$ unless $(a)$ and $(b)$ are proportional. This inequality can be generalised as follows:

$$\begin{vmatrix} \sum a^2& \sum ab & ... & \sum al \\ ... & ... & ... & ... \\ \sum la& \sum lb & ... &\sum l^2 \end{vmatrix} > 0$$ unless the sets $(a)$, $(b)$, …, $(l)$ are linearly dependent, i.e., unless there are numbers $x$, $y$, …, $w$ not all zero, such that $xa_\nu + yb_\nu + ... + wl_\nu=0$ for every $\nu$.

Some inequalities useful in elementary analysis:

If $\xi>0$, $0<m<n$, then

$$\left(1+\frac{\xi}{m}\right)^m < \left(1+\frac{\xi}{n}\right)^n$$ .
If also $\xi<m$, then
$$\left(1-\frac{\xi}{m}\right)^{-m} < \left(1-\frac{\xi}{n}\right)^{-n}$$
$\S$
If $\xi>0$, $\xi\neq1$, $0<m<n$, then
$$n(\xi^{1/n}-1) < m(\xi^{1/m}-1)$$
$\S$
$$a_1^{p_1}a_2^{p_2} ... a_n^{p_n} < \left(\frac{p_1a_1+ ... +p_na_n}{p_1+ ... +p_n}\right)^{p_1+ ... + p_n}$$
$\S$
$$a_1^{q_1}a_2^{q_2} ... a_n^{q_n} < \sum qa$$ where as usually $\sum q =1$. The simplest case is: $a^\alpha+b^\beta < a\alpha+b^\beta$ with $\alpha+\beta=1$.
$\S$
$$a_1^\alpha b_1^\beta + a_2^\alpha b_2^\beta < (a_1+a_2)^\alpha(b_1+b_2)^\beta$$ with $\alpha+\beta=1$.
$\S$
$$\left((a_1+b_1)^r+(a_2+b_2)^r\right)^{1/r} < (a_1^r+b_1^r)^{1/r} + (b_1^r+b_2^r)^{1/r}$$ with $r>1$. When $r<1$, the inequality is reversed.
$\S$
If $\alpha+\beta+ ...+ \lambda=1$, then
$$a_1^\alpha b_1\beta ... l_1^\lambda + a_2^\alpha b_2\beta ... l_2^\lambda < (a_1+a_2)^\alpha (b_1+b_2)^\beta ... (l_1+l_2)^\lambda$$ , unless $a_1/a_2 = b_1/b_2= ... = l_1/l_2$ or one of the sets is nul.