Inequalities - Holder's inequality

$\S$ Suppose that $(a)$, $(b)$, …, $(l)$ are $m$ sets each of $n$ numbers.
Then $\mathcal{G}(a) + \mathcal{G}(b) + ... + \mathcal{G}(l) < \mathcal{G}(a+b+ ... +l)$, unless either (1) every two of $(a)$, $(b)$, …, $(l)$ are proportional, or (2) there is a $\nu$ such that $a_\nu = b_\nu = ... = l_\nu = 0$.

$\S$ If $\alpha, \beta, ..., \lambda$ are positive and $\alpha + \beta + ...+\lambda = 1$, then

$$\sum a^\alpha b^\beta ... l^\lambda < \left(\sum a\right)^\alpha \left(\sum b\right)^\beta ... \left(\sum l\right)^\lambda,$$ unless either (1) the sets $(a), (b), ..., (l)$ are all proportional, or (2) one set is nul.

$\S$ If $r, \alpha, \beta, ..., \lambda$ are positive and $\alpha+\beta+...+\lambda =1$, then

$$\mathcal{\bar{A}}_r(ab...l) < \mathcal{\bar{A}}_{r/\alpha}(a) \mathcal{\bar{A}}_{r/\beta}(b) ... \mathcal{\bar{A}}_{r/\lambda}(l)$$ unless $(a^{1/\alpha}), (b^{1/\beta}), ..., (l^{1/\lambda})$ are proportional or one of the factor on the right-hand side is zero. If $r<0$, the inequality is reversed.

$\S$ Any two real numbers $k$ and $k'$ are conjugate if they are satisfied any one of the three following equalities:

1. $k' = \frac{k}{k-1}$, with $k\neq 1$
2. $(k-1)(k'-1)=1$
3. $\frac{1}{k} + \frac{1}{k'} = 1$ with $k\neq0$, and $k'\neq 0$

Suppose that $k\neq0$, $k'\neq0$, and that $k'$ is conjugate to $k$. Then:

$$\sum ab < \left(\sum a^k\right)^{1/k}\left(\sum b^{k'}\right)^{1/k'}$$ holds with $(k>1)$, unless $(a^k)$ and $(b^{k'})$ are proportional; and

$$\sum ab > \left(\sum a^k\right)^{1/k}\left(\sum b^{k'}\right)^{1/k'}$$ holds with $(k<1)$, unless $(a^k)$ and $(b^{k'})$ are proportional or $(ab)$ is nul. Furthermore, those two inequalies can be combined in a single inequality:
$$\left( \sum ab \right)^{kk'} < \left(\sum a^k\right)^{k'}\left(b^{k'}\right)^k$$ with $k\neq0$ and $k\neq1$.

General property of mean values:

$\S$ If $r<s$ then

$$\mathcal{\bar{A}}_r(a) <\mathcal{\bar{A}}_s(a)$$ , unless the $a$ are all equal, or $s\le 0$ and ana is zero.