Inequalities - Minkowski's inequality

$\S$ Suppose that $r$ is finite and not equal to $1$. Then

$$\mathcal{\bar{A}}(a) + \mathcal{\bar{A}}(b) + ... + \mathcal{\bar{A}}(l) > \mathcal{\bar{A}}(a+b+ ... + l)$$ holds with $r>1$; and
$$\mathcal{\bar{A}}(a) + \mathcal{\bar{A}}(b) + ... + \mathcal{\bar{A}}(l) < \mathcal{\bar{A}}(a+b+ ... + l)$$ holds with $r<1$. The inequality holds unless $(a), (b), ..., (l)$ are proportional, or $r\le 0$ and $a_\nu = b_\nu = ... = l_\nu=0$ for some a $\nu$.

The main result remains true when $r=+\infty$ or $r=-\infty$, except that the conditions for equality require a restatement.

$\S$ If $r$ is finite and not equal to $0$ or $1$, then

$$\left(\sum(a+b+ ... + l)^r\right)^{1/r} < \left(\sum a^r\right)^{1/r} + ... +\left(\sum l^r\right)^{1/r}$$ holds with $r>1$, and

$$\left(\sum(a+b+ ... + l)^r\right)^{1/r} > \left(\sum a^r\right)^{1/r} + ... +\left(\sum l^r\right)^{1/r}$$ holds with $r<1$.

$\S$ If $r$ is positive and not equal to $1$, then

$$\sum(a+b+ ... + l)^r > \sum a^r + ... +\sum l^r$$ holds with $r>1$, and
$$\sum(a+b+ ... + l)^r < \sum a^r + ... +\sum l^r$$ holds with $r<1$.

In a nutshell, what is usually required in practice is as follows:
If $r>0$ then,

$$\left(\sum(a+b+ ... + l)^r\right)^{R} < \left(\sum a^r\right)^{R} + ... +\left(\sum l^r\right)^{R}$$ where $R=1$ if $0<r\le1$ and $R=\frac{1}{r}$ if $r>1$.

The following inequality is often useful for the prupose of determining an upper bound for $\sum a^k$.

$\S$ Suppose that $k>1$, that $k'$ is conjugate to $k$, and that $B>0$. Then a necessary and sufficient condition that $\sum a^k \le A$ is that $\sum ab \le A^{1/k}B^{1/{k'}}$ for all $b$ for which $\sum b^{k'}\le B$.

The geometrical interpretations are illustrated as follows. When $k=2$, $A=l^2$, and $B=1$ holds, we take rectangular coordinats. The theorem asserts that, if the length of the projection of a vector along an arbitrary direction does not exceed $l$, the length of the vector does not exceed $l$.