Inequalities - Minkowski's inequality
§ \S Suppose that r r is finite and not equal to 11. Then
¯(a)+¯(b)+...+¯(l)>¯(a+b+...+l)\mathcal{\bar{A}}(a) + \mathcal{\bar{A}}(b) + ... + \mathcal{\bar{A}}(l) > \mathcal{\bar{A}}(a+b+ ... + l) holds with r>1 r>1; and
¯(a)+¯(b)+...+¯(l)<¯(a+b+...+l)\mathcal{\bar{A}}(a) + \mathcal{\bar{A}}(b) + ... + \mathcal{\bar{A}}(l) holds with r<1 r. The inequality holds unless (a),(b),...,(l) (a), (b), ..., (l) are proportional, or r≤0 r\le 0 and aν=bν=...=lν=0 a_\nu = b_\nu = ... = l_\nu=0 for some a ν \nu.
The main result remains true when r=+∞ r=+\infty or r=−∞ r=-\infty, except that the conditions for equality require a restatement.
§ \S If r r is finite and not equal to 00 or 1 1, then
(∑(a+b+...+l)r)1/r<(∑ar)1/r+...+(∑lr)1/r\left(\sum(a+b+ ... + l)^r\right)^{1/r} holds with r>1 r>1, and
(∑(a+b+...+l)r)1/r>(∑ar)1/r+...+(∑lr)1/r\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 r.
§ \S If r r is positive and not equal to 11, then
∑(a+b+...+l)r>∑ar+...+∑lr\sum(a+b+ ... + l)^r > \sum a^r + ... +\sum l^r holds with r>1 r>1, and
∑(a+b+...+l)r<∑ar+...+∑lr\sum(a+b+ ... + l)^r holds with r<1 r.
In a nutshell, what is usually required in practice is as follows:
If r>0 r>0 then,
\left(\sum(a+b+ ... + l)^r\right)^{R} where R=1 R=1 if 0<r≤1 0 and R=1r R=\frac{1}{r} if r>1 r>1.
The following inequality is often useful for the prupose of determining an upper bound for ∑ak \sum a^k.
§ \S Suppose that k>1 k>1, that k′ k' is conjugate to k k, and that B>0B>0. Then a necessary and sufficient condition that ∑ak≤A \sum a^k \le A is that ∑ab≤A1/kB1/k′ \sum ab \le A^{1/k}B^{1/{k'}} for all b b for which ∑bk′≤B\sum b^{k'}\le B.
The geometrical interpretations are illustrated as follows. When k=2 k=2, A=l2 A=l^2, and B=1 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 l, the length of the vector does not exceed ll.
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