Talk:Minkowski's Inequality

In the general form, is p restricted to the natural numbers, or can it be any real, or something else? --Cynic (talk) 23:32, 3 February 2009 (UTC)

One assumes it can't be complex because Complex Numbers cannot be Ordered Compatibly with Ring Structure. One assumes it can't be negative because negative roots are not generally defined for negative $p$. I suspect it also needs to be $p \ge 1$. Apart from that (because of a general unwritten law of aggregation: "results for integers tend to be able to be extended to real numbers") I assume one should be able to prove it for all real $p \ge 1$. Not sure, I haven't researched it properly yet. --prime mover (talk)] 06:25, 4 February 2009 (UTC)

... okay I've looked it up. When $p < 1, p \ne 0$ the inequality is reversed. For $p < 0$ you need all $x_i, y_i > 0$. For $p=1$ you've obviously got an equality. Nowhere can I find anyone restricting this to integral $p$. --prime mover (talk) 06:32, 4 February 2009 (UTC)