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 Can Not be Ordered‎. 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. --Matt Westwood 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 $$p$$. --Matt Westwood 06:32, 4 February 2009 (UTC)