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 Totally 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)