most recent 30 from http://mathoverflow.net 2013-05-20T03:23:31Z http://mathoverflow.net/feeds/question/97540 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97540/a-polynomial-inequality-proof profiles111166771186296351341 2012-05-21T08:34:41Z 2012-05-21T11:37:14Z

<p>Given $x_1,x_2, \ldots, x_n \ge 0, \alpha \ge 1$, show that</p> <p>$\sum_{i}\alpha x_i(\sum_{j \le i}{x_j})^{\alpha-1} \ge (\sum_{i}x_i)^\alpha$</p> <p>We're pretty sure the ineuqality holds for the given precondition. It can be validated using a small piece of matlab code. Now we post it for rigorous proof. Its continuous version is easy:</p> <p>$\int^{n}\alpha x(\tau)(\int^{\tau}{x(t)}dt)^{\alpha-1}d\tau= (\int^{n}x(t)dt)^\alpha$</p> <p>When writing down the proof, there are more details to be finalized, so we're stuck for a rigorous proof. Hope there's anyone who can help or point us to some resource.</p>

http://mathoverflow.net/questions/97540/a-polynomial-inequality-proof/97546#97546 js 2012-05-21T11:37:14Z 2012-05-21T11:37:14Z

<p>Setting $s_i = \sum_{j \leq i} x_j$, just write $$ \sum_{i \leq n} \alpha x_i s_i^{\alpha-1} = \sum_{i \leq n} \int_{s_{i-1}}^{s_i} \alpha s_i^{\alpha-1} dt \geq \sum_{i \leq n} \int_{s_{i-1}}^{s_i} \alpha t^{\alpha-1} dt = \int_{0}^{s_n} \alpha t^{\alpha-1} dt = s_n^{\alpha} $$</p>