Given $x\_1,x\_2, \ldots, x\_n \ge 0, \alpha \ge 1$, show that


$\sum\_{i}\alpha x\_i(\sum\_{j \le i}{x\_j})^{\alpha-1} \ge (\sum\_{i}x\_i)^\alpha$


We're pretty sure the inequality 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:


$\int^{n}\alpha x(\tau)(\int^{\tau}{x(t)}dt)^{\alpha-1}d\tau= (\int^{n}x(t)dt)^\alpha$


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.



A polynomial inequality
Setting $s\_i = \sum\_{j \leq i} x\_j$, just write
$$ \sum\_{i \leq n} \alpha x\_i s\_i^{\alpha-1} = \sum\_{i \leq n} \int\_{s\_{i-1}}^{s\_i} \alpha s\_i^{\alpha-1} dt \geq \sum\_{i \leq n} \int\_{s\_{i-1}}^{s\_i} \alpha t^{\alpha-1} dt = \int\_{0}^{s\_n} \alpha t^{\alpha-1} dt = s\_n^{\alpha}
$$


