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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.

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  • $\begingroup$ Here's a sketch: * associate continuous $x(t)$ with $x_i$ * construct an s(t) being integral of $x(t)$ and equal to or less than sum of $x_i$ Intuitively, it should work in this way but the construction seems not trivial. $\endgroup$
    – user23849
    Commented May 21, 2012 at 8:38

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

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  • $\begingroup$ thanks very much. That's accurate now!! What a dull head I get.. I was all stucking thinking how to construct a continuous x(t)... $\endgroup$
    – user23849
    Commented May 21, 2012 at 12:21

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