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Let $n_i$, $i=1,\ldots,m+1$ be nonnegative natural numbers, sum of which $\sum_{i=1}^{m+1}n_i=N$.

I woul like to find an upper bound for the following$ \sum_{i=1}^{m+1}\frac{\sqrt n_i}{2^{i-1}}$

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    @David: I think it is: $\sqrt{n_i} \le \sqrt{N}$.2012-04-13

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Using $x=(\sqrt{n_1},...,\sqrt{n_{m+1}})$, $y=(1,\frac{1}{2},...,\frac{1}{2^m})$, the Cauchy-Schwarz inequality ( \leq ||x|| ||y||) gives $\sum_{i=1}^{m+1}\frac{\sqrt n_i}{2^{i-1}} \leq \sqrt{N} \frac{2}{\sqrt{3}} \sqrt{1-\frac{1}{4^{m+1}}}$