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Let $n_1,...,n_m$ be non-negative natural numbers, such that their sum $n_1+\dots+n_m=2n$.

Fix $m$. I am wondering how to compute, or bound from above $\sum_{i=1}^m \sqrt{1+\frac{1}{2n_i-1}}. $

(One can assume that the $n_i$ are positive, since any term with $n_i=0$ doesn't contribute to the sum.)

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    @ThomasAndrews yes that is right. I overlooked that fact. So $m$ could be a larger number than $n$. I can't wait to see how someone solves this.2012-04-12

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Assuming each $n_i$ is positive: the largest any term can be is $\sqrt 2$, and there are at most $2n$ terms, so the sum is less than or equal to $(2\sqrt2)n$. This upper bound can be achieved by setting $m=2n$ and $n_1=\cdots=n_{2n}=1$.

Since the $n_i$ are bounded above by $2n$, the smallest any term can be is $\sqrt{1+1/(4n-1)}$, and so the sum is greater than or equal to $\sqrt{1+1/(4n-1)}$. This lower bound can be achieved, as Thomas Andrews pointed out, by setting $m=1$ and $n_1=2n$.

I'm not sure what more can be said about the distribution of the possible values of the sum, since we have the best possible upper and lower bounds.

If you mean to ask this question with $m$ fixed instead of variable, then it's a bit harder. Presumably the smallest it can be is when $n_1=2n$ and all the rest equal zero, while the largest it can be is when the $n_i$ are as equal as possible.

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    Thank you. But I;ve meant that m is fixed.2012-04-12
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Here is an answer for the problem where $m$ is fixed. For any fixed positive integer $s$, note that the function $ f(t) = \sqrt{1+\frac1{2t-1}} + \sqrt{1+\frac1{2(s-t)-1}} $ is decreasing on the interval $[1,s/2]$ and increasing on the interval $[s/2,s-1]$ (this is a simple calculus exercise). In particular, if n_1 are positive integers with $n_1+n_4 = s = n_2+n_3$, then $ \sqrt{1+\frac1{2n_1-1}} + \sqrt{1+\frac1{2n_4-1}} > \sqrt{1+\frac1{2n_2-1}} + \sqrt{1+\frac1{2n_3-1}}. $

This is enough to imply that the sum $ \sum_{i=1}^m \sqrt{1+\frac1{2n_i-1}} $ is maximized when $n_1=\cdots=n_{m-1}=1$ and $n_m = 2n-(m-1)$: for any other configuration, there are $n_i$ and $n_j$ both strictly larger than 1, in which case replacing them by $n_i-1$ and $n_j+1$ results in a larger sum.

By similar reasoning, it implies that the sum is minimized when the $n_i$ are all as equal as possible: specifically, with $k=[n/m]$ and $r=n\mod m$ (so that 0\le r), set $n_1=\cdots=n_r=k+1$, $n_{r+1}=\cdots=n_m=k$.