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Here is another tricky question: Suppose $m_1,\cdots,m_j$ and $m'_1,\cdots,m'_j$ are positive integers such that $\sum\limits_{k=1}^j m_k = \sum\limits_{k=1}^j m'_k$ and $\sum\limits_{k=1}^j m^2_k > \sum\limits_{k=1}^j m'^2_k$. What can we say about the sums of the reciprocals $\sum\limits_{k=1}^j \frac{1}{m_k}$ and $\sum\limits_{k=1}^j \frac{1}{m'_k}$. Which is larger than the other?

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The graphs of $f(x)=\sum\limits_{k=1}^j m_k^x$ and $g(x)=\sum\limits_{k=1}^j m'_k^x$ cross at $x=1$ and at $x=0$. –  GEdgar May 5 '11 at 15:07
    
I have a feeling that $\sum\limits_{k=1}^j \frac{1}{m'_k}< \sum\limits_{k=1}^j \frac{1}{m_k}$ –  pebox11 May 5 '11 at 15:18

2 Answers 2

up vote 4 down vote accepted

Try some examples. $1 + 3 = 2 + 2$ with $1^2 + 3^2 > 2^2 + 2^2$ and $1/1 + 1/3 > 1/2 + 1/2$. On the other hand, $2 + 5 + 12 = 9 + 9 + 1$ with $2^2 + 5^2 + 12^2 > 9^2 + 9^2 + 1^2$ and $1/2 + 1/5 + 1/12 < 1/9 + 1/9 + 1/1$.

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yes the counter-example works! –  pebox11 May 5 '11 at 16:44

This isn't a solution but rather an interesting point that I observed while thinking about your problem. The Arithmetic-Harmonic Mean combined with Jensen's Inequality with $f(x) = x^{2}$ and $\mu_{k} = \frac{1}{j}$ gives \begin{align} \left( \frac{j}{\sum_{k = 1}^{j} \frac{1}{m_k}} \right)^{2} \leq \left( \tfrac{1}{j} \sum_{k = 1}^{j} m_{k} \right)^{2} \leq \tfrac{1}{j} \sum_{k = 1}^{j} m_{k}^{2}. \end{align}

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