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We let $n\geq 3$.

How we can find $n$ positive real numbers $x_1, x_2, \ldots x_n$, such that \begin{align}x_1&>\displaystyle\sum_{i\neq 1} x_i, \\ x_2&>\displaystyle\sum_{i\neq 2} x_i,\\ &\vdots \\ x_n&>\displaystyle\sum_{i\neq n} x_i. \end{align}

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  • $\begingroup$ It should be fairly easy to find out that you can't. $\endgroup$ – T. Bongers Dec 19 '17 at 20:29
  • $\begingroup$ Any thoughts? How about taking a crack at the case $n=3$? $\endgroup$ – lulu Dec 19 '17 at 20:30
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Say $n=3$ and your numbers are $a,b,c$. Then, $a > b+c$ hence $c-a > b$ and $$c - a > b > a + c,$$ which implies $c-a > a+c$, or equivalently, $a < 0$, which contradicts the requirement $a,b,c>0$.

This is impossible.

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Guide:

$$x_j > \sum_{i \neq j} x_i, j=1, \ldots, n$$

Sum over all $j$ and see what you get.

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Let note $\displaystyle \bar x=\frac 1n\sum\limits_{i=1}^n x_i\quad$ the average of all values.

Your condition is $x_i > n\bar x -x_i\iff 2x_i > n\bar x\iff x_i>\frac n2 \bar x$

Since $n\ge 3$ then $\frac n2> 1$ and your condition becomes $$\forall i=1..n:\ x_i>\bar x$$

You are requesting that all elements should be strictly above the average, this is impossible !

By definition of the average there are elements below it and elements above it, or at the worst all equal, but not all can be strictly greater that the average.

Just sum up the inequality over all $x_i$ and this gives $n\bar x>n\bar x$ which is an obvious contradiction.

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