Reals with absolute sum 1, permutation smaller than n+1/2

Valentin Vornicu

Let $x_1$ , $x_2$ , $\ldots$ , $x_n$ be real numbers satisfying the conditions:
$\left\{\begin{array}{cccc} |x_1 + x_2 + \cdots + x_n | & = & 1 & \ \\ |x_i| & \leq & \displaystyle \frac ...$
Show that there exists a permutation $y_1$ , $y_2$ , $\ldots$ , $y_n$ of $x_1$ , $x_2$ , $\ldots$ , $x_n$ such that
$| y_1 + 2 y_2 + \cdots + n y_n | \leq \frac {n + 1}{2}.$

**Posted:** Thu Oct 27, 2005 1:37 pm

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We all use math everyday: to forecast weather, to tell time, to handle money; we also use math to analyze crime, reveal patterns, predict behavior. Using numbers we can solve the biggest mysteries we know.

Peter

$|(y_1+...+ny_n)+(ny_1+...+y_n)|=(n+1)(1)=n+1$ --> not to fulfil our conditon, $|y_1+...+ny_n|$ and $|ny_1+...+y_n|$ must differ by more $n+1$ .

Now if we switch 2 neighbouring elements, then $|y_1+...+ny_n|$ changes by $k.(y_k-y_{k-1})+(k-1)(y_{k-1}-y_k) = 1(y_{k}-y_{k-1})$ , which is at most $n+1$ .

And this is a contradiction, since there would be no way to cross the gap, but both ends can be attained by switching.

The idea in this problem helps a lot in finding the solution.

**Posted:** Fri Oct 28, 2005 1:21 am

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Boo!

sexynsmartjenny

I did the hard way by adding all possible permutations together to get their average, which is $(n+1)/2$ . And since they are all equal, or at least one of the permutations is greater than average, which means there exist a permutation less than average. Am I right?

**Posted:** Tue Jul 01, 2008 12:13 pm

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Ulanbek_Kyzylorda KTL · Poincare Conjecture Offline Joined: 24 Jan 2009 Posts: 134

who has another solution?

**Posted:** Wed Nov 04, 2009 11:43 pm