MathLabs

Problem 3

Let x1,x2,…,xn x_1,x_2,\ldots,x_n be real numbers satisfying ∣x1+x2+⋯+xn∣=1|x_1+x_2+\cdots+x_n|=1 and ∣xi∣≤(n+1)/2|x_i|\le(n+1)/2 for every i i . Show that there is a permutation y1,…,yn y_1,\ldots,y_n of the xi x_i such that ∣y1+2y2+⋯+nyn∣≤(n+1)/2|y_1+2y_2+\cdots+ny_n|\le(n+1)/2.
Step 2 of 4: Compare an ordering with its reverse
In plain words

Compare an ordering with its reverse

W+Wrev=n+1W+W^{\rm rev}=n+1
Detailed analysis

For any ordering define W=y1+2y2+⋯+nyn W=y_1+2y_2+\cdots+ny_n . Its reverse has weighted sum Wrev W^{\rm rev}. Pairing positions gives W+Wrev=(n+1)(x1+⋯+xn)=n+1 W+W^{\rm rev}=(n+1)(x_1+\cdots+x_n)=n+1. Hence either one is already in [−(n+1)/2,(n+1)/2][-(n+1)/2,(n+1)/2], or one is above and the other below this interval.