IMC 1994 · Day 2 · P11

We define $\pi$ inductively. Set $\pi(1) = 1$. Assume $\pi$ is defined for $i = 1,2,\dots,n$ and also $$\tag{1} \biggl\| \sum_{i=1}^{n} x_{\pi(i)} \biggr\|^2 \le \sum_{i=1}^{n} \| x_{\pi(i)} \|^2.$$ Note (1) is true for $n = 1$. We choose $\pi(n+1)$ in a way that (1) is fulfilled with $n+1$ instead of $n$. Set $y = \sum\limits_{i=1}^{n} x_{\pi(i)}$ and $A = \{1,2,\dots,k\} \setminus \{ \pi(i) : i = 1,2,\dots,n \}$. Assume that $(y, x_r) > 0$ for all $r \in A$. Then $\left( y, \sum\limits_{r \in A} x_r \right) > 0$ and in view of $y + \sum\limits_{r \in A} x_r = 0$ one gets $-(y,y) > 0$, which is impossible. Therefore there is $r \in A$ such that $$\tag{2} (y, x_r) \le 0.$$ Put $\pi(n+1) = r$. Then using (2) and (1) we have $$\biggl\| \sum_{i=1}^{n+1} x_{\pi(i)} \biggr\|^2 = \| y + x_r \|^2 = \|y\|^2 + 2 (y, x_r) + \| x_r \|^2 \le \|y\|^2 + \| x_r \|^2 \le \sum_{i=1}^{n} \| x_{\pi(i)} \|^2 + \| x_r \|^2 = \sum_{i=1}^{n+1} \| x_{\pi(i)} \|^2,$$ which verifies (1) for $n+1$. Thus we define $\pi$ for every $n = 1,2,\dots,k$. Finally from (1) we get $$\biggl\| \sum_{i=1}^{n} x_{\pi(i)} \biggr\|^2 \le \sum_{i=1}^{n} \| x_{\pi(i)} \|^2 \le \sum_{i=1}^{k} \| x_i \|^2.$$