OC633 from Crux Mathematicorum Vol. 49, No. 5

OC633 (from Crux Mathematicorum Vol. 49, No. 5)

Let \(n \in N\), \(n \ge 2\). Prove that for all complex numbers \(a_1, a_2, .\ldots , a_n\) and \(b_1, b_2, \ldots , b_n\) the following statements are equivalent:

1. \(\sum^{n}_{k=1} |z - a_k|^2 \le \sum^{n}_{k=1} |z - b_k|^2\) for all \(z \in C\);

2. \(\sum^{n}_{k=1} a_k = \sum^{n}_{k=1} b_k\) and \(\sum^{n}_{k=1} |a_k|^2 \le \sum^{n}_{k=1} |b_k|^2\).

Proof

Let \(d\) denote the complex number \(\sum^{n}_{k=1} (b_k-a_k)\). Starting from the statement (a), we have the following:

\[\sum^{n}_{k=1} |z - a_k|^2 \le \sum^{n}_{k=1} |z - b_k|^2\] \[ \iff \sum^{n}_{k=1} |a_k|^2 - (z \sum^{n}_{k=1}\overline{a_k} + \overline{z} \sum^{n}_{k=1} a_k) \le \sum^{n}_{k=1} |b_k|^2 - (z \sum^{n}_{k=1} \overline{b_k} + \overline{z}\sum^{n}_{k=1}b_k)\] \[ \iff z \sum^{n}_{k=1} (\overline{b_k} - \overline{a_k}) + \overline{z} \sum^{n}_{k=1} (b_k-a_k) \le \sum^{n}_{k=1} |b_k|^2 - \sum^{n}_{k=1} |a_k|^2 \] \[ \iff z \overline{d} + \overline{z} d \le \sum^{n}_{k=1} |b_k|^2 - \sum^{n}_{k=1} |a_k|^2 \] \[ \iff 2Re(z \overline{d}) \le \sum^{n}_{k=1} |b_k|^2 - \sum^{n}_{k=1} |a_k|^2. \] Since both \(d\) and \(\sum^{n}_{k=1} |b_k|^2 - \sum^{n}_{k=1} |a_k|^2\) are bounded constants, the only possibility for the above inequalities being true for all \(z \in C\) is that \(d=0\) and \(\sum^{n}_{k=1} |b_k|^2 - \sum^{n}_{k=1} |a_k|^2 \ge 0\) which is exactly statement (b).

It is also obvious that if \(\sum^{n}_{k=1} a_k = \sum^{n}_{k=1} b_k\) and \(\sum^{n}_{k=1} |a_k|^2 \le \sum^{n}_{k=1} |b_k|^2\), then \(\sum^{n}_{k=1} |z - a_k|^2 \le \sum^{n}_{k=1} |z - b_k|^2\) for all \(z \in C\) by working backwards.