Baltic Way 1997 · Problem 4

Solution:

Denote $y_{i}=x_{i}-a$. Then $y_{1}+y_{2}+\cdots+y_{n}=0$. We can assume $y_{1} \leqslant y_{2} \leqslant \cdots \leqslant y_{k} \leqslant 0 \leqslant y_{k+1} \leqslant \cdots \leqslant y_{n}$. Let $y_{1}+y_{2}+\cdots+y_{k}=-z$, then $y_{k+1}+\cdots+y_{n}=z$ and

$$\begin{aligned} y_{1}^{2}+y_{2}^{2}+\cdots+y_{n}^{2} & =y_{1}^{2}+y_{2}^{2}+\cdots+y_{k}^{2}+y_{k+1}^{2}+\cdots+y_{n}^{2} \leqslant \\ & \leqslant\left(y_{1}+y_{2}+\cdots+y_{k}\right)^{2}+\left(y_{k+1}+\cdots+y_{n}\right)^{2}=2 z^{2}= \\ & =\frac{1}{2}(2 z)^{2}=\frac{1}{2}\left(\left|y_{1}\right|+\left|y_{2}\right|+\cdots+\left|y_{n}\right|\right)^{2} . \end{aligned}$$

Alternative solution. The case $n=1$ is trivial (then $x_{1}-a=0$ and we get the inequality $0 \leqslant 0$ ). Suppose now that $n \geqslant 2$. Consider a square of side length $\left|x_{1}-a\right|+\left|x_{2}-a\right|+\ldots+\left|x_{n}-a\right|$ and construct squares of side lengths $\left|x_{1}-a\right|,\left|x_{2}-a\right|, \ldots,\left|x_{n}-a\right|$ side by side inside it as shown on Figure 1. Since none of the side lengths of the small squares exceeds half of the side length of the large square, then all the small squares are contained within the upper half of the large square, i.e. the sum of their areas does not exceed half of the area of the large square, q.e.d.

Figure 1