Problem 1242

I. solution. Based on the conditions,

$$0=c \cdot 1-b^{2}=\left(\sum_{i=1}^{n} i^{2} a_{i}\right)\left(\sum_{i=1}^{n} a_{i}\right)-\left(\sum_{i=1}^{n} i a_{i}\right)^{2}$$

from which, after performing the operations, we get

$$0=\sum_{i=1}^{n-1} \sum_{j=i+1}^{n}\left(i^{2}-2 i j+j^{2}\right) a_{i} a_{j}$$

On the right side, non-negative terms stand, so their sum can only be zero if each term is individually zero. This means that for every pair $1 \leqq i \leqq j \leqq n$, $a_{i} a_{j}=0$, which implies that at least $n-1$ of the numbers $a_{1}, a_{2}, \ldots, a_{n}$ are zero. The $n$-th value, based on $\sum_{i=1}^{n} a_{i}=1$, is 1, let this be $a_{k}$. Then $b=\sum_{i=1}^{n} i a_{i}=k \cdot a_{k}=k$, an integer. This proves the statement of the problem.

II. solution. If we place masses of $a_{1}, a_{2}, \ldots, a_{n}$ units at the points $1,2, \ldots, n$ on the number line, the center of mass of the system will be at $b$. The extent to which the mass is spread around its center of mass is generally measured by

$$d=\sum_{i=1}^{n}(i-b)^{2} a_{i}$$

This quantity is non-negative by definition and can only be 0 if each term in the above sum is 0. Since the sum of the $a_{i}$'s is 1, not all of them can be 0. However, among the factors $(i-b)^{2}$, at most one can be 0, and this only if $b$ is an integer and $i$ is equal to it. Therefore, if $d=0$, then $b$ is an integer, so it is sufficient to prove that $d=0$. If we perform the squaring, we get

$$d=\sum_{i=1}^{n} i^{2} a_{i}-2 b \sum_{i=1}^{n} i a_{i}+b^{2} \sum_{i=1}^{n} a_{i}=c-b^{2}$$

if therefore $c=b^{2}$, then the value of $d$ is indeed 0.

Remark. Since $d \geqq 0$, it also follows from our solution that $c \geqq b^{2}$ always holds, that is,

$$\sum_{i=1}^{n} i^{2} a_{i} \geqq\left(\sum_{i=1}^{n} i a_{i}\right)^{2}$$

Similarly, it can generally be shown that if $a_{i} \geqq 0, \sum_{i=1}^{n} a_{i}=1$, and $x_{1}, \ldots, x_{n}$ are arbitrary, then

$$\sum_{i=1}^{n} x_{i}^{2} a_{i} \geqq\left(\sum_{i=1}^{n} x_{i} a_{i}\right)^{2}$$