Problem 1766

To prove the inequality
$$\sum_{i=1}^n a_i^2 - \sum_{i=1}^n a_i a_{i+1} \le \left\lfloor \frac{n}{2} \right\rfloor (M - m)^2,$$
where $a_{n+1} = a_1$, $M = \max_{1 \le i \le n} a_i$, and $m = \min_{1 \le i \le n} a_i$, we will use mathematical induction.

1. Base Case:
- For $n = 1$:
$$\sum_{i=1}^1 a_i^2 - \sum_{i=1}^1 a_i a_{i+1} = a_1^2 - a_1 a_1 = 0.$$
Since $\left\lfloor \frac{1}{2} \right\rfloor = 0$, the inequality holds as $0 \le 0$.

- For $n = 2$:
$$\sum_{i=1}^2 a_i^2 - \sum_{i=1}^2 a_i a_{i+1} = a_1^2 + a_2^2 - (a_1 a_2 + a_2 a_1) = a_1^2 + a_2^2 - 2a_1 a_2 = (a_1 - a_2)^2.$$
Since $\left\lfloor \frac{2}{2} \right\rfloor = 1$, the inequality holds as $(a_1 - a_2)^2 \le (M - m)^2$.

2. Inductive Step:
- Assume the statement is true for $n = 2m - 1$, i.e.,
$$\sum_{i=1}^{2m-1} a_i^2 - \sum_{i=1}^{2m-1} a_i a_{i+1} \le (m-1)(M - m)^2.$$
- We need to show it holds for $n = 2m$:
$$\sum_{i=1}^{2m} a_i^2 - \sum_{i=1}^{2m} a_i a_{i+1}.$$
Consider the expression:
$$\sum_{i=1}^{2m} a_i^2 - \sum_{i=1}^{2m} a_i a_{i+1} = \sum_{i=1}^{2m-1} a_i^2 + a_{2m}^2 - \left( \sum_{i=1}^{2m-1} a_i a_{i+1} + a_{2m} a_1 \right).$$
This can be rewritten as:
$$\left( \sum_{i=1}^{2m-1} a_i^2 - \sum_{i=1}^{2m-1} a_i a_{i+1} \right) + a_{2m}^2 - a_{2m} a_1.$$
By the inductive hypothesis:
$$\sum_{i=1}^{2m-1} a_i^2 - \sum_{i=1}^{2m-1} a_i a_{i+1} \le (m-1)(M - m)^2.$$
Therefore:
$$\sum_{i=1}^{2m} a_i^2 - \sum_{i=1}^{2m} a_i a_{i+1} \le (m-1)(M - m)^2 + a_{2m}^2 - a_{2m} a_1.$$
Since $a_{2m} \in [m, M]$, we have:
$$a_{2m}^2 - a_{2m} a_1 \le (M - m)^2.$$
Thus:
$$\sum_{i=1}^{2m} a_i^2 - \sum_{i=1}^{2m} a_i a_{i+1} \le (m-1)(M - m)^2 + (M - m)^2 = m(M - m)^2.$$
Hence, the statement holds for $n = 2m$.

3. **Inductive Step for Odd $n$:**
- Assume the statement is true for $n = 2m$, i.e.,
$$\sum_{i=1}^{2m} a_i^2 - \sum_{i=1}^{2m} a_i a_{i+1} \le m(M - m)^2.$$
- We need to show it holds for $n = 2m + 1$:
$$\sum_{i=1}^{2m+1} a_i^2 - \sum_{i=1}^{2m+1} a_i a_{i+1}.$$
Consider the expression:
$$\sum_{i=1}^{2m+1} a_i^2 - \sum_{i=1}^{2m+1} a_i a_{i+1} = \sum_{i=1}^{2m} a_i^2 + a_{2m+1}^2 - \left( \sum_{i=1}^{2m} a_i a_{i+1} + a_{2m+1} a_1 \right).$$
This can be rewritten as:
$$\left( \sum_{i=1}^{2m} a_i^2 - \sum_{i=1}^{2m} a_i a_{i+1} \right) + a_{2m+1}^2 - a_{2m+1} a_1.$$
By the inductive hypothesis:
$$\sum_{i=1}^{2m} a_i^2 - \sum_{i=1}^{2m} a_i a_{i+1} \le m(M - m)^2.$$
Therefore:
$$\sum_{i=1}^{2m+1} a_i^2 - \sum_{i=1}^{2m+1} a_i a_{i+1} \le m(M - m)^2 + a_{2m+1}^2 - a_{2m+1} a_1.$$
Since $a_{2m+1} \in [m, M]$, we have:
$$a_{2m+1}^2 - a_{2m+1} a_1 \le (M - m)^2.$$
Thus:
$$\sum_{i=1}^{2m+1} a_i^2 - \sum_{i=1}^{2m+1} a_i a_{i+1} \le m(M - m)^2 + (M - m)^2 = (m+1)(M - m)^2.$$
Hence, the statement holds for $n = 2m + 1$.

By mathematical induction, the statement is true for all $n \ge 1$.

$\blacksquare$