Circular arrangement of numbers with constraint on 3 consecutive triples of them gives a unique solution

Question

Around a circle write 1000 numbers, such that each three consecutive numbers A, B, C (B is between A and C) satisfy $A^2+C^2 \leq B-(1/8)$ . Find the maximum and the minimum value for the sum of the numbers around the circle. The answer is 250.

Answer 1

Let us denote $x_k \ (k=1,2,...1000)$ these numbers ; the constraint being written :

$$\text{for all } \ k, \ \ \ \ \ x_{k-1}^2+x_{k+1}^2 \leq x_{k}-\dfrac18 \tag{1}$$

(conventionally, $x_{1001}=x_1$ and $x_{1002}=x_2$.)

1) If one takes $x_k = \dfrac14$ for all $k$, (1) is fullfilled (with an equal sign), giving indeed $\sum x_k=250$.

2) We are now going to establish that the solution above is the only solution. Let us take it as a "point of departure" and set :

$$x_k=\dfrac14+y_k\tag{2}$$

and show that all the $y_k$ are zero. Constraint (1) becomes :

$$\left(\dfrac14+y_{k-1}\right)^2+\left(\dfrac14+y_{k+1}\right)^2 \leq \dfrac14+y_{k}-\dfrac18$$

i.e., after cancellations :

$$y_{k-1}^2+y_{k+1}^2 + \dfrac12 \left(y_{k-1}+y_{k+1}\right) \leq y_{k}\tag{3}$$

Let $S=\sum y_k$ and $T=\sum y_k^2$.

Adding up the 1000 inequalities (3) gives :

$$2T+\dfrac12 \left(S+S\right)\leq S \tag{4}$$

(4) boils down to $T\le 0$ which is possible if and only if all $y_k$ are zero, our objective !

Conclusion : $\sum x_k$ is constant ; therefore, its min and its max are both equal to $250$.