I recommend drawing the real line to understand this solution. Choose $ a_i = i/4$. Put the numbers of form $ a_i - k_i$ in increasing order and call them $ b_1, b_2, b_3$ and $ b_4$. We have $ b_4 \ge b_3 + 0.25 \ge b_2 + 0.5 \ge b_1 + 0.75$, because the fractionary parts of the $ b_i$'s are distinct and are multiples of $ 0.25$. Therefore, $ a \ge (b_1 - b_2)^2 + (b_2 - b_3)^2 + (b_3 - b_4)^2 + (b_1 - b_3)^2 + (b_2 - b_4)^2 + (b_1 - b_4)^2 \ge 3.(0.25)^2 + 2.(0.5)^2 + 0.75^2 = 1.25$ Now we can prove $ a = 1.25$ satisfies the condition (and now I strongly recommend the real line). We may assume w.l.o.g. $ 0 \le a_1 \le a_2 \le a_3 \le a_4 \le 1$ (the idea here and forward is that the $ k_i$'s allow us to consider any number of the form $ a_i + t$ (if $ t \in \mathbb{Z}$) instead of $ a_i$). Now we consider the differences $ a_2 - a_1, a_3 - a_2, a_4 - a_3$ and $ a_1 + 1 - a_4$ (observe that $ a_1 + 1$ is another "representation" of $ a_1$). Let $ M$ be the greatest of such differences. I will assume $ M = a_3 - a_2$ to avoid notation. We take $ k_1 = k_2 = - 1$ and $ k_3 = k_4 = 0$, so that $ x_1 = a_1 + 1$, $ x_2 = a_2 + 1$, $ x_3 = a_3$ and $ x_4 = a_4$. Back to the $ b_i$'s, we have $ b_1 = a_3$, $ b_2 = a_4$, $ b_3 = a_1 + 1$ and $ b_4 = a_2 + 1$. Now we define $ x = b_2 - b_1$, $ y = b_3 - b_2$ and $ z = b_4 - b_3$. It is easy to see that $ x + y + z + M = 1$. Moreover, $ M = \text{max}\{M, x, y, z\}$ by definition. We want to prove, given these two constraints (and the restriction $ x, y, z, M \ge 0$), that $ 1.25 \ge \sum_{1 \le i < j \le 4} (b_i - b_j)^2 = x^2 + y^2 + z^2 + (x + y)^2 + (y + z)^2 + (x + y + z)^2$ The worst case is $ y = \text{max}\{x, y, z\}$. Now, since $ f(t) = t^2$ is a convex function, keeping $ x + y$ fixed and increasing $ y$, the sum above rises. This also holds for $ y$ and $ z$. On the other hand, keeping $ x + z$ fixed and increasing $ |x - z|$, the sum also rises. So we just need to test some cases ($ 1 \ge M \ge 1/2$, $ 1/2 > M \ge 1/3$ and $ 1/3 > M \ge 1/4$), each one leading to some variables being equals to $ 0$ or $ M$. It is boring, but it works. CommentMy first idea was choosing $ M$ equals to a difference $ a_{i + 1} - a_{i} \ge 1/4$ (not necessarily the greatest). But $ M = 1/4$, $ y = 3/4$ and $ x = z = 0$ was a counter example to $ 1,25 \ge$ ugly sum. The problem was having a big $ b$. But we could avoid this interchanging $ b$ and $ M$. Or more directly: choose a maximal $ M$. Another thing: sometimes I am scared by the possibility that nobody reads my texts. I'd be happy with a private message or a reply saying "don't worry, you're not talking the the ether".