Iranian Mathematical Olympiad

First consider the concave sequence $a_i = (1389 - i)d$, $0 \le i \le 1389$, $d > 0$. Note that the maximum possible value of $c$ for this sequence is $c_0 = \frac{\sum_{i=0}^{1389} i a_i^2}{\sum_{i=0}^{1389} a_i^2} = \frac{1389^2 - 5 \times 1389}{4 \times 1389 - 2}$.

Now we claim that for every concave sequence $a_i$, we have $\sum_{i=0}^{1389} i a_i^2 \ge c_0 \sum_{i=0}^{1389} a_i^2$, and since we have the above equality, $c_0$ must be the answer.
Consider a concave sequence $a_0, a_1, \dots, a_{1389}$, we define
$$b_i = (1389 - i) \frac{a_{\lfloor c_0 \rfloor}}{1389 - \lfloor c_0 \rfloor}$$
Indeed we are trying to make the sequence linear. Now we must prove that:
$$\sum_{i=\lfloor c \rfloor}^{1389} (i-c)a_i^2 \ge \sum_{i=0}^{\lfloor c \rfloor} (c-i)a_i^2$$
Considering the definition of $b_i$ and its linearity, and the fact that $a_{\lfloor c_0 \rfloor} = b_{\lfloor c_0 \rfloor}$, $a_{1389} \ge b_{1389}$, easily it can be proved that if $i \ge \lfloor c_0 \rfloor$ then $a_i \ge b_i$, and if $i \le \lfloor c_0 \rfloor$ then $a_i \le b_i$. So we have:
$$\sum_{i=\lfloor c_0 \rfloor}^{1389} (i - c_0)a_i^2 \ge \sum_{i=\lfloor c_0 \rfloor}^{1389} (i - c_0)b_i^2 \ge \sum_{i=0}^{\lfloor c_0 \rfloor} (c_0 - i)b_i^2 \ge \sum_{i=0}^{\lfloor c_0 \rfloor} (c_0 - i)a_i^2$$
So the desired $c$ equals to $c_0 = \frac{1389^2 - 5 \times 1389}{4 \times 1389 - 2}$.