Problem 1693

To solve the problem, we need to find the largest constant $c$ such that for every concave sequence of non-negative real numbers $a_0, a_1, \ldots, a_{1389}$, the following inequality holds:
$$\sum_{i=0}^{1389} i a_i^2 \geq c \sum_{i=0}^{1389} a_i^2.$$

1. Lemma: If $a, b, c \geq d > 0$ and $2c \geq a + b$, then $2\sqrt{c^2 - d^2} \geq \sqrt{a^2 - d^2} + \sqrt{b^2 - d^2}$.

Proof: Since the inequality is homogeneous, we can set $d = 1$ without loss of generality. We need to prove:
$$2\sqrt{\left(\frac{a+b}{2}\right)^2 - 1} \geq \sqrt{a^2 - 1} + \sqrt{b^2 - 1}.$$
Squaring both sides, we get:
$$4\left(\left(\frac{a+b}{2}\right)^2 - 1\right) \geq (\sqrt{a^2 - 1} + \sqrt{b^2 - 1})^2.$$
Simplifying, we need to show:
$$(a+b)^2 - 4 \geq a^2 - 1 + b^2 - 1 + 2\sqrt{(a^2 - 1)(b^2 - 1)}.$$
This reduces to:
$$ab - 1 \geq \sqrt{(a^2 - 1)(b^2 - 1)}.$$
This can be seen due to Aczel's inequality. Note that $a, b > 1$.

2. Sum Identities:
$$\sum_{i=0}^{n} i = \frac{n(n+1)}{2}, \quad \sum_{i=0}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{i=0}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2.$$

3. Main Problem: We will prove the problem for $c = c_n = \frac{n(n-1)}{2(2n+1)}$. The equality case is $a_i = n - i$.

Define:
$$S = \sum_{i=0}^{n} i a_i^2 - c_n \sum_{i=0}^{n} a_i^2.$$
If we rearrange the $a_i$'s such that $a_0 \geq a_1 \geq \cdots \geq a_n$, the RHS remains the same, but the LHS decreases. Thus, we can assume $a_0 \geq a_1 \geq \cdots \geq a_n$.

4. Transformation: Set $a_i \rightarrow \sqrt{a_i^2 - a_n^2}$. It is clear that:
$$S \rightarrow S - \left(\frac{n(n+1)}{2} - c\right) a_n^2 \leq S.$$
By the lemma, the new $a_i$'s are still concave. Thus, we can assume $a_n = 0$.

5. Rearrangement: Set $x_i = a_{n-i}$. We need to prove:
$$T = d_n \sum_{i=1}^{n} x_i^2 - \sum_{i=1}^{n} i x_i^2 \geq 0,$$
where $d_n = \frac{3n(n+1)}{2(2n+1)}$. Note that the sequence $x_i$ is still concave and $2x_1 \geq x_2$.

6. Monotonicity: We prove that $\frac{x_i}{i} \geq \frac{x_{i+1}}{i+1}$. Let $k$ be the smallest number such that this is not true. Then:
$$2kx_k \geq k(x_{k+1} + x_{k-1}) \geq (k+1)x_k + kx_{k-1} \Rightarrow \frac{x_{k-1}}{k-1} \geq \frac{x_k}{k},$$
which is a contradiction. Call such a sequence "nice".

7. Induction: We will prove $T \geq 0$ for all "nice" sequences $x_i$ by induction on $n$. The base case is clear.

8. Splitting: Let $k$ be the smallest number such that $x_k < kx_1$. Define:
$$A = d_n \sum_{i=1}^{k-1} x_i^2 - \sum_{i=1}^{k-1} i x_i^2, \quad B = d_n \sum_{i=k}^{n} x_i^2 - \sum_{i=k}^{n} i x_i^2.$$
Since $d_n > d_{n-1}$, it is clear that $A \geq 0$. If $B \geq 0$, then $T = A + B \geq 0$ and we are done. Otherwise, let $B < 0$.

9. Scaling: Set $x_k, \ldots, x_n \rightarrow a x_k, \ldots, a x_n$ where $a = \frac{k}{x_k} > 1$. We can see that:
$$T' = A + a^2 B < A + B = T.$$
The sequence being nice, one can easily conclude the new sequence defined as $x_1, \ldots, x_{k-1}, a x_k, \ldots a x_n$ is nice as well.

10. Conclusion: Either the process stops when $B \geq 0$, proving $T \geq 0$, or we reach the point where $a_i = i a_1$. At this point, using the sum identities, we get $T = 0$, so the initial $T$ was also nonnegative. Thus, we are done.

The final answer is $\boxed{\frac{1389 \cdot 1388}{2 \cdot (2 \cdot 1389 + 1)}}$.