China National Team Selection Test

The largest possible value of $\lambda(n)$ is $\frac{n(n+1)^2}{4}$.

Let $a_1 = a_2 = \dots = a_n = 1$. Then we have $\lambda(n) \le \frac{n(n+1)^2}{4}$.

We shall show that for any real numbers $a_0, a_1, a_2, \dots, a_n$ satisfying the indicated property in the problem, the following inequality holds:
$$\left(\sum_{i=1}^{n} ia_i\right)^2 \geqslant \frac{n(n+1)^{2}}{4}\left(\sum_{i=1}^{n} a_{i}^{2}\right). \qquad \textcircled{1}$$
First, we notice that
$$a_1 \ge \frac{a_2}{2} \ge \dots \ge \frac{a_n}{n}.$$
Indeed, by assumption, $2ia_i \ge i(a_{i+1} + a_{i-1})$ holds for $i = 1, 2, \dots, n-1$. For any given positive integer $1 \le l \le n-1$, summing the above inequality over $i = 1, 2, \dots, l$, we have $(l+1)a_l \ge la_{l+1}$, i.e.
$$\frac{a_l}{l} \ge \frac{a_{l+1}}{l+1} \text{ for } l = 1, 2, \dots, n-1.$$
In what follows, we show that for any $i, j, k \in \{1, 2, \dots, n\}$, if $i > j$, then
$$\frac{2ik^2}{i+k} > \frac{2jk^2}{j+k}.$$
Indeed, the above inequality is equivalent to $2ik^2(j+k) > 2jk^2(i+k)$, i.e. $(i-j)k^3 > 0$, which is clearly true.

Now, we are going to show inequality ①. We shall start by estimating the lower bound of $a_i a_j$ for $1 \le i < j \le n$.
By previous results, we have $\frac{a_i}{i} \ge \frac{a_j}{j}$, i.e. $ja_i - ia_j \ge 0$. Since $a_i - a_j \le 0$, we have $(ja_i - ia_j)(a_j - a_i) \ge 0$, i.e. $a_i a_j \ge \frac{i}{i+j}a_j^2 + \frac{j}{i+j}a_i^2$.
Thus, we have
$$\begin{aligned} \left(\sum_{i=1}^{n} ia_i\right)^2 &= \sum_{i=1}^{n} i^2 a_i^2 + 2 \sum_{1 \le i < j \le n} ija_i a_j \\ &\ge \sum_{i=1}^{n} i^2 \times a_i^2 + 2 \sum_{1 \le i < j \le n} \left( \frac{i^2 j}{i+j} a_j^2 + \frac{ij^2}{i+j} a_i^2 \right) \\ &= \sum_{i=1}^{n} \left( a_i^2 \times \sum_{k=1}^{n} \frac{2ik^2}{i+k} \right). \end{aligned}$$
Let $b_i = \sum_{k=1}^{n} \frac{2ik^2}{i+k}$. We see from previous results that $b_1 \le b_2 \le \dots \le b_n$.
Since $a_1^2 \le a_2^2 \le \dots \le a_n^2$, by the Chebyshev inequality, we have
$$\sum_{i=1}^{n} a_i^2 b_i \ge \frac{1}{n} \left( \sum_{i=1}^{n} a_i^2 \right) \left( \sum_{i=1}^{n} b_i \right).$$
Hence
$$\left(\sum_{i=1}^{n} ia_i\right)^2 \ge \frac{1}{n} \left(\sum_{i=1}^{n} a_i^2\right) \left(\sum_{i=1}^{n} b_i\right).$$
Since
$$\begin{align*} \sum_{i=1}^{n} b_i &= \sum_{i=1}^{n} \sum_{k=1}^{n} \frac{2ik^2}{i+k} = \sum_{i=1}^{n} i^2 + 2 \sum_{1 \le i < j \le n} \left( \frac{i^2 j}{i+j} + \frac{ij^2}{i+j} \right) \\ &= \sum_{i=1}^{n} i^2 + 2 \sum_{1 \le i < j \le n} ij = \left( \sum_{i=1}^{n} i \right)^2 = \frac{n^2(n+1)^2}{4}, \end{align*}$$
we find that $(\sum_{i=1}^{n} ia_i)^2 \ge \frac{n(n+1)^2}{4} \sum_{i=1}^{n} a_i^2$, which proves inequality ①.

We conclude that the maximum possible value of $\lambda(n)$ is
$$\frac{n(n+1)^2}{4}$$