China National Team Selection Test

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

Let $a_1=a_2=\cdots =a_n=1$. 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: $${\left(\sum _{i=1}^{n}i{a}_i\right)}^2⩾\frac{n(n+1{)}^2}{4}\left(\sum _{i=1}^n{a}_i^2\right).\stackrel{◯}{1}$$ First, we notice that $$a_1\ge \frac{a_2}{2}\ge \cdots \ge \frac{a_n}{n}.$$ Indeed, by assumption, $2i{a}_i\ge i(a_{i+1}+a_{i-1})$ 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 $(l+1)a_l\ge l{a}_{l+1}$, 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 $$\frac{2i{k}^2}{i+k}>\frac{2j{k}^2}{j+k}.$$ Indeed, the above inequality is equivalent to $2i{k}^2(j+k)>2j{k}^2(i+k)$, 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. $j{a}_i-i{a}_j\ge 0$. Since $a_i-a_j\le 0$, we have $(j{a}_i-i{a}_j)(a_j-a_i)\ge 0$, i.e. $a_i{a}_j\ge \frac{i}{i+j}{a}_j^2+\frac{j}{i+j}{a}_i^2$. Thus, we have $$\begin{array}{rl}{\left(\sum _{i=1}^{n}i{a}_i\right)}^2& =\sum _{i=1}^n{i}^2{a}_i^2+2\sum _{1\le i<j\le n}ij{a}_i{a}_j\\ & \ge \sum _{i=1}^n{i}^2\times a_i^2+2\sum _{1\le i<j\le n}\left(\frac{i^{2}j}{i+j}{a}_j^2+\frac{i{j}^2}{i+j}{a}_i^2\right)\\ & =\sum _{i=1}^n\left(a_i^2\times \sum _{k=1}^n\frac{2i{k}^2}{i+k}\right).\end{array}$$ Let $b_i={\sum }_{k=1}^n\frac{2i{k}^2}{i+k}$. We see from previous results that $b_1\le b_2\le \cdots \le b_n$. Since $a_1^2\le a_2^2\le \cdots \le a_n^2$, 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 $${\left(\sum _{i=1}^{n}i{a}_i\right)}^2\ge \frac{1}{n}\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i\right).$$ Since $$\begin{array}{rl}\sum _{i=1}^n{b}_i& =\sum _{i=1}^n\sum _{k=1}^n\frac{2i{k}^2}{i+k}=\sum _{i=1}^n{i}^2+2\sum _{1\le i<j\le n}\left(\frac{i^{2}j}{i+j}+\frac{i{j}^2}{i+j}\right)\\ & =\sum _{i=1}^n{i}^2+2\sum _{1\le i<j\le n}ij={\left(\sum _{i=1}^{n}i\right)}^2=\frac{n^2(n+1{)}^2}{4},\end{array}$$ we find that $({\sum }_{i=1}^{n}i{a}_i{)}^2\ge \frac{n(n+1{)}^2}{4}{\sum }_{i=1}^n{a}_i^2$, which proves inequality ①.

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