AustriaMO2013

The assertion can be rewritten as $$\sum _{k=1}^n\sum _{j=1}^k{a}_j\le \sum _{k=1}^n{k}^3.$$ It therefore suffices to prove $$\sum _{j=1}^k{a}_j\le k^3(6)$$ for every $k=1,\dots ,n$. In order to prove (6) we fix $k$ and set $$x_i=\frac{1}{k}-\frac{i}{N}\text{for }i=1,\dots ,k$$ and $$x_i=\frac{n+1-i}{N^2}\text{for }i=k+1,\dots ,n$$ for some integer $N>0$ to be determined as follows: The conditions $x_1>x_2>\cdots >x_n>0$ are certainly fulfilled in the case $k=n$ and for $k<n$ the only non-trivial relation is $x_k>x_{k+1}$ that is $\frac{1}{k}-\frac{k}{N}>\frac{n-k}{N^2}$. Hence we choose $N>kn$, in order to have $\frac{1}{k}-\frac{k}{N}>\frac{n-k}{N}\ge \frac{n-k}{N^2}$. The condition $x_1+\cdots +x_n<1$ means $1-\frac{k(k+1)}{N}+\frac{(n-k)(n-k+1)}{N^2}<1$ and will be satisfied for $$N>\frac{(n-k)(n-k+1)}{k(k+1)}$$ For $N>\max \{kn,\frac{(n-k)(n-k+1)}{k(k+1)}\}$ the numbers $x_1,\dots ,x_n$ fulfill the relevant conditions and we conclude $$\sum _{i=1}^n{a}_i{x}_i^3<1.$$ By taking the limit $N\to \infty$ we get ${\sum }_{i=1}^n{a}_i{\lim }_{N\to \infty }{x}_i^3\le 1$, which gives ${\sum }_{i=1}^k{a}_i\frac{1}{k^3}\le 1$, hence the desired estimate (6). $\square$