I'm assuming $n\ge 2$ since $n=1$ gives $1\le 0$ which is false.

arkanm wrote: I'm assuming $n\ge 2$ since $n=1$ gives $1\le 0$ which is false. Yes $n \geq 2$ because for $n=1$ we cannot have at the same time $a_1=0$ (the first condition) and $|a_1|=1$ the second one.

Hint :

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To prove $|a_k+a_{k+1}+\cdots +a_n|\le \frac{1}{2}\ \forall\ k$, notice that by the Triangle Inequality: $1=|a_1|+|a_2|+\cdots +|a_n|$ $\ge |a_1+a_2+\cdots +a_{k-1}|$ $+|a_k+a_{k+1}+\cdots +a_n|$ $=|a_k+a_{k+1}+\cdots +a_n|$ $+|a_k+a_{k+1}+\cdots +a_n|$ $=2|a_k+a_{k+1}+\cdots +a_n|$ $\Leftrightarrow 1\ge 2|a_k+a_{k+1}+\cdots +a_n|$ $\Leftrightarrow |a_k+a_{k+1}+\cdots +a_n|\le \frac{1}{2}$.

Extension. If $a_1=0$, $|a_k|=|a_{k-1}+1|$ for $k\in \{2,3,...,n\}$, minimize: $\frac{1}{n}\sum_{k=1}^n a_k$.

beautiful ! I'll think to your generalization later!!!!

Attachments:

arkanm wrote: Extension. If $a_1=0$, $|a_k|=|a_{k-1}+1|$ for $k\in \{2,3,...,n\}$, minimize: $\frac{1}{n}\sum_{k=1}^n a_k$. For $n$ even then from $a_{k-1}=\frac{a_k^2-a_{k-1}^2}{2}-\frac{1}{2}$ we get $\sum_{k=1}^n a_k=\frac{a_n^2}{2}-\frac{n-2}{2}+a_n=\frac{(a_n+1)^2}{2}-\frac{n-1}{2} \geq -\frac{n-1}{2}$ The equality occurs when $a_1=0, a_2=-1,a_3=0,…,a_n=-1$. For $n$ odd, $a_n$ must be an even number and $(a_n+1)^2$ is non negative and odd integer, thus $(a_n+1)^2 \geq 1$ with equality when $a_1=0,a_2=-1,…,a_{n-2}=0, a_{n-1}=1,a_n=-2$ Thus $\sum_{k=1}^n a_k \geq 1--\frac{n-1}{2}$ in this case.

why I am thought that this problem was set somewhere

One-liner \[2\left|\sum_{i=1}^n ia_i\right|=\left|\sum_{i=1}^n (i-1)a_i\right|+\left|\sum_{i=1}^n\sum_{j=i}^n a_j\right|=\left|\sum_{i=1}^n (i-1)a_i\right|+\left|\sum_{i=1}^n\sum_{j=1}^{i-1} a_j\right|\leq\sum_{i=1}^n(i-1)|a_i|+\sum_{i=1}^n\sum_{j=1}^{i-1}|a_j|=\sum_{i=1}^n ((n-i)+(i-1))|a_i|=\frac{n-1}{2}\]