### 2015 IMO SL #A1

Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies$a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}$for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.

We shall use induction on $n$. Taking $k=1$ gives $a_1a_2 \ge 1$. Therefore, by AM-GM, we have $a_1+a_2\ge 2$, which establishes the $n=2$ case. Rewrite the given condition as$$\frac{k}{a_{k+1}}\le a_k+\frac{k-1}{a_k}.$$Summing this from $k=1$ to $k=n$, we see$$a_1+a_2+\cdots +a_n\ge \frac{n}{a_{n+1}} \implies a_{n+1}=\frac{n}{a_1+a_2+\cdots+a_n}.$$Hence,$$a_1+a_2+\cdots +a_{n+1}\ge a_1+a_2+\cdots+a_n+\frac{n}{a_1+a_2+\cdots+a_n} \ge n+1,$$where the last step follows from the inductive step $a_1+a_2+\cdots +a_n\ge n$. $\square$