An algebra problem of olympiad Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies the relation:
 $$a_{k+1} \geq \frac{ka_k}{a_k^2 + (k-1)}$$
 for every positive integer $k$. Prove that:
 $$a_1+a_2+\cdots+a_n⩾n \text{ for } n⩾2.$$ 
This is an indian olympiad problem.Can you guys help me solve this out. 
edit:
I have tried it by first taking $k = 1$,
then we get $a_2 \geq \frac{1}{a_1}$. By putting $k=2$ we get
 $a_3 \geq \frac{2a_2}{a_2^2 + 1}$ and similarly, $a_4 \geq \frac{3a_3}{a_3^2 +2}$.
But I couldn't find any relation between them. 
 A: 
Lemma
Let $\{ a_n \} _ {n=1} ^{\infty} $ be as stated in the problem. For
all $n \geq 2$, the following inequality holds: $$ \sum_{1 \leq k < n}
 a_k \geq \frac{n-1}{a_n} $$

proof. Use induction on $n$. First of all, we know $a_2 \geq \frac{a_1}{a_1^2 + (1-1)}=\frac{1}{a_1}$. Thus $a_1 \geq \frac{1}{a_2}$. Now assume the claim holds for $n\geq 2$. Then $$a_1 + \cdots + a_n = (a_1 + \cdots + a_{n-1})+a_n \geq \frac{n-1}{a_n}+a_n = \frac{(n-1) + a_n^2}{a_n} \geq \frac{n}{a_{n+1}} $$ The last equality is because the sequence $\{ a_n \} _ {n=1} ^{\infty} $ satisfies the relation $a_{n+1} \geq \frac{na_n}{a_n^2 + (n-1)}$.

Let's prove $$a_1 + \cdots + a_n \geq n$$ by induction on $n$.
If $n=2$, we have $a_1 + a_2 \geq a_1 + \frac{1}{a_2} \geq 2$ by the AM-GM inequality.
Assume the inequality holds for some $n \geq 2$. If $a_{n+1} \geq 1$, it is immediate that $a_1 + \cdots + a_{n+1} \geq n+1$. Let's suppose $0< a_{n+1} < 1$. Observe that $$ X:=\frac{(n-1)+ a_{n}^2 }{a_{n}} \geq \frac{n}{a_{n+1}} \geq n $$
and that $f(x) = x + \frac{n}{x}$ is an increasing function on $[\sqrt{n}, \infty)$. Now  \begin{align*} a_1 + \cdots + a_{n+1} &= (a_1 + \cdots + a_{n-1}) + a_{n} + a_{n+1}  
\\ &\geq  \frac{n-1}{a_{n}} + a_{n}  + \frac{n}{  a_{n} + \frac{n-1}{a_{n}} } 
\\ &= f(X) 
\\ &\geq f(n) = n + 1. \end{align*} so the induction step is achieved.

About a year after answering, I found that this is a problem from the 2015 IMO shortlist.
