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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.

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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.

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    $\begingroup$ Please use MathJax in future posts. Also, we would appreciate if you show your efforts to solve this problem. $\endgroup$ – Virtuoz Sep 15 at 8:52
  • $\begingroup$ I would try induction! $\endgroup$ – Dr. Sonnhard Graubner Sep 15 at 8:57
  • $\begingroup$ Please go ahead $\endgroup$ – Aditya Saran Sep 15 at 8:58
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    $\begingroup$ How is difficult offensive for people? $\endgroup$ – infinitezero Sep 15 at 9:05
  • $\begingroup$ I am a high school student and don't know asymptotic behaviour of the inequality. Please write an answer. $\endgroup$ – Aditya Saran Sep 15 at 9:06
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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.

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