# 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.

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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.

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

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.