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I was solving this question:

Let $A_1,A_2,A_3\cdots A_n$ be a sequence of real numbers satisfying $A_{i+j} \leq A_i + A_j ,\forall i, j \in\mathbb{N}$. Prove that $A_1 + \frac{A_2}2 + \frac{A_3}{3} +\cdots + \frac{A_n}n \geq A_n$.

I was able to prove that $A_1 \geq \frac{A_n}n$ for all $n$ being natural using induction. But after that I am unable to use this fact to prove the required statement.

I don't think that from given inequality, any other inequality will actually prove to be useful, so what am I missing?

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  • $\begingroup$ Thank you for edit. I didn't know how to write like this in math notations.Sorry $\endgroup$ Commented Sep 28, 2020 at 13:40
  • $\begingroup$ Let $n=2$. If $A_1=1$ and $A_2=30$ then $A_1+\frac{1}{2}A_2\geq A_2$. But this is the same as $1+\frac{1}{2}30\geq 30$. $\endgroup$ Commented Sep 28, 2020 at 16:48
  • $\begingroup$ Does this answer your question? Prove that for any $n$, we have $\sum_{i=1}^n \frac{a_i}i \ge a_n $. $\endgroup$
    – Martin R
    Commented Nov 4, 2023 at 18:22

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For $n=1$ it's true.

Let $a_1+\frac{a_2}{2}+...+\frac{a_n}{n}\geq a_n$ for $n\in\{1,2,...,n\}$.

After summing of these $n$ inequalities we obtain: $$na_1+(n-1)\frac{a_2}{2}+...+\frac{a_n}{n}\geq a_1+a_2+...+a_n,$$ which gives $$na_1+a_1+(n-1)\frac{a_2}{2}+a_2+...+\frac{a_n}{n}+a_n\geq$$ $$\geq(a_1+a_n)+(a_2+a_{n-1})+...+(a_n+a_1)\geq na_{n+1},$$ which gives $$(n+1)\left(a_1+\frac{a_2}{2}+...+\frac{a_n}{n}\right)\geq na_{n+1}$$ or $$a_1+\frac{a_2}{2}+...+\frac{a_n}{n}+\frac{a_{n+1}}{n+1}\geq\frac{na_{n+1}}{n+1}+\frac{a_{n+1}}{n+1}$$ or $$a_1+\frac{a_2}{2}+...+\frac{a_n}{n}+\frac{a_{n+1}}{n+1}\geq a_{n+1}$$ and we are done by induction.

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  • $\begingroup$ There was nothing to end. Just one more line to add.I feel like an idiot.But,Thanks for your help. $\endgroup$ Commented Sep 28, 2020 at 13:52
  • $\begingroup$ @Combat Miners You are welcome! Good luck! $\endgroup$ Commented Sep 28, 2020 at 13:54
  • $\begingroup$ So how do you end it? $\endgroup$ Commented Sep 28, 2020 at 18:18
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    $\begingroup$ @lockedscope I added something. See now. $\endgroup$ Commented Sep 28, 2020 at 18:36

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