# Prove the inequality using simple induction.

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?

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

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.