If $f(n)\geq\beta \sum_{i=0}^{n-1}f(i)$,then $f(n)?$

Question

If $f(n)\geq\beta \sum_{i=0}^{n-1}f(i)$, then what can i say about $f(n)$ in terms of $f(1)$?

I tried to find by taking equality and see how it progress, but i couldn't get. Any help is appreciated.

Answer 1

Does the following help?
Let $g(n) = \sum_{i=0}^{n-1} f(i)$.

Then we have $f(n) \ge \beta g(n)$ and $g(n) = f(n-1) + g(n-1)$. So

$$f(n) \ge \beta g(n) = \beta f(n-1) + \beta g(n-1) \ge \beta^2 g(n-1)+\beta g(n-1) = \beta(\beta+1)g(n-1)$$

$$\implies f(n) \ge \beta(\beta+1)\left(f(n-2)+g(n-2) \right) $$ $$\implies f(n) \ge \beta(\beta+1)^2g(n-2)$$ $$...$$ $$\implies f(n) \ge \beta (\beta+1)^{n-1} g(1)$$

As $g(1) = f(0)$, this means $f(n) \ge \beta (\beta+1)^{n-1} f(0)$