We have the following problem
Let a function $f:R\rightarrow R$ be a thrice differentiable function such that $f(x),f'(x),f''(x),f'''(x)>0 \,\forall x\in R$. Given $f'''(x)<f(x),\forall x\in R$. Then $f'(x)<nf(x)\,\forall x\in R$. Find $n$
This question is quite mind-boggling, and I can't figure out where to start.
I tried to do the following:
First I multiplied $f''(x)$: $$f''(x)f'''(x)<f(x)f''(x)$$ since $(f'(x))^2>0$ $$f''(x)f'''(x)<f(x)f''(x)+(f'(x))^2$$ We get $$f''(x)f'''(x)<(f(x)f'(x))'$$ Integrating: $$\frac{1}{2}(f''(x))^2<f(x)f'(x)$$ However, now I am stuck and I can't think of another way to tackle this problem. Please help, thanks in advance.
