Lower bound on $F$ under the assumption $\theta F(s)\le sF'(s)$

Question

Let $F(s)=\displaystyle \int_0^{s}f(t)\,\mathrm dt$. We suppose that there exists $\theta>2$ such that $\theta F(s)\le f(s)s$ for all $s\in \mathbb{R}$ and that $F(s)>0$ for all $s\in(0,+\infty)$.

How to prove that $F(s)\ge F(1)s^\theta$?

Answer 1

Since $F(t)>0$ for $t>0$ we have $$ \frac{F'(t)}{F(t)}\ge \frac{\theta}{t} \quad \forall t>0. $$ Hence: $$ \ln\frac{F(s)}{F(1)}=\int_1^s\frac{F'(t)}{F(t)}\,dt \ge \int_1^s\frac{\theta}{t}\,dt=\theta\ln s=\ln s^\theta. $$ It follows that $$ F(s)\ge F(1)s^\theta \quad \forall s>0. $$