This might help, (by definition):
$$\cosh x = \frac{e^x + e^{-x}}{2}$$
Replace x with $\ln x$ and simplify, note that $e^{\ln x} = x$
Use a similar idea for your second expression and note that:
$$ \sinh x = \frac{e^x - e^{-x}}{2}$$
Let $G(r):=1+\left(\int_1^r \frac{s}{V(s)}\, \mathrm{d}s\right)_+$. First, suppose $r\geq 1.$ Then since the exponential is bounded by $1,$ $F(r)\leq G(r)+\int_0^1 \frac{se^{-\frac{1}{s}}}{V(s)}\,\mathrm{d}s$. This means $\frac{F(r)}{G(r)}\leq 1+\frac{\int_0^1 \frac{se^{-\frac{1}{s}}}{V(s)}\,\mathrm{d}s}{G(r)}$. Since $G(r)\geq 1,$ we have that
$$
\frac{F(r)}{G(r)}\leq 1+\int_0^1 \frac{se^{-\frac{1}{s}}}{V(s)}\,\mathrm{d}s=F(1).
$$
Since everything is positive, this also means
$$
\frac{G(r)}{F(r)}\geq \frac{1}{1+\int_0^1 \frac{se^{-\frac{1}{s}}}{V(s)}\,\mathrm{d}s}=\frac{1}{F(1)}.
$$
Now, suppose $r<1.$ Then $G(r)\equiv 1$ since
$$
\int_1^r \frac{s}{V(s)}\, \mathrm{d}s=-\int_r^1 \frac{s}{V(s)}\, \mathrm{d}s<0.
$$
It is then trivial to see that $\frac{F(r)}{G(r)}=F(r),$ which is increasing so $\frac{F(r)}{G(r)}\leq F(1)$ and obviously,
$$
\frac{G(r)}{F(r)}\geq \frac{1}{F(1)}.
$$
EDIT: An easy, one line proof of the claim that you proposed is that $G(r)\geq 1$ so $\frac{F(r)}{G(r)}\leq F(r)< F(\infty)$ and $\frac{G(r)}{F(r)}\geq \frac{1}{F(r)}> \frac{1}{F(\infty)}$. The advantage of my solution above is that $F(1)$ is a better bound.
Best Answer
Use the property that $\displaystyle \frac{e^a}{e^b}=e^{a-b}$.
Then, everything follows as such.