Proof using asymptotic notion with fractional function

Hey I am stuck on this one: Let $f,g,h: \mathbb{N} \to\mathbb{R}^{+}\setminus \{ 0 \}$ be given such that: $f(n)=O(h(n))$, $g(n)=O(h(n))$, and $h(n)=o(1).$ I already proved that $f(n)+g(n)=O(h(n))$ and $f(n)g(n)=O(h(n))$ and now I'm left off with $1/(1+f(n))=1+O(h(n))$. I tried to reformulate that a bit in order to get: $f(n)/(1+f(n)) = O(h(n))$, but now I find no way to prove that. Any help is appreciated.

Hint: $$\left|\frac{1}{1+f(n)}-1\right|=\frac{f(n)}{1+f(n)}\leq f(n)$$ since $f(n)\geq 0$ and $1+f(n)\geq 1$.

Hint. A useful thing to know: If $f(n) = o(1)$, then $\frac{1}{1+f(n)}$ is approximately $1-f(n)$. Perhaps prove this by expanding $(1+f(x))(1-f(x))$ to find an actual $o$-estimate instead of merely saying "approximately".