My answer on MO:
If we are allowed to consider somewhat weaker bounds, both answers depend only on the number of unitary divisors of $k$, which is $2^{\omega(k)}$. By the quoted result of O. Izhboldin and L. Kurliandchik (see Fedor's and Myerson's comments here), for any set of $n$ positive integers {$a_{1}, \dots, a_{n}$} such that $\sum_{i = 1}^{n} \frac{1}{a_{i}} <1$, we have
\begin{eqnarray}
\sum_{i = 1}^{n} \frac{1}{a_{i}} \leq \sum_{i = 1}^{n} \frac{1}{d_{i}} = \frac{d_{n+1} - 2}{d_{n+1} -1} < 1,
\end{eqnarray}
where $d_{i}$ is the $i^{\text{th}}$-Euler number, which satisfies the quadratic recurrence $d_{i} = d_{1} \cdots d_{i-1} + 1$ with $d_{1} = 2$. The first few terms of the sequence are 2, 3, 7, 43, 1807.... (A000058). It is relatively straightforward to show that the aforementioned recurrence is equivalent to $d_{i} = d_{i-1}(d_{i-1}-1) + 1$, and it is known from the recurrence that $d_{i} = \lfloor \theta^{2^{i}} + \frac{1}{2} \rfloor$, where $\theta \approx 1.2640$.... Thus,
\begin{eqnarray}
\sum_{p \mid k} \frac{1}{p + 1} \leq \sum_{i = 1}^{\omega(k)} \frac{1}{d_{i}} = \frac{d_{\omega(k) + 1} - 2}{d_{\omega(k)+1} - 1} = \frac{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 2}{\lfloor \theta^{2^{\omega(k) + 1}} + \frac{1}{2} \rfloor - 1} < 1.
\end{eqnarray}
One can also show that $d_{i} - 1 = \lfloor \vartheta^{2^{i-1}} - \tfrac{1}{2} \rfloor$, where where $\vartheta \approx 1.5979$...., so we have
\begin{eqnarray}
1 - \sum_{p \mid k} \frac{1}{p + 1} \geq 1 - \frac{d_{\omega(k)+1} - 2}{d_{\omega(k)+1} - 1} = \frac{1}{d_{\omega(k)+1} - 1} = \lfloor \vartheta^{2^{\omega(k)}} - \tfrac{1}{2} \rfloor^{-1} > 0.
\end{eqnarray}
Remark E. Deutsche points out that the sequence {$d_{i} - 1$} (A007018) counts the number of ordered rooted trees with out-going degree up to 2 with all leaves at top level.