We will use the following notation:
$$H_i = \sum_{k=1}^{i} \frac{1}{k} \qquad P_i(m) = \sum_{k=m}^{N} \frac{H_{k+i}}{k(k+1)}.$$
Lemma 1. For all $i \ge 1$ and all $1 \le M \le N$ we have
$$\sum_{k=M}^{N} \frac{1}{k(k+i)} < \frac{1}{i} \sum_{k=M}^{M+i-1} \frac{1}{k} .$$
Proof. Using $\frac{1}{k(k+i)} = \frac{1}{i}\left(\frac{1}{k} - \frac{1}{k+i}\right)$ we see that
$$\sum_{k=M}^{N} \frac{1}{k(k + i)} = \frac{1}{i} \sum_{k=M}^{N} \left( \frac{1}{k} - \frac{1}{k + i} \right) = \frac{1}{i} \sum_{k=M}^{M+i-1} \frac{1}{k} - \frac{1}{i} \sum_{k=N+1}^{N+i} \frac{1}{k} < \frac{1}{i} \sum_{k=M}^{M+i-1} \frac{1}{k}.$$
If $M + i - 1 \le N$ the terms $1/k$ for $M + i \le k \le N$ cancel out as they appear in both sums. When $M + i - 1 > N$, we have introduced extra terms which appear in both sums. In this case, the sum on the right hand side would only need to go up to $k = N$, but we don't need this stronger inequality. $\square$

$$i = 2 \qquad \sum_{k=M}^{N} \frac{1}{k(k+2)} < \frac{1}{2M} + \frac{1}{2(M+1)} \qquad (32)$$
$$M = 1 \qquad \sum_{k=1}^{N} \frac{1}{k(k+i)} < \frac{1}{i} \sum_{k=1}^{i} \frac{1}{k} = \frac{H_i}{i} \qquad (33)$$

Lemma 2. For all $N > m \ge 1$ and $i \ge 1$ we have
$$P_i(m) < \frac{1}{m} H_{m+i} + \frac{1}{i} \sum_{k=m+1}^{m+i} \frac{1}{k}.$$
$$\begin{align*} P_i(m) &= \sum_{k=m}^{N} \frac{H_{k+i}}{k(k+1)} \\ &= \sum_{k=m}^{N} \left( \frac{1}{k} - \frac{1}{k+1} \right) H_{k+i} = \sum_{k=m}^{N} \frac{1}{k} H_{k+i} - \sum_{k=m+1}^{N+1} \frac{1}{k} H_{k+i-1} \\ &= \frac{1}{m} H_{m+i} + \sum_{k=m+1}^{N} \frac{1}{k} (H_{k+i} - H_{k+i-1}) - \frac{1}{N+1} H_{N+i} \\ &< \frac{1}{m} H_{m+i} + \sum_{k=m+1}^{N} \frac{1}{k(k+i)} < \frac{1}{m} H_{m+i} + \frac{1}{i} \sum_{k=m+1}^{m+i} \frac{1}{k} \quad \text{using Lemma 1.} \end{align*}$$
$\square$
In particular, we obtain
$$P_1(m) < \frac{1}{m+1} + \frac{1}{m} H_{m+1} \qquad (34)$$
$$P_2(m) < \frac{1}{2(m+1)} + \frac{1}{2(m+2)} + \frac{1}{m} H_{m+2}. \quad (35)$$

$$\begin{align*} S &= \sum_{a,b,c=1}^{N} \frac{1}{abc(a+b+c+1)} = \sum_{a,b=1}^{N} \frac{1}{ab} \sum_{c=1}^{N} \frac{1}{c(a+b+c+1)} \\ &< \sum_{a,b=1}^{N} \frac{H_{a+b+1}}{ab(a+b+1)} && \text{using (33) with } i = a+b+1 \\ &= \sum_{a,b=1}^{N} \frac{a+b}{ab} \cdot \frac{H_{a+b+1}}{(a+b)(a+b+1)} \\ &= \sum_{k=2}^{2N} \sum_{a+b=k} \left(\frac{1}{a} + \frac{1}{b}\right) \frac{H_{k+1}}{k(k+1)} = 2 \sum_{k=2}^{2N} \sum_{m=1}^{k-1} \frac{1}{m} \cdot \frac{H_{k+1}}{k(k+1)}. \end{align*}$$
$$$$
\sum_{k=2}^{2N} \sum_{m=1}^{k-1} \frac{1}{m} \cdot \frac{H_{k+1}}{k(k+1)} = \sum_{m=1}^{2N-1} \frac{1}{m} \sum_{k=m+1}^{2N} \frac{H_{k+1}}{k(k+1)}.
$$Therefore,$$
$$\begin{align*} S < 2 \sum_{m=1}^{2N-1} \frac{1}{m} \sum_{k=m+1}^{2N} \frac{H_{k+1}}{k(k+1)} &= 2 \sum_{m=1}^{2N-1} \frac{1}{m} P_1(m+1) \\ &< 2 \sum_{m=1}^{2N-1} \frac{1}{m} \left( \frac{1}{m+2} + \frac{1}{m+1} H_{m+2} \right) && \text{using (34)} \\ &= 2 \sum_{m=1}^{2N-1} \frac{1}{m(m+2)} + 2 \sum_{m=1}^{2N-1} \frac{H_{m+2}}{m(m+1)} \\ &= 2 \sum_{m=1}^{2N-1} \frac{1}{m(m+2)} + 2P_2(1) \\ &< 1 + \frac{1}{2} + 2 \left( \frac{1}{4} + \frac{1}{6} + H_3 \right) = 6. && \text{using (32) and (35).} \end{align*}$$
$$$$