For any $r\geq 1$ and any positive integers $d_1, \dots d_r\in \mathbb Z_{>0}$ define $L(d_1,\dots , d_r)$ (think logarithmic lcm) to be \begin{equation*} L(d_1,\dots , d_r) ~:=~ \sum_{i=1}^r (-1)^{i+1}\Big(\sum_{1\leq k_1 < ... < k_i\leq r} \text{gcd}(d_{k_1}, \dots, d_{k_i}) \Big). \end{equation*} It is straightforward to check that $L$ is symmetric, homogeneous of degree 1 and that
\begin{equation*} (i) ~~~~~ L(d_1,\dots , d_r) = L(d_1,\dots , d_{r-1}) + L(d_1,\dots , d_{r-2}, d_n) - L(d_1,\dots , d_{r-2}, \text{gcd}( d_{r-1},d_r)). \end{equation*} Furthermore it follows directly from symmetry and property $(i)$ that \begin{equation*} (ii) ~~~~\text{if} ~~ d_r \big \vert d_i~~ \text{for some}~ 1\leq i \leq r-1 \text{, then} ~~L(d_1,\dots , d_r) = L(d_1,\dots , d_{r-1}). \end{equation*} The third property we want to establish is that \begin{eqnarray*} (iii) ~~~L(d_1,\dots , d_r) %&=& \sum_{i=1}^r (-1)^{i+1}\Big(\sum_{1\leq k_1 < ... < k_i\leq r} \text{gcd}(d_{k_1}, \dots, d_{k_i}) \Big) %\\ &\leq& \prod_{i=1}^r \Big(\prod_{1\leq k_1 < ... < k_i\leq r} \text{gcd}(d_{k_1}, \dots, d_{k_i}) \Big)^{(-1)^{i+1}} = \text{lcm}( d_1 ,\dots d_r), \end{eqnarray*} with equality if and only if for some $1\leq i \leq r$, $L(d_1,\dots , d_r) $ can be reduced to $ L(d_i)$ in the sense of property $(ii)$, i.e. $ d_j \big \vert d_i$ for all $j\neq i$.
To see this let $G$ be the cyclic group of order $\text{lcm}( d_1 ,\dots, d_r)$ and pick elements $c_1, \dots, c_r\in G$ of order $d_1, \dots, d_r$ respectively. One obviously has $ \# \Big ( \bigcup_{i=1}^r ~ \langle c_i \rangle \Big )~\leq ~ \# G. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(\star) $ Since $G$ is cyclic one has $\# \big( \langle c_{k_1} \rangle \cap \dots \cap \langle c_{k_i} \rangle \big) = \text{gcd}(d_{k_1}, \dots, d_{k_i})$. Hence by the inclusion exclusion principal the left hand side of equation ($\star$) is equal to $L(d_1,\dots, d_r) $ and the right hand side is equal to $\text{lcm}( d_1 ,\dots, d_r)$. Observe that equality holds iff one of the $c_i$ has order $\text{lcm}( d_1 ,\dots, d_r)$, which proves the claim.