Inclusion Exclusion and lcm I would like to show that for any positive integers $d_1, \dots, d_r$ one has
$$
\sum_{i=1}^r (-1)^{i+1}\biggl( \sum_{1\leq k_1 < \dots < k_i \leq r} \gcd(d_{k_1}, \dots , d_{k_i})\biggr) ~\leq~  \prod_{i=1}^r\biggl(  \prod_{1\leq k_1 < \dots < k_i \leq r} \gcd(d_{k_1}, \dots , d_{k_i}) \biggl)^{(-1)^{i+1}}. 
$$
Note that the rhs of the upper inequality is exactly $\operatorname{lcm}(d_1,\dots,d_r)$. Also note that if we denote the lhs of the upper equation by $L(d_1, \dots, d_r)$, then one has that 
$$
L(d_1, \dots, d_r) = L(d_1, \dots, d_{r-2}, d_{r-1}) + L(d_1, \dots, d_{r-2}, d_{r}) - L(d_1, \dots, d_{r-2}, \text{gcd}(d_{r-1},d_r)).
$$
Thanks for the help!
 A: 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. 
