Rewriting the LCMs as GCDs is a good first step. We finish by removing these GCD constraints. Define $k_0 = \gcd(a,b,c)$. Then we can write $a=k_0a'$, $b=k_0b'$, and $c=k_0c'$ for integers $a'$, $b'$, and $c'$. Substituting and cancelling factors of $k_0$ gives us
$$ a'b'c'=k_0\gcd(a',b')\gcd(b',c')\gcd(c',a'). $$
Now let $k_1=\gcd(a',b')$, $k_2=\gcd(b',c')$ and $k_3=\gcd(c',a')$. We can write $a'=a''k_1k_3$, $b'=b''k_1k_2$ and $c'=c''k_2k_3$, since $\gcd(a',b',c')=1$. (Note that $a''$, $b''$, and $c''$ are integers.) Substituting and cancelling again, we have
$$ k_1k_2k_3a''b''c''=k_0, $$
where $a''$, $b''$, and $c''$ are pairwise coprime. Assume without loss of generality that $a''\le b''\le c''$.
Case 1: $a''\neq 1$
We then have $a''\ge 2$, $b''\ge 3$, and $c''\ge 5$. Furthermore, our answer is monotonically increasing in each of the $k_i$. Thus we obtain our minimal solution when we set $a''=2$, $b''=3$, $c''=5$, and all the $k_i=1$:
$$ a+b+c = k_0(k_1k_3a'' + k_1k_2b'' + k_2k_3c'') = 300. $$
Case 2: $a''=1$.
Then $b''\ge 2$, $c''\ge 3$. Because none of $a,b,c$ divide each other, $k_1,k_3\ge 2$. Plugging in these inequalities, we have:
$$ a + b + c\ge 336. $$
Thus the first case gives $300$ as our answer.
