What is going on at the top is finding factors common to all the numbers. The black $2,2,3$ are those. The product of them is the greatest common divisor of all the numbers. Once we get to $4,6,9$ there are no longer any common factors, so the greatest common divisor has been determined. Now we are using the fact that the least common multiple of $an, bn$ is $n\operatorname{lcm}(\frac an, \frac bn)$. We use that until there are no pairs of numbers with a common factor, here $2,1,3$. Finally we use the fact that $\operatorname{lcm}(a,b)=ab$ whenever $a$ and $b$ are coprime.
The first key is:
Theorem 1: If $a_1,\cdots,a_k$ are positive integers which are pair-wise relatively prime (that is, $\gcd(a_i,a_j)=1$ when $i\neq j$) we have that $$\operatorname{lcm}(a_1,\dots,a_k)=\prod_{i=1}^{k}a_i$$
The second key is:
Theorem 2: if $a_1,\cdots,a_k$ are positive integers then there are pair-wise relatively prime integers $a_1',\dots,a_k'$ with $1\leq a_i'\leq a_i$ for all $i$ and $$\prod_{i=1}^{n}a_i'=\operatorname{lcm}(a_1',\dots,a_k')=\operatorname{lcm}(a_1,\dots,a_k)$$
(You can actually get $a_i'\mid a_i.$)
So the set of least common multiples of $k$ numbers from $1$ to $n$ is the same as the set of products of $k$ pairwise relatively prime numbers from $1$ to $n.$
Then, in order to find the maximum LCM, we need to find the maximum product of three pairwise relatively prime numbers numbers.
When $n$ is odd, $n,n-1,$ and $n-2$ are pairwise relatively prime, so we get a least common multipe of $n(n-1)(n-2)$ which is the maximum product of three numbers from $1$ to $n,$, at least if $n>2.$
When $n$ is even, but not divisible by $3,$ then $n,n-1$ and $n-3$ are pairwise relatively prime. This is because the only product of three different numbers from $1$ to $n$ is $n(n-1)(n-2),$ but $n$ and $n-2$ are not relatively prime when $n$ is even.
When $n$ is even and divisible by $3$, then we have that $n-1$ is odd, so $(n-1)(n-2)(n-3)$ is a lest common multiple. On the other hand, any other product of three pairwise relatively prime integers with $n$ involved is at most $n(n-1)(n-5).$ But $n(n-5)<(n-2)(n-3)$ so $n(n-1)(n-5)<(n-1)(n-2)(n-3).$ So when $n$ is divisible by $6,$ the maximal value is $(n-1)(n-2)(n-3).$
You have to take care when $n\leq 2.$ In those cases, multiple values can be equal to $1.$ . When $n=2,$ you have a maximal LCM is $2.$ When $n=1,$ the maximum LCM is $1.$
Example of theorem 2: If $a_1=12,a_2=45,a_3=30,$ then we have:
$$a_1=2^23^15^0\\a_2=2^03^25^1\\a_3=2^13^15^1,$$ then we can take:
$$a_1'=2^2\\a_2'=3^2\\a_3'=5^1.$$
The $a_i'$ are pairwise relatively prime and the LCM is the product which is $2^23^25^1=180,$ and this is the least common multiple of the original triple $12,45,30.$
Best Answer
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.