GCD and LCM

Note the notations $[a,b]=lcm(a,b)$ and $(a,b)=gcd(a,b)$ .

Using the relation $[a,b](a,b)=a.b$ one can re-write the equality, in the question, as below.

$[a,b][b,c][a,c]=\frac{a^2b^2c^2}{(a,b)(b,c)(a,c)}=abc(a,b,c) \implies$

$abc=(a,b)(b,c)(a,c)(a,b,c) \ (*)$

Few facts (not hard to prove):

1- $a,b,c \neq 0,1$ , because non of them should divide the other one

2- $a \neq b \neq c$

We know that $(a,b,c) \neq 1$ , because, if $a<b<c$ , then the RHS of $*$ would be $a.b$ at most, which forces $c$ to be $1$ . Therefore $(a,b,c)=t \neq 1$ .

Also, one can notice that, given $(a,b)=d$ and $a=a'd,b=b'd$ and $(a',b')=1$ , $a'|(a,c) \implies a'|c$ . In a similar manner, $b'|c$ . Therefore, $a'b'|c$ and, by symmetry, we conclude that the parts of any of two integers $x,y \in \{a,b,c\} \ , x \neq y$ , that are not included in $(x,y)$ , should be included in $z \in \{a,b,c\} z\neq x ,z\neq y$ .

The first structure, that comes to mind, for the integers is $a=pqt , b=prt , c=rqt$ , such that $p,q,r$ are pairwise coprimes. if we substitute in $*$ , we get

$pqr=t$

Hence, $a=p^2q^2r , b=p^2qr^2 , c=pq^2r^2$ . However, it is possible to come with yet better structure that leads to smaller integers: the following structure $a=p^2qr=pt \ , \ b=pq^2r=qt \ , \ c=pqr^2=rt$ , where $t=pqr$ , such that what is discussed till this point is satisfied. So, we have $a+b+c=pqr(p+q+r)$ We just need to take $p,q,r$ to be the lowest possible, mutually coprime, positive integers ( $p=2,q=3,r=5$ ).

let $p$ be any prime, and suppose $p^r,p^s,p^t$ are the maximum prime powers that are factors of $a,b,c$ respectively. Suppose $r\le s\le t$ and isolating the powers of $p$ yields:

$(p^s)(p^t)(p^t)=(p^{r+s+t})(p^r)\implies s+2t=2r+s+t\implies t=2r$

Therefore, for any prime that appears as a factor in $a,b,c$ , the greatest exponent of that prime (in $a,b,\text {or } c$ ) must be exactly twice the exponent of that same prime in $a,b,\text {or } c$ . In particular, if any prime divides one of $a,b,\text { and } c$ , then it must divide all of them. We can't have $a,b,\text {or } c$ all be a power of the same prime, since then clearly the smallest one would divide the largest one.

$\textbf{case 1}$ : Suppose that each of $a,b,c$ are divisible by exactly two primes namely $p \text { and } q$ .

We can not have one of $a,b,c$ have the largest power of both primes simultaneously, since otherwise the other two numbers would divide it. So without loss of generosity, suppose that $a$ has the largest power of $p$ and $b$ has the largest power of $q$ . Since $c$ does not divide $a$ or $b$ , we must have $c$ has a higher power of $q$ than $a$ and a higher power of $p$ than $b$ . In other words, we can not have $a=p^2q \text { and } b=pq^2$ which would imply that $c=p^2q^2\implies a|c \text { and } b|c$ . The next smallest possibility is:

$a=p^4q^2, b=p^2q^4, \text { and } c=p^3q^3$ . Plugging in the smallest possible primes $p=2 \text { and } q=3 \implies a =2^43^2=144, b=2^23^4=324, c=2^33^3=216$ . The sum of $a+b+c=684$ . This is the smallest possible sum with two primes.

$\textbf{case 2}$ : Now suppose that each of $a,b,c$ are divisible by exactly three primes $p,q, \text { and } r\implies a=p^2qr, b=pq^2r \text { and } c=pqr^2$ . That is the smallest possibility that for each prime, two of the numbers are divisible by the prime and the third is divisible by the prime squared. Plugging in the smallest primes $p=2, q=3, \text { and } r=5 \implies a=2^2.3.5=60, b=2.3^2.5=90, c=2.3.5^2=150$ , giving a sum of $a+b+c=60+90+150=300$ , indicating the smallest sum so far.

$\textbf{case 3}$ : If there are 4 or more primes, then each number must be at least $2.3.5.7=210$ , so the sum must at least 630, which is larger than 300.

Thus, the smallest possible sum is $\boxed{300}$ .