Here is long and weird solution that I found during contest (There is shorter and simpler solution): Taking $(a,b,c)=(1,2,4)$ gives that given expression (Let's say it $T$) can get value $\frac{5}{2}$. Now let's show that $T\geq \frac{5}{2}$. Let $a=dxyp, b=dyzq, c=dzxr$, where $d=\gcd(a,b,c), dx=\gcd(a,c), dy=\gcd(a,b), dz=\gcd(b,c)$. So $T=\frac{dxyz(pq+qr+rp)}{d(x+y+z)}=\frac{xyz(pq+qr+rp)}{x+y+z}$. $\bullet$ If $x,y,z\geq 2 $ Then we can easily show that $xyz>x+y+z \implies T>pq+qr+rp>3>\frac{5}{2}$. $\bullet$ If only one of $x,y,z$ is $1$, WLOG $z=1$, then if $xy>4$ we can easily show that $xy\geq x+y+1 \implies T\geq pq+qr+rp>3>\frac{5}{2}$ and it's ok. If $xy=4$ we get $x=y=2$. Since $b\ne c$ we get $q\ne r$,so at least one of $q,r>2$ , so $T=\frac{4(pq+qr+rp)}{5}\geq\frac{4\cdot 5}{5}=4>\frac{5}{2}$ and it's also ok. $\bullet$ Now assume at least $2$ of $x,y,z=1$, WLOG $y=z=1$. Note that then at least $1$ of $p,r>1$,and let $x>1$. Then $T=\frac{x(pq+qr+rp)}{x+2}\geq \frac{pq+qr+rp}{2}\geq \frac{5}{2}$. Now let $x=1$. Since $x=y=z=1$, we get $p,q,r$ are distinct. So $pq+qr+rp\geq 1\cdot 2+2\cdot 3+3\cdot 1=11$. So $T\geq \frac{11}{2}>\frac{5}{2}$. And it finishes the proof.

Easy diophonte equation

Solved with bluelinfish, CyclicISLscelesTrapezoid, DottedCaculator, ETS1331, amuthup, squareman, v4913. The answer is $\frac{5}{2}$, an equality case is $(1,2,4)$. Notice that we can scale $a$, $b$, and $c$ by a constant without changing the value of the expression, so we can assume WLOG that $\text{gcd} (a,b,c) = 1$. Moreover, if $\not\exists x,y\in \{a,b,c\}$ such that $x = 2y$, then all three quotients of the form $\frac{\text{lcm} (x,y)}{\text{gcd} (x,y)}$ are at least $3$ and hence the expression is at least $3$. Hence, we can assume $b = 2a$. But using $\text{gcd} (a,b,c) = 1$ we can bound $\frac{\text{lcm} (a,b) + \text{lcm} (b,c) + \text{lcm} (c,a)}{\text{gcd} (a,b) + \text{gcd}(b,c) + \text{gcd}(c,a)} \ge \frac{2a + 2a + a}{a + 2 + 1}$ so we are done if $a\ge 3$. If $a = 1$ we have $c\ge 3$. If $c = 3$ the expression is greater than $\frac{5}{2}$, if $c = 4$ we have equality, and if $c\ge 5$ we can bound $\frac{\text{lcm} (a,b) + \text{lcm} (b,c) + \text{lcm} (c,a)}{\text{gcd} (a,b) + \text{gcd} (b,c) + \text{gcd} (c,a)} \ge \frac{2+5+5}{1+2+1}$. If $a = 2$ we know $c$ is odd and we can bound $\frac{\text{lcm} (a,b) + \text{lcm} (b,c) + \text{lcm} (c,a)}{\text{gcd} (a,b) + \text{gcd}(b,c) + \text{gcd}(c,a)} \ge \frac{4+4+2}{2+1+1}$ and we're done.

Sol: $a\times b=\text{lcm}(a,b)\times \gcd(a,b)$ $\text{lcm}(a,b)=\frac{ab}{\gcd(a,b)}$ Since, $a$ and $b$ are distinct, therefore the least value for $\text{lcm}(a,b)$ is when $\gcd(a,b)$ is close to $ab.$ Closest to get is for $b=2a,\quad \gcd(a,b)=a,\quad \text{lcm}(a,b)=2a.$ Repeat this for $b$ and $c.$ Now, we have $(a,2a,4a)$ $\therefore x=\frac{(2a+4a+4a)}{(a+2a+a)}=\frac{10a}{4a}\implies x=\frac{5}{2}.$

After supposing that minimum value 5/2, we can solve with easy inequalities. Denoting a,b,c by their pairwise greatest common divisor and by m,n and p, like this (I will show how to denote a, for ex): a=(gcd(a,b)×gcd(a,c)×m)/gcd(b,c), denote b and c like this, then use that lcm(a,b)=ab/gcd(a,b). After using these properties, we can finish the problem with easy diophantine inequalities.