Find all positive integers $a,b,c,x,y,z$ satisfying $a+b+c=xyz$ and $x+y+z=abc$. 
Find all positive integers $a,b,c,x,y,z$ satisfying $$a+b+c=xyz,\tag{1}$$and$$x+y+z=abc,\tag{2}$$ where $a\ge b\ge c\ge 1$ and $x\ge y\ge z\ge 1$.


My try: I think this problem is unique in a way that there are 6 variables and 2 equations , in that way, there may be lot of cases. Also unlike other diophantine equations factoring is not possible. Here is something i did.
Obviously $xyz\ge 3$ and similarly $abc\ge 3$ ,starting with the equality $xyz=3$ or  $x=3,y=1,z=1$,and it happens when $a=b=c=1$. Naturally it does not satisfy equation $(2)$ .I tried randomly setting variables some values to see if some pattern popped up,but all efforts were futile.
Next  i tried setting $y=1 ,z=1$  which implies $a+b+c=abc-2$ again one could get many triplets.
i am totally  stuck.Could anyone nudge me to the right track
 A: If all the integers are greater than or equal to $2$, then
$$
a+b+c < abc = x+y+z < xyz = a+b+c.
$$
Contradiction. Therefore, assume WLOG that $z=1$. You get
$$
a+b+c = xy, \qquad x+y+1 = abc.
$$
Now assume again, $a, b, c, x, y\geq2$, you get
$$
a+b+c < abc = x+y+1 \leq xy + 1 = a+b+c+1.
$$
Therefore, $abc=a+b+c+1$ and $x+y=xy$. Since $x, y\geq 2$, this implies $x=y=2$. Easy to find $a, b, c$ from here (if there are any).
The remaining cases are when $c=1$ or $y=1$.
$(a, b, c) = (3, 2, 1)$ and $(x, y, z) = (3, 2, 1)$ is one example.
A: Claim : at least one of
$$ bc, yz$$ is less than 3.
Assume bc=3, then b=3, c=1.
$$ a+b+c<3a=abc$$
If bc>3, then $$ abc>3a≥a+b+c =xyz.$$ Thus for bc≥3,  we have $$abc>a+b+c, $$
And $$3x≥x+y+z=abc>a+b+c=xyz, $$ or
$$ 3x>xyz $$
$$ 3>yz.$$
This proves our claim. $$ $$
WLOG, suppose that yz= 1, or 2.$$ $$
Case 1:$ $if  $yz=1$. We have: $$ abc=x+2=xyz+2=a+b+c+2$$
Again, if c≥2, then bc≥4, and$$ $$ $$4a ≤abc=a+b+c+2≤4a;$$
Thus, $$ a=b=c=2.$$ We obtain the solution$$ (2,2,2,6,1,1) and (6,1,1,2,2,2). $$
Similarly we get solutions at c=1.
Case 2 is when yz=2.$$ $$
...$$ $$
Hope you got the main idea.
