first, a symetry exists among a, b, c, d

so, suppose a=<b=<c=<d

in this way, 4/a>=1

a>=1

well, you can suppose a=1 or a=2 or a=3 or a=4

* if a=1

there is no solutions

* if a=2

then,

or b=<c=<d

so 3/b>=1/2

b=<6

in this way, b=1 or b=2 or b=3 or b=4 or b=5 or b=6

**if b=1 : no solutions

**if b=2 : no solutions

**if b=3 :

or c=<d

so 2/c=>1/6

c<=12

in this way, c=1 or c=2 or c=3 or c=4 or c=5 or c=6 or c=7 or c=8 or c=9 or c=10 or c=11 or c=12

***if c=1 or c=2 or c=3 or c=4 or c=5 or c=6 : no solutions

***if c=7:

so, d=42

solution 1 : a=2, b=3, c=7, d=42

***if c=8 :

so, d=24

solution 2 : a=2, b=3, c=8, d=24

***if c=9 :

so, d=18

solution 3 : a=2, b=3, c=9, d=18

***if c=10 :

so, d=15

solution 4 : a=2, b=3, c=10, d=15

***if c=11 : no solution

***if c=12 :

so, d=12

solution 5 : a=2, b=3, c=12, d=12

**if b=4:

.....

it seems too long....but for sure it's an exhaustive way to solve this problem.