Let the integers be x-1, x and x+1.
(x+1)/5-(x-1)/3≥3.
Multiply through by 15:
3x+3-5x+5≥45.
-2x+8≥45,
8-45≥2x,
-37≥2x,
-18.5≥x, x≤-18.5. There are many solutions for x so a minimum cannot be defined.
However, if (x-1)/3-(x+1)/5≥3,
5x-5-3x-3≥45
2x-8≥45,
x-4≥22.5
x≥26.5. Minimum value is x=27, the integers are 26, 27, 28.
CHECK
28/5=5.6, 26/3=8.67. Difference is 3.07 approx.