The Euclidean algorithm repeatedly uses the division algorithm until the gcd pops out. Example, to find the gcd of 12 and 93:
93=12*7+9
12=9*1+3
9=3*3+0
This tells you that gcd(93,12)=3. We can work backwards to write 3 as a linear combination of 93 and 12:
3=12-9*1
from the second line. The first line says 9=93-12*7, sub this in for 9:
3=12-(93-12*7)*1=12*8-93*1
and we have 3 as a linear combination of 12 and 93.
This works in general, and you can prove this way that given a,b you can always find s and t so that a*s+b*t=gcd(a,b).
Proving that any other linear combination (that is positive) will be larger will follow from properties of the gcd. It divides both a and b, so what about a linear combination? Maybe look at the gcd=1 case first, if as+bt=1, what if d divides both a and b?