[tex]\dfrac ab=\dfrac38\implies a=\dfrac38b[/tex]
[tex]\dfrac bc=\dfrac6{11}\implies c=\dfrac{11}6b[/tex]
[tex]\implies a+b+c=\left(\dfrac38+1+\dfrac{11}6\right)b=\dfrac{77}{24}b[/tex]
Since each of [tex]a,b,c[/tex] are integers, so must be [tex]a+b+c=\dfrac{77}{24}b[/tex]. The only way for this to be the case is if [tex]b[/tex] is a multiple of 24 because 77 and 24 share no common divisors. The smallest multiple would be [tex]b=24[/tex]. So the smallest value of [tex]a+b+c[/tex] is 77.
