We are given:

1/a = 1+b

1/b = 3+c

1/c = 4+b

If we invert the third equation we have an expression for c that we can use in the second equation.

c=1/(4+b)

1/b = 3+1/(4+b)

1/b = (3(4+b)+1)/(4+b) = (13+3b)/(4+b)

Cross multiplying we get a quadratic:

3b^2+12b-4 = 0

The value of b that lies between 0 and 1 is (4*sqrt(3)-6)/3

We can put this into the first equation:

1/a = 1+(4*sqrt(3)-6)/3 = (3+4*sqrt(3)-6)/3 = (4*sqrt(3)-3)/3

a = 3/(4*sqrt(3)-3), which does fulfil the requirements of the question, but we’d prefer to not have irrationals in the square root:

a = 3/(4*sqrt(3)-3)*(4*sqrt(3)+3)/(4*sqrt(3)+3)=

(12*sqrt(3)+9)/(48-9) = (4*sqrt(3)+3)/13