
Algorithms:

0<a1<a2<a3

F()= ( b1/a1-) +  (b2/a2-) + (b3/a3-) -1

F()= 2/a1-(b1/a1-) + 2/a2-(b2/a2-) + 2/a3-(b3/a3-)

  is always < a1 wich is positive
 then initial guess o=0.



The exact result at a third iteration:

= -0.000110081
f()=2.442E-15
f'()=2.482880