Integrate[1/(1+x^2)^2,x]

F1[x_]:=(1/2) ((x/(1+(x^2)))+ArcTan[x])

Integrate[1/(1+x^2)^3,x]

F2[x_]:=(1/8) (((x (5+3 (x^2)))/(((1+(x^2)))^2))+3 ArcTan[x])

ga[b_,y_]:=(F2[b/Sqrt[2y]]+F2[Sqrt[1-y]/Sqrt[y]])/y/(F1[b/Sqrt[2y]]+F1[Sqrt[1-y]/Sqrt[y]])

pl[y_]:=Graphics[Text[y,{Sqrt[2(1-y)],ga[0,y]-0.3}]]

pp[y_]:=Plot[Evaluate[ga[b,y]],{b,0,Sqrt[2(1-y)]}]

Table[pp[y],{y,0.1,0.9,0.1}];

Table[pl[y],{y,0.1,0.9,0.1}];

Show[%%,%,PlotRange->All,AxesOrigin->{0,0},AxesLabel->{β,γ}]

(*γ versus β for different values of η. Note that the value of γ at β=α is the same as that at β=0.
In each case there is a range of values of γ for which there are two positive roots for β.*)

(* Now let η=1-ϵ, β=(ϵ^(1/2))b and γ=1+ϵΓ/3, where ϵ<<1.*)

Series[Evaluate[ga[Sqrt[e]b,1-e]],{e,0,1}]//FullSimplify

%-1-e Γ/3

Normal[%/e]

Solve[%==0,b]

b/.%[[1]]

(*The smaller root is the physically relevant solution. We need Γ>2 so that this root is positive, and Γ<9/4 so that it is real*)