Calling #f(x,y) = sqrt(y + sqrt(2xy - x^2)) + sqrt(y - sqrt(2xy - x^2))#
If #x > y# then prove that #sqrt(y + sqrt(2xy - x^2)) + sqrt(y - sqrt(2xy - x^2)) = sqrt(2x)#?
If #x > y# then #y = x-delta^2# with #delta in RR, delta ne 0#
Now substituting #y# into #f(x,y)# we have
#(f @ (y-x=delta^2)) = sqrt[x-delta^2 - sqrt[x (x-2 delta^2)]] + sqrt[x-delta^2 + sqrt[
x (x-2 delta^2)]]#
Now squaring
#(f @ (y-x=delta^2))^2=2 x + 2 sqrt[x-delta^2 - sqrt[x (x-2 e^2 )]]sqrt[x-delta^2 + sqrt[x (x-2 e^2 )]]-2 delta^2=#
#=2x-2delta^2+2sqrt((x-delta^2)^2-x(x-2delta^2))=#
#2x-2delta^2+2sqrt(delta^4) = 2x# so we workout
#(f @ (y-x=delta^2))^2=2x#
which resumes the demonstration.
