If x > y then prove that #sqrt(y + sqrt(2xy - x^2)) + sqrt(y - sqrt(2xy - x^2)) = sqrt(2x)#?

1 Answer
May 8, 2017

See below.

Explanation:

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.