Answer :
given:
[tex]x>0,\ \ y \ge 1,\ \ \ a=\frac{2xy}{y^2+1}\\\\x+a=x[1+\frac{2y}{y^2+1}]=x[\frac{(1+y)^2}{y^2+1}]\\\\x-a=x[1-\frac{2y}{y^2+1}]=x[\frac{(1-y)^2}{y^2+1}]\\\\x^2-a^2=\frac{x^2(1+y)^2(1-y)^2}{(y^2+1)^2}\\\\\frac{(\sqrt{x+a}-\sqrt{x-a})}{(\sqrt{x+a}+\sqrt{x-a})}\\\\=\frac{(\sqrt{x+a}-\sqrt{x-a})(\sqrt{x+a}-\sqrt{x-a})}{(\sqrt{x+a}+\sqrt{x-a})(\sqrt{x+a}-\sqrt{x-a})}\\\\=\frac{x+a+x-a-2\sqrt{(x+a)(x-a)}}{x+a-(x-a)}\\\\=\frac{1}{2a}[2x-2\sqrt{x^2-a^2}]\\\\=\frac{x}{a}[1-\frac{(1-y^2)}{1+y^2}][/tex]
[tex]=\frac{x}{a}[\frac{1+y^2-1+y^2}{1+y^2}]\\\\=\frac{1}{a}\frac{2xy^2}{(1+y^2)}\\\\=\frac{1}{a}a*y\\\\=y[/tex]
So answer is y >= 1.
[tex]x>0,\ \ y \ge 1,\ \ \ a=\frac{2xy}{y^2+1}\\\\x+a=x[1+\frac{2y}{y^2+1}]=x[\frac{(1+y)^2}{y^2+1}]\\\\x-a=x[1-\frac{2y}{y^2+1}]=x[\frac{(1-y)^2}{y^2+1}]\\\\x^2-a^2=\frac{x^2(1+y)^2(1-y)^2}{(y^2+1)^2}\\\\\frac{(\sqrt{x+a}-\sqrt{x-a})}{(\sqrt{x+a}+\sqrt{x-a})}\\\\=\frac{(\sqrt{x+a}-\sqrt{x-a})(\sqrt{x+a}-\sqrt{x-a})}{(\sqrt{x+a}+\sqrt{x-a})(\sqrt{x+a}-\sqrt{x-a})}\\\\=\frac{x+a+x-a-2\sqrt{(x+a)(x-a)}}{x+a-(x-a)}\\\\=\frac{1}{2a}[2x-2\sqrt{x^2-a^2}]\\\\=\frac{x}{a}[1-\frac{(1-y^2)}{1+y^2}][/tex]
[tex]=\frac{x}{a}[\frac{1+y^2-1+y^2}{1+y^2}]\\\\=\frac{1}{a}\frac{2xy^2}{(1+y^2)}\\\\=\frac{1}{a}a*y\\\\=y[/tex]
So answer is y >= 1.