Помогите, пожалуйста!

$(\sqrt{a+b}+\sqrt{a-b})^2=a-b+2\sqrt{(a+b)(a-b)}+a-b=\\=2a+2\sqrt{a^2-b^2}\\(\sqrt{a+b}-\sqrt{a-b})^2=2a-2\sqrt{a^2-b^2}\\\frac{\sqrt {2a+2\sqrt{a^2-b^2}}-\sqrt{a-b}}{\sqrt{2a-2\sqrt{a^2-b^2}}+\sqrt{a-b}}=\frac{{\sqrt{(\sqrt{a+b}+\sqrt{a-b})^2}-\sqrt{a-b}}}{\sqrt{(\sqrt{a+b}-\sqrt{a-b})^2}+\sqrt{a-b}}=\\=\frac{|\sqrt{a+b}+\sqrt{a-b}|-\sqrt{a-b}}{|\sqrt{a+b}-\sqrt{a-b}|+\sqrt{a-b}}=\frac{\sqrt{a+b}+\sqrt{a-b}-\sqrt{a-b}}{\sqrt{a+b}-\sqrt{a-b}+\sqrt{a-b}}=\\=\frac{\sqrt{a+b}}{\sqrt{a+b}}=1$
a+b>=0, a-b>=0
a>=-b, a>=b --->
$\sqrt{a+b} \geq \sqrt{a-b}$