2020/003) If 1 < x < 2, Then Prove that $\frac{1}{\sqrt{x+2\sqrt{x-1}}}+ \frac{1}{\sqrt{x-2\sqrt{x-1}}} = \frac {2}{2-x}$

Solution

Let $\sqrt{x+2\sqrt{x-1}} = \sqrt{m} + \sqrt{n}$
Hence  $\sqrt{x-2\sqrt{x-1}} = \sqrt{m} - \sqrt{n}$
Without loss of generality let us assume $m >=n$
Square both sides to get  $x+2\sqrt{x-1} = m + n + 2\sqrt{mn}$
Comparing rational and surds on both sides we get
$x=m +n\cdots(1)$
$x-1= mn\cdots(2)$
From (1) and (2)
$(m-n)^2 = (m+n)^2-4mn = x^2-4(x-1) = (2-x)^2$ we chose 2-x as $x < 2$
or $m-n=2-x\cdots(3)$
From (3) and (1) we get $m=1, n = x-1$
Hence we get
$\frac{1}{\sqrt{x+2\sqrt{x-1}}}+ \frac{1}{\sqrt{x-2\sqrt{x-1}}}$
$= \frac{1}{\sqrt{m}+\sqrt{n}}+ \frac{1}{\sqrt{m}-\sqrt{n}}$
$= \frac{\sqrt{m}+\sqrt{n}+ \sqrt{m}-\sqrt{n}} {m-n}$
$= \frac{2\sqrt{m}} {m-n}$
$= \frac{2\sqrt{1}} {2-x}$
$= \frac{2} {2-x}$