Question

Solve the following inequality:
$\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}>\frac{3}{2}.$

Answer 1

$\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}>\frac{3}{2}$ - Equation 1

1. For definition of square roots,

$x-1\ge 0$ - Equation 2

$\sqrt{x+2\sqrt{x-1}}\ge 0$

$\Rightarrow x+2\sqrt{x-1}\ge 0$(On squaring)

$\Rightarrow (x-1)\ge \frac{x^2}{4}$(On squaring again)

$\Rightarrow 0\ge x^2-4x+4$

$\Rightarrow 0\ge (x-2{)}^2$ - Equation 3

$\sqrt{x-2\sqrt{x-1}}\ge 0$

$\Rightarrow x-2\sqrt{x-1}\ge 0$(On squaring)

$\Rightarrow \frac{x^2}{4}\ge x-1$(On squaring again)

$\Rightarrow x^2-4x+4\ge 0$

$\Rightarrow (x-2{)}^2\ge 0$ - Equation 4

According to Equation 3& Equation 4, only one such point can exist and i.e. $x=2$.

And let's check it by putting in Equation 1,

$\sqrt{2+2\sqrt{2-1}}+\sqrt{2-2\sqrt{2-1}}$

$\Rightarrow \sqrt{4}$

$\Rightarrow 2>\frac{3}{2}$

$\Rightarrow x=2$