2022/023) Solve for x $\sqrt{3-x} - \sqrt{1+x} >= \frac{1}{2}$

We first note that $3-x \ge 0$ and $1+x <=0$ as both are under square root and so we have $-1 \le x \le 3$

 Secondly we have the LHS is monotonically decreasing as x increases

so let uf have $f(x) = \sqrt{3-x} - \sqrt{1+x}$

We have the LHS is monotonically decreasing as x increases and if is defined in $[-1,3]$

now $f(-1) = \sqrt{3-(-1)} - \sqrt{1+{-1}} = 2$ and $f(3) = \sqrt{3-3} - \sqrt{1+3} = -2$

so we need to solve for  $\sqrt{3-y} - \sqrt{1+y} = \frac{1}{2}$

and then $y \le x \le -3$ 

we need to solve 

$\sqrt{3-y} - \sqrt{1+y} = \frac{1}{2}$

square both sides to get $3 - y + 1 + y - 2\sqrt{(3-y)(1+y)} = \frac{1}{4}$

or $4 - \frac{1}{4} = 2\sqrt{(3-y)(1+y)}$

square both sides to get $\frac{225}{64} = (3-y)(1+y)$

or $\frac{225}{64} = (3-y)(1+y)$

putting y = z + 1 we get $\frac{225}{64} = (2-z)(2+z) = 4 - z^2$

or $z^2 = \frac{31}{64}$

so $z = \pm \frac{\sqrt{31}}{8}$

so $y = 1 \pm \frac{\sqrt{31}}{8}$

now whether we show take the plus or minus we check  that $f(1) = \sqrt{2} - \sqrt{2} = 0$ wich does not sattisfy the condition so we must have -ve numebr and hence

$y =    1  - \frac{\sqrt{31}}{8}$

or $-1 \le x \le  1  - \frac{\sqrt{31}}{8}$

or