Question

Solve the following inequality:
$\sqrt{5+x}-\sqrt{-x-3}<1+\sqrt{(x+5)(-x-3)}.$

Answer 1

$\sqrt{5+x}-\sqrt{-x-3}<1+\sqrt{(x+5)(-x-3)}$ - Equation 1

For definition of square roots,

1. $5+x\ge 0\Rightarrow x\ge -5$ -Equation 2

2. $-x-3\ge 0\Rightarrow x\le -3$ - Equation 3

3. $(x+5)(-x-3)\ge 0\Rightarrow (x+5)(x+3)\le 0 -Equation 4

From Equation 4,

$(5+x)+(-x-3)-2\sqrt{(x+5)(-x-3)}<1+(x+5)(-x-3)+2\sqrt{(x+5)(-x-3)}$

$\Rightarrow 1+(x+5)(x+3)<4\sqrt{(x+5)(-x-3)}$

$\Rightarrow 1+(x+5{)}^2(x+3{)}^2+2(x+5)(x+3)<16(x+5)(-x-3)$

$\Rightarrow \left[\right(x+5\left)\right(x+3){]}^2+18(x+5\left)\right(x+3)+1<0$

$\Rightarrow \left[\right(x+5\left)\right(x+3\left)\right]-\frac{-18\pm \sqrt{3}20}{2}]<0$

$\Rightarrow [x^2+8x+15+9-\frac{\sqrt{3}20}{2}][x^2+8x+15+9+\frac{\sqrt{3}20}{2}]$

$\Rightarrow [x^2+8x+24-\frac{\sqrt{3}20}{2}][x^2+8x+24+\frac{\sqrt{3}20}{2}]$ - Equation 5

For $[x^2+8x+24-\frac{\sqrt{3}20}{2}=f(x)$

$\forall x\in R,ifD<0,a>0\Rightarrow$ it is always positive implies that $(x^2+8x+24-\frac{\sqrt{3}20}{2})$ is negative.

$\Rightarrow (x^2+8x+24-4\sqrt{5})<0$

$\Rightarrow x-\frac{-8\pm \sqrt{32+16\sqrt{5}}}{2}<0$

$\Rightarrow (x+4+2\sqrt{2+\sqrt{5}})\left(\right(x+4-2\sqrt{2+\sqrt{5}})<0$ - Equation 6

In Figure 1, Range is $x\in [-5,-3]$