What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(x+1)({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{2}+3x+18}{({x+1})^{2}}$
$\frac{{x}^{2}(x+1)}{({\sqrt{x+1}}-1)^{2}}< {x}^{2}+3x+18$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{({x}^{2}+3x+18)({\sqrt{x+1}-1})^{2}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{3}+5{x}^{2}+24x+36-(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}-{x}^{3}-5{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4(x+3)^{2}+2({x}^2+3x+18)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
Does anyone have a hint for the solution?
 A: We need
$$\left\{\begin{aligned} & x+1 > 0  \\& x+1 \ne \sqrt{x+1} \end{aligned}\right. \Rightarrow  \left\{\begin{aligned} & x > -1  \\& x \ne 0\end{aligned}\right. \quad (1)$$
Setting $t = \sqrt{x+1} > 0,$ then $x=t^2-1,$ now
$$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2}  = \frac{(t+1)^2}{t^2},$$
and
$$\frac{{x}^{2}+3x+18}{({x+1})^{2}} = \frac{t^4+t^2+16}{t^4}.$$
Inequality become
$$\frac{(t+1)^2}{t^2}<\frac{t^4+t^2+16}{t^4},$$
or
$$\frac{2(2-t)[(t+1)^2+3]}{t^4}>0.$$
Therefore
$$0 < t < 2\Rightarrow 0 < \sqrt{x+1} < 2,$$
or
$$-1 < x < 3. \quad (2)$$
From $(1)$ and $(2)$ we get $x \in (-1,0)$ or $x \in (0,3).$
A: We have that
$$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}
\iff \frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2}-\frac{{x}^{2}+3x+18}{({x+1})^{2}}<0 $$
$$\iff \frac{({x}^{2}(x+1)^2-({x}^{2}+3x+18)(({x+1-\sqrt{x+1}})^2) }{({x+1-\sqrt{x+1}})^2(x+1)^2} < 0$$
$$\iff ({x}^{2}(x+1)^2-({x}^{2}+3x+18)(({x+1-\sqrt{x+1}})^2)<0$$
with the condition
$$x+1>0 \implies {x+1-\sqrt{x+1}}\neq 0$$
then by $x+1=y>0$
$$\iff ((y-1)^2y^2-((y-1)^2+3(y-1)+18)(({y-\sqrt{y}})^2)<0$$
$$\iff 2y(\sqrt y-1)^2(y+2\sqrt y+4)(\sqrt y-2)<0$$
A: Note that $$x=(x+1)-1=(\sqrt{x+1}-1)(\sqrt{x+1}+1)$$  Thus the left-hand side simplifies to $$(\sqrt{x+1}+1)^2<\frac{x^2+3x+18}{x+1}$$  This doesn't look promising, but let's see if we can get an approximate solution by pulling out the leading term on the right-hand side.  Well, $x^2+3x+18=(x+1)(x+2)+16$.  Thus partial fractions gives $$(\sqrt{x+1}+1)^2<x+2+\frac{16}{x+1}$$
Next, if we can isolate the square root, then we can square both sides and turn this into a polynomial problem.  When we expand out the square on the left-hand side, we discover a miracle: $$(\sqrt{x+1}+1)^2=(x+1)+1+2\sqrt{x+1}=x+2+2\sqrt{x+1}$$  The $x+2$s cancel!  So $$2\sqrt{x+1}<\frac{16}{x+1}$$
I trust you can solve it from here.
A: The domain gives $x>-1$ and we need to solve $$\frac{x^2}{(\sqrt{x+1}-1)^2}<\frac{x^2+3x+18}{x+1}$$ or
$$x^2(x+1)<(x+2-2\sqrt{x+1})(x^2+3x+18)$$ or
$$2(x+3)^2>(x^2+3x+18)\sqrt{x+1}$$
or $$4(x+3)^4>(x^2+3x+18)^2(x+1)$$ or after factoring
$$x^2(3-x)(x^2+6x+21)>0,$$ which gives $$(-1,3)\setminus\{0\}$$
