<s>[INST] 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?
 [/INST] 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 $$(-1,3)\setminus\{0\}$$
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