<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] 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.
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