<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] 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 $x \in (0,3).$
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