Find real solutions to the inequality with nested radicals

[anemone]

Find all the real values of $x$ which satisfy $\dfrac{x^2}{(x+1-\sqrt{x+1})^2}<\dfrac{x^2+3x+18}{(x+1)^2}$.

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[anemone]

Congratulations to lfdahl for his correct solution!:)

lfdahl's solution:

Spoiler

We're asked to find the range of $x$ for:

\[\frac{x^2}{\left (x+1-\sqrt{x+1} \right )^2}<\frac{x^2+3x+18}{(x+1)^2}\]

Before we proceed, note that

1. The denominator from the RHS tells us $x+1\ne 0$, this says $x \ne -1$

2. The square root of $x+1$ in the denominator of the LHS, we must have $x+1\ge 0$, this will give us $x\ge -1$ but from point above, we know that $-1$ must be excluded so in this case we have $x >-1$.

3. \[\lim_{x\rightarrow 0}\left \{ \frac{x^2}{\left (x+1-\sqrt{x+1} \right )^2} \right \}=-4,\]
\[ thus \: \: \: x =0\: \:also \: \:belongs \: \:to \: \:the \: \:set \: \: of \: \: real \: \: solutions \]

Now, if we
\[Let \: \: \sqrt{x+1}\doteq u:\\\\ \frac{(u^2-1)^2}{(u^2-u)^2}< \frac{(u^2-1)^2+3(u^2-1)+18}{u^4}\\\\ \Rightarrow (u^2-1)^2u^4< \left ( (u^2-1)^2+3(u^2-1)+18 \right )(u^2-u)^2\\\\ \Rightarrow u^5-2u^4+u^3-8u^2+16u-8<0.\]
\[P(u)\: \: has\: \: simple\: \: roots.\: Using \: \: polynomial\: \: division\: \: gives:\\\\ (u-1)^2(u-2)(u^2+2u+4)<0.\]
\[So\: \:the \: \: real \: \: solutions\: \: are: \sqrt{x+1}< 2\Rightarrow x <3\]

But since $x >-1$, we can conclude that \[x \in (-1,3).\]