$\frac{2x^2}{x + 1 - \sqrt{2x + 1}} - (2x + 9) < 0$ $\frac{\sqrt{2x + 1}(2x + 9) - (11x + 9)}{x + 1 - \sqrt{2x + 1}} < 0$ Solving this we get $x < \frac{45}{8}$ with $x \neq 0$.

We also have $2x+1\geq 0$ $x\geq -\frac12$ So the solution is $-\frac12\leq x<0$ or $0<x<\frac{45}{8}$

Let $2x+1 = y^{2}$, where $y\in\mathbb{R}^{+}$. The given inequality is equivalent with \[(y+1)^{2}= \frac{(y-1)^{2}(y+1)^{2}}{(y-1)^{2}}= \frac{(y^{2}-1)^{2}}{(1-y)^{2}}= \frac{(2x)^{2}}{\left(1-\sqrt{2x+1}\right)^{2}}< 2x+9 = y^{2}+8.\] Therefore, we must have that $2y+1<8$, i.e. $y<\frac{7}{2}$. And now it's easy to proceed and get $x\in\left[\frac{-1}{2},\frac{45}{8}\right[\setminus\{0\}$.

First of all, $ x>-\frac{1}{2}$ Multiply both sides by $ (1-\sqrt{2x+1})^{2}$, we get: $ (1+\sqrt{2x+1})^{2}< 2x+9$ $ 1+2x+1+2\sqrt{2x+1}< 2x+9$ $ 2\sqrt{2x+1}< 7$ $ \sqrt{2x+1}< \frac{7}{2}$ $ 2x+1< \frac{49}{4}$ $ 2x< \frac{45}{4}$ $ x< \frac{45}{8}$ Therefore, $ -\frac{1}{2}< x< \frac{45}{8}$

Note that the function $f(x)=(1+\sqrt{1+2x})^2-2x-9=2\sqrt{1+2x}-7$ is increasing function. Also $f(\frac{45}{8})=0$

We know that $2x+1\ge0$ so $x\ge-\frac12$. We also know that $1-\sqrt{2x+1}\ne0$ so $x\ne0$. We can multiply both sides of $\frac{4x^2}{(1-\sqrt{2x+1})^2}<2x+9$ by $\frac{(1+\sqrt{2x+1})^2}{(1+\sqrt{2x+1})^2}$ for a difference of squares to get $(1+\sqrt{2x+1})^2<2x+9$. We can now solve the inequality to get \begin{align*} 1+2\sqrt{2x+1}+2x+1 &< 2x+9 \\ 2\sqrt{2x+1} &< 7 \\ \sqrt{2x+1} &< \frac72 \\ 2x+1 &< \frac{49}4 \\ 2x &< \frac{45}4 \\ x &< \frac{45}8 \end{align*}so our answer is $-\frac12\le x<0$ and $0<x<\frac{45}8$.

OlympusHero wrote:

x cannot be 0

Henceforth assume $x\geq-\tfrac{1}{2}$ and $x\neq0$ to have $\sqrt{1+2x}\in\mathbb{R}$ and $(1-\sqrt{1+2x})^2\neq0$, respectively. As $(1-\sqrt{1+2x})^2$ is positive, we can multiply by it to get $$4x^2<(1-\sqrt{1+2x})^2(2x+9)=(2+2x-2\sqrt{1+2x})(2x+9)=4x^2+22x+18-(4x+18)\sqrt{1+2x},$$or, equivalently, $2x^2<2x^2+11x+9-(2x+9)\sqrt{1+2x}\iff(2x+9)\sqrt{1+2x}<11x+9$. Since all terms are positive as $x\geq-\tfrac{1}{2}$, we can square both sides to obtain an equivalent formulation: $(2x+9)^2(1+2x)<(11x+9)^2$. Now, the LHS equals $$(2x+9)^2(2x+1)=(4x^2+36x+81)(2x+1)=8x^3+76x^2+198x+81,$$while the RHS equals $121x^2+198x+81$, so we have $$8x^3+76x^2+198x+81<121x^2+198x+81\iff8x^3-45x^2<0.$$Dividing by $x^2$ since it is positive (recall the assumption $x\neq0$) gives $8x-45<0\iff x<\tfrac{45}{8}$. Therefore, the solutions to the inequality are given by $\boxed{x\in[-\tfrac{1}{2},\tfrac{45}{8})/\{0\}}$. $\blacksquare$

Rearranging we have that $(2x+9)(2x-2\sqrt{2x+1}+2)>4x^2.$ If we expand and simplify we get $\sqrt{2x+1}(4x+18)<22x+18$, now we square both sides to get that $(2x+1)(4x+18)^2<(22x+18)^2 \implies 32x^3-180x^2<0$, so $\boxed{-\frac{1}{2} \leq x < 0, 0 < x < \frac{45}{8}}.$