It's Mathematics Olympiad! - 3

To solve this easily, rationalize LHS.

$\displaystyle \frac {4x^2}{(1-\sqrt {2x+1} )^2} \cdot \frac {(1+\sqrt {2x+1} )^2}{(1+\sqrt {2x+1} )^2} \le 2x+9 \implies (1+\sqrt {2x+1} )^2 \le 2x+9$

Hence,

$\displaystyle 0 \le \sqrt {2x+1} \le \frac {7}{2} \implies x \in \Big [ - \frac {1}{2} , \frac {45}{8} \Big ]$

Since $\sqrt{1+2x}\ge 0$ , the minimum value of $x$ is $-\dfrac{1}{2}$ .

In order to eliminate the square root in the denominator, let $x=\dfrac{a^2-1}{2}$ and $x\ge -\dfrac{1}{2}\implies a\ge 0$ .

So, we have, $\begin{aligned} \frac{4\left(\dfrac{a^2-1}{2}\right)^2}{\left(1-\sqrt{1+2\left(\dfrac{a^2-1}{2}\right)}\right)^2}&\le 2\left(\dfrac{a^2-1}{2}\right)+9 \\ \frac{(a^2-1)^2}{(1-a)^2}&\le a^2+8 \\ (a+1)^2&\le a^2+8 \\ a^2+2a+1&\le a^2+8 \\ a&\le \frac{7}{2} \\ \therefore \; x & \le \dfrac{\left(\dfrac{7}{2}\right)^2-1}{2}=\dfrac{45}{8} \end{aligned}$ Therefore, $a=-\dfrac{1}{2},\; b=\dfrac{45}{8}\implies a+b=\boxed{\dfrac{41}{8}}$