Maximum Value

Given that

$\begin{aligned} \sqrt{x-3} & \ge a - \sqrt{6-x} \\ \sqrt{x-3} + \sqrt{6-x} & \ge a \end{aligned}$

The inequality will have a solution if $a \le \max(\sqrt{x-3} + \sqrt{6-x})$ . Let us find the $\max(\sqrt{x-3} + \sqrt{6-x})$ using Cauchy-Schwarz inequality .

$\begin{aligned} (\sqrt{x-3} + \sqrt{6-x})^2 & \le (1+1)(x-3+6-x) = 6 \\ \sqrt{x-3} + \sqrt{6-x} & \le \sqrt 6 \end{aligned}$

Therefore, $a \le \max(\sqrt{x-3} + \sqrt{6-x}) = \sqrt 6 \approx \boxed{2.4495}$ .

When there are no solutions, it means that there exist some values $a$ such that

$\sqrt{x-3}+\sqrt{6-x}\geq a$

does not hold. Using $AM-QM$ inequality we get that

$\frac{\sqrt{x-3}+\sqrt{6-x}}{2}\geq \sqrt{\frac{3}{2}}$

always hold, with equality when $x=\frac{9}{2}$ The greatest value for $a$ is then $\sqrt{\frac{3}{2}}*2$