<s>[INST] Inequality with paramater Find all values of $a$ such that the inequality
$$\frac{a-\left(\log_3x+2\sqrt6\log_x3-5\right)}{\left(3\cos\sqrt{x-9}-4\right)-a}\le0$$
has no solution.
My work:
I used $\frac bc< 0$ if only if $bc< 0$ and considered case $a=0, b\not=0$ but I don't know how to get answer.
 [/INST] First of all, we have to have
$$x\ge 9\tag1$$
Under $(1)$, we have, by AM-GM inequality,
$$\begin{align}&\left(\log_3x+2\sqrt6\log_x3-5\right)-\left(3\cos\sqrt{x-9}-4\right)\\\\&=\log_3x+\frac{2\sqrt 6}{\log_3x}-3\cos\sqrt{x-9}-1
\\\\&\ge 2\sqrt{2\sqrt 6}-3-1\gt 0\end{align}$$
from which we see that 
$$\left(\log_3x+2\sqrt6\log_x3-5\right)\gt\left(3\cos\sqrt{x-9}-4\right)\tag2$$
holds for all $x\ge 9$.
Considering $(2)$, we see that $$\frac{a-\left(\log_3x+2\sqrt6\log_x3-5\right)}{\left(3\cos\sqrt{x-9}-4\right)-a}\le0$$
is equivalent to
$$a\ge \left(\log_3x+2\sqrt6\log_x3-5\right)\quad\text{or}\quad a\lt \left(3\cos\sqrt{x-9}-4\right)$$
Note here that the former has at least one real solution if and only if $a\ge 2\sqrt{2\sqrt 6}-5$, and that the latter has at least one real solution if and only if $a\lt -1$.
Hence, the answer is
$$\color{red}{-1\le a\lt 2\sqrt{2\sqrt 6}-5}$$
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