<s>[INST] Solve for $x: \dfrac{x+6}{x-6}\left(\dfrac{x-4}{x+4}\right)^2+\dfrac{x-6}{x+6}\left(\dfrac{x+9}{x-9}\right)^2\lt\dfrac{2x^2+72}{x^2-36}$ 
Solve for $x: \dfrac{x+6}{x-6}\left(\dfrac{x-4}{x+4}\right)^2+\dfrac{x-6}{x+6}\left(\dfrac{x+9}{x-9}\right)^2\lt\dfrac{2x^2+72}{x^2-36}$

I tried taking LCM on LHS, but it's getting quite unwieldy: $$\frac{(x+6)^2(x-4)^2(x-9)^2+(x-6)^2(x+9)^2(x+4)^2}{(x^2-36)(x+4)^2(x-9)^2}\lt\frac{2(x^2+36)}{x^2-36}$$
Maybe $x^2-36$ can be cancelled from the denominators, provided $x\lt-6$ or $x\gt6$.
For $x\in(-6,6)$, the sign of inequality would change. But still the inequality remains in an unmanageable form. The symmetry of the question suggests that there should be some neater approach to it. But not able to figure that out.
EDIT: @Parasseux Nguyen gave the hint that $2(x^2+36)=(x+6)^2+(x-6)^2$. Here is my try after that:
$$\frac{(x+6)^2(x-4)^2(x-9)^2+(x-6)^2(x+9)^2(x+4)^2}{(x^2-36)(x+4)^2(x-9)^2}-\frac{(x^2+6)^2+(x^2-6)^2}{x^2-36}\lt0$$
$$\frac{(x+6)^2(x-4)^2(x-9)^2+(x-6)^2(x+9)^2(x+4)^2-(x^2+6)^2(x+4)^2(x-9)^2-(x^2-6)^2(x+4)^2(x-9)^2}{(x^2-36)(x+4)^2(x-9)^2}\lt0$$
$$\frac{8x(x+6)^2(x-9)^2+18x(x-6)^2(x+4)^2}{(x^2-36)(x+4)^2(x-9)^2}\lt0$$
Taking $2x$ common in the numerator still gives quartic in the bracket, with some not so neat coefficients. I wonder if this is the only way to go about it.
 [/INST] The inequality is satisfied if
$$\begin{align} x & \;<\; -6\tag{1a}\\
\text{or}\quad\frac 65\big(1-\sqrt{26}\,\big)\,=\, -4.92 \;<\; x & \;<\; 0\;\;\text{but }\; x\neq -4\tag{1b}\\
\text{or}\quad 6 \;<\; x & \;<\; 7.32\,=\,\frac 65\big(1+\sqrt{26}\,\big)\,.\tag{1c}
\end{align}$$
To arrive at this result, exploit Paresseux Nguyen's helpful spot in the comments concerning the RHS, rescale by substituting $s=\frac x6$, and regroup all terms on one side:
$$\frac{s+1}{s-1}\:\left(1-\left(\frac{\frac 32s-1}{\frac 32s+1}\right)^2\right)\ +\ \frac{s-1}{s+1}\:\left(1-\left(\frac{\frac 23s+1}{\frac 23 s-1}\right)^{2}\right)\tag{2}$$
We have to find out for which values of $s$ the preceding function is strictly positive.
Let $t=\frac{s+1}{s-1}$ (which is a Möbius transformation).
It defines a homeomorphism from $\,\mathbb R\backslash\{1\}$ to itself, and
it is its own inverse, thus $s=\frac{t+1}{t-1}$. By courtesy of desmos.com, here is its graph including the asymptotes in orange:

Changing the variable in $(2)$ from $s$ to $t$ yields
$$f(t) := t\left(1-\left(\frac{t+5}{5t+1}\right)^{2}\right)\ +\ \frac{1}{t}\left(1-\left(\frac{5t-1}{t-5}\right)^{2}\right)\tag{3}$$
The function $f$ has two properties which are our friends:
Firstly
$$f\left(-\frac 1t\right) \:=\: -f(t)\;\text{ for all }
t\in\mathbb R\backslash\{-\tfrac 15,0,5\}\tag{4}$$
This is checked in a straightforward way, and can be used to cut down the analysis to values $t>0$.
And secondly, using $\,p^2-q^2=(p+q)(p-q)$ repeatedly, $f(t)$ can be completely factorised:
$$\begin{align} f(t) & \;=\; t\cdot\frac{(5t+1+t+5)(5t+1-t-5)}{(5t+1)^2}
 +\frac 1t\cdot\frac{(t-5+5t-1)(t-5-5t+1)}{(t-5)^2}\\[2ex]
 & \;=\;\frac{6\cdot 4(t+1)}{t\,(5t+1)^2}(t-1)
\underbrace{\left(t+\frac{5t+1}{t-5}\right)\left(t-\frac{5t+1}{t-5}\right)}_{\frac{(t^2+1)\left((t-5)^2-26\right)}{(t-5)^2}}
\\[2ex]
 & \;=\;\frac{24(t+1)(t^2+1)(t+\sqrt{26}-5)}{t\,(5t+1)^2}
\cdot\frac{(t-1)\big(t-5-\sqrt{26}\big)}{(t-5)^2}
\tag{5}\end{align}$$
Assume $t>0$, then the first factor is positive, and from the second one we read off that $\,0<f(t)\,$ if
$$\begin{align} 0 \;<\; & t\;<\; 1\tag{6a}\\
\text{or}\quad 5+\sqrt{26} \;<\; & t\,.\tag{6c}
\end{align}$$
If $\,1<t<5+\sqrt{26}\,$ but $\,t\neq 5$, then $\,f(t)<0$. By $(4)$ we conclude that $\,0<f(t)\,$ also holds if
$$\,-1<t<5-\sqrt{26}=-\frac 1{5+\sqrt{26}}\quad\text{and excluding }\:t=-\frac 15\,.\tag{6b}$$
Finally, rollback the variables from $t$ via $s$ to $x$ to obtain the result given at the outset. The letters in the tags indicate how the intervals w.r.to the different variables correspond to each other.
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