$\frac{2(1+4a^2)}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$ for $0\leq a<\frac13$ and $x>\frac{(1+a)^2}8$ 
How to prove the inequality below？
$$\frac{2(1+4a^2)}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$$ holds for all $0\leqslant a<\frac{1}{3}$ and $x>\frac{(1+a)^2}{8}$.

This question has been bothering me for a while. Using numerical experiment, we found that $\frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}-\frac{2(1+4a^2)}{(12x-1)^3}$ is strictly increasing in $x\in (\frac{(1+a)^2}{8},+\infty)$, but it seems not easy to prove it.

Thanks to everyone!


Note that
$$(1 - a)^4 + (1 + a)^4 - 2(1 + 4a^2) = 4a^2 + 2a^4 \ge 0.$$
It suffices to prove that
$$\frac{(1 - a)^4 + (1 + a)^4}{(12x-1)^3}\leqslant \frac{(1-a)^4}{[12x-(1-a)^2]^3}+\frac{(1+a)^4}{[12x-(1+a)^2]^3}$$
or
\begin{align*}
 &\frac{1}{(12x - 1)^3}\\
 \le\,& \frac{(1 - a)^4}{(1 - a)^4 + (1 + a)^4}\cdot \frac{1}{[12x - (1 - a)^2]^3}
 + \frac{(1 + a)^4}{(1 - a)^4 + (1 + a)^4}\cdot \frac{1}{[12x - (1 + a)^2]^3}.
\end{align*}
Note that $u\mapsto \frac{1}{(12x - u)^3}$ is convex on $0 \le u \le 8x$.
By Jensen's inequality,
we have
$$\mathrm{RHS} \ge \frac{1}{(12x - A)^3} \ge \frac{1}{(12x - 1)^3}$$
where
$$A = \frac{(1 - a)^4}{(1 - a)^4 + (1 + a)^4}(1 - a)^2
+ \frac{(1 + a)^4}{(1 - a)^4 + (1 + a)^4}(1 + a)^2,$$
and we have used
$$A - 1 = \frac{(1 - a)^6 + (1 + a)^6 - (1 - a)^4 - (1 + a)^4}{(1 - a)^4 + (1 + a)^4} \ge 0.$$
We are done.
