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.