I don't think application of C-S is the right approach, at least not directly. However, note the following (I changed your variable from $x$ to $t$ since I feel $x$ is used for two different meanings):
$$A(t)+B(t)+C(t)=t^{-d/3}\sum_{k\in M}t^k.$$
If we replace $t$ by $\zeta_l=e^{\frac{2\pi i l}{n}}$, then it follows that
$$A(\zeta_l)+B(\zeta_l)+C(\zeta_l)=0$$
for $l=1,2,\ldots,n-1$ (recalling that $M=\{1,2,\ldots,n\}$). From the identity $$p^3+q^3+r^3-3pqr=(p+q+r)(p^2+q^2+r^2-qr-rp-pq),$$ it follows that $p^3+q^3+r^3=3pqr$ if $p+q+r=0$, and so
$$\big(A(\zeta_l)\big)^3+\big(B(\zeta_l)\big)^3+\big(C(\zeta_l)\big)^3=3\cdot A(\zeta_l)\cdot B(\zeta_l)\cdot C(\zeta_l).$$
On the other hand,
$$\big(A(\zeta_0)\big)^3+\big(B(\zeta_0)\big)^3+\big(C(\zeta_0)\big)^3=\big(A(1)\big)^3+\big(B(1)\big)^3+\big(C(1)\big)^3$$
satisfies $A(1),B(1),C(1)\geq 0$. Therefore, we can apply AM-GM (or C-S) to get
$$\big(A(\zeta_0)\big)^3+\big(B(\zeta_0)\big)^3+\big(C(\zeta_0)\big)^3\geq 3\cdot A(1)\cdot B(1)\cdot C(1)=3\cdot A(\zeta_0)\cdot B(\zeta_0)\cdot C(\zeta_0).$$
Therefore,
$$\sum_{l=0}^{n-1}\Big(\big(A(\zeta_l)\big)^3+\big(B(\zeta_l)\big)^3+\big(C(\zeta_l)\big)^3\Big)\geq \sum_{l=0}^{n-1}\Big(3\cdot A(\zeta_l)\cdot B(\zeta_l)\cdot C(\zeta_l)\Big),$$
which is equivalent to the required result.
Indeed, it can be shown that
\begin{align}2|S_1|-|S_2|&=2\Big(\big(A(1)\big)^2+\big(B(1)\big)^2+\big(C(1)\big)^2-B(1)\cdot C(1)-C(1)\cdot A(1)-A(1)\cdot B(1)\Big)\\&=\big(B(1)-C(1)\big)^2+\big(C(1)-A(1)\big)^2+\big(A(1)-B(1)\big)^2.\end{align}
You are asking to maximize
$$\begin{equation}\begin{aligned}
f(a_1,a_2,...,a_n) & = \vert a_1-a_2\vert+\vert a_2-a_3\vert+...+\vert a_{n-1}-a_n\vert \\
& = \sum_{i=1}^{n-1}\vert a_{i} - a_{i+1} \vert
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
If $a_i \ge a_{i+1}$, then $\vert a_i - a_{i+1} \vert = a_i - a_{i+1}$, else $\vert a_{i} - a_{i+1} \vert = a_{i+1} - a_{i}$. In either case, when you remove the absolute value signs to get the corresponding equivalent value, you have one sequence term being added and another one subtracted. Thus, over the $n - 1$ absolute value terms, you would have $n - 1$ sequence terms being added and $n - 1$ sequence terms being subtracted.
Note that $a_1$ and $a_n$ are used just once each. Also, $a_i$, for $i \le 2 \le n - 1$, are being used twice. The maximum possible value of \eqref{eq1A} would occur if all of the $n - 1$ sequence terms being added were the largest ones and with the $n - 1$ sequence terms being subtracted being the smallest ones, and with the special case of the $2$ terms of $a_1$ and $a_n$, as they are used just once, being the $2$ largest terms among those being subtracted.
This can be done. In particular, as stated above, have the $2$ "middle" (with ones at either side if $n$ is even) values be at the start and end of the sequence. Then at the odd indices from the start, i.e., $3,5,7,\ldots$ and the even indices back from the end, i.e., $n-2,n-4,n-6,\ldots$, you alternate back & forth placing the remaining smallest values from the smallest up. Also, at the even indices from the start, i.e., $2,4,6,\ldots$, and the odd indices back from the end, i.e., $n-1,n-3,n-5,\ldots$, you alternate back & forth placing the largest values from the largest down.
The top "half" of the values will always be beside a value from the bottom "half" of the values on either side of them in the expression in \eqref{eq1A} and, thus, the top half values will be the ones added and the bottom half values will be the ones subtracted. I'm using "half" in quotes because the $2$ ones near the middle are handled specially and there's a small issue with odd versus even values of $n$ to deal with.
As you say you used to be a contest math lover as a high school student and have majored in math now, I trust you can finish the rest yourself.
Best Answer
To demonstrate the inequality is wrong, all you need is a counterexample ... so given your analysis, can you come up with one? That should show it pretty quickly.