[proofplan]
The proof is an algebraic comparison between the quadratic cost inequality and the classical dot-product inequality. For a finite cycle in $\Gamma$, expand each quadratic term into $|x_i|^2$, $|y_i|^2$, and $x_i \cdot y_i$. The cyclic permutation of the second coordinates preserves the total $|x_i|^2$ and $|y_i|^2$ contributions, so the cost inequality reduces exactly to a dot-product inequality. A final reindexing identifies that dot-product inequality with classical cyclical monotonicity.
[/proofplan]
[step:Fix a finite cycle and expand the quadratic cost]
Fix $N \in \mathbb{N}$ and a finite family $(x_i,y_i)_{i=1}^N \subset \Gamma$. Throughout this proof, indices are taken cyclically, so $x_{N+1}:=x_1$ and $y_{N+1}:=y_1$.
For each $i \in \{1,\dots,N\}$, the Euclidean norm identity gives
\begin{align*}
c(x_i,y_i)=\frac{1}{2}|x_i-y_i|^2=\frac{1}{2}|x_i|^2+\frac{1}{2}|y_i|^2-x_i\cdot y_i.
\end{align*}
Similarly,
\begin{align*}
c(x_i,y_{i+1})=\frac{1}{2}|x_i-y_{i+1}|^2=\frac{1}{2}|x_i|^2+\frac{1}{2}|y_{i+1}|^2-x_i\cdot y_{i+1}.
\end{align*}
[/step]
[step:Cancel the terms unchanged by cyclic permutation]
Summing the two identities over $i \in \{1,\dots,N\}$ gives
\begin{align*}
\sum_{i=1}^N c(x_i,y_i)
=
\frac{1}{2}\sum_{i=1}^N |x_i|^2
+
\frac{1}{2}\sum_{i=1}^N |y_i|^2
-
\sum_{i=1}^N x_i\cdot y_i
\end{align*}
and
\begin{align*}
\sum_{i=1}^N c(x_i,y_{i+1})
=
\frac{1}{2}\sum_{i=1}^N |x_i|^2
+
\frac{1}{2}\sum_{i=1}^N |y_{i+1}|^2
-
\sum_{i=1}^N x_i\cdot y_{i+1}.
\end{align*}
Because $(y_{i+1})_{i=1}^N$ is only a cyclic reordering of $(y_i)_{i=1}^N$, we have
\begin{align*}
\sum_{i=1}^N |y_{i+1}|^2=\sum_{i=1}^N |y_i|^2.
\end{align*}
Therefore the cost inequality
\begin{align*}
\sum_{i=1}^N c(x_i,y_i) \leq \sum_{i=1}^N c(x_i,y_{i+1})
\end{align*}
is equivalent to
\begin{align*}
-\sum_{i=1}^N x_i\cdot y_i \leq -\sum_{i=1}^N x_i\cdot y_{i+1}.
\end{align*}
Multiplying by $-1$ reverses the inequality, so this is equivalent to
\begin{align*}
\sum_{i=1}^N x_i\cdot y_i \geq \sum_{i=1}^N x_i\cdot y_{i+1}.
\end{align*}
[guided]
The point of using the quadratic cost is that its expansion separates into terms depending only on $x_i$, terms depending only on $y_i$, and one interaction term. For the original matching $(x_i,y_i)$, we have
\begin{align*}
c(x_i,y_i)=\frac{1}{2}|x_i|^2+\frac{1}{2}|y_i|^2-x_i\cdot y_i.
\end{align*}
For the cyclically shifted matching $(x_i,y_{i+1})$, the same identity gives
\begin{align*}
c(x_i,y_{i+1})=\frac{1}{2}|x_i|^2+\frac{1}{2}|y_{i+1}|^2-x_i\cdot y_{i+1}.
\end{align*}
Now sum over $i$. The $x$-terms are identical on both sides:
\begin{align*}
\frac{1}{2}\sum_{i=1}^N |x_i|^2
\end{align*}
appears in both sums. The $y$-terms are also identical after summation, because the list $y_2,\dots,y_N,y_1$ is just a cyclic reordering of $y_1,\dots,y_N$. Hence
\begin{align*}
\sum_{i=1}^N |y_{i+1}|^2=\sum_{i=1}^N |y_i|^2.
\end{align*}
Thus the only part of the cost comparison that survives is the interaction term. Substituting the expansions into
\begin{align*}
\sum_{i=1}^N c(x_i,y_i) \leq \sum_{i=1}^N c(x_i,y_{i+1})
\end{align*}
and cancelling the common $x$- and $y$-sums gives
\begin{align*}
-\sum_{i=1}^N x_i\cdot y_i \leq -\sum_{i=1}^N x_i\cdot y_{i+1}.
\end{align*}
Finally, multiplying both sides by $-1$ reverses the inequality, yielding
\begin{align*}
\sum_{i=1}^N x_i\cdot y_i \geq \sum_{i=1}^N x_i\cdot y_{i+1}.
\end{align*}
This is the exact algebraic form in which the quadratic cost inequality becomes a monotonicity inequality.
[/guided]
[/step]
[step:Reindex the dot-product inequality into the classical form]
The inequality
\begin{align*}
\sum_{i=1}^N x_i\cdot y_i \geq \sum_{i=1}^N x_i\cdot y_{i+1}
\end{align*}
is equivalent to
\begin{align*}
\sum_{i=1}^N x_i\cdot y_i-\sum_{i=1}^N x_i\cdot y_{i+1} \geq 0.
\end{align*}
Reindex the second sum by replacing $i+1$ with $i$, using the cyclic convention. Then
\begin{align*}
\sum_{i=1}^N x_i\cdot y_{i+1}=\sum_{i=1}^N x_{i-1}\cdot y_i,
\end{align*}
where $x_0:=x_N$. Hence the preceding inequality is equivalent to
\begin{align*}
\sum_{i=1}^N y_i\cdot (x_i-x_{i-1}) \geq 0.
\end{align*}
Replacing the cyclic index $i$ by $i+1$ gives the equivalent inequality
\begin{align*}
\sum_{i=1}^N y_i\cdot (x_{i+1}-x_i) \leq 0.
\end{align*}
Writing $p_i:=y_i$ for each $i \in \{1,\dots,N\}$, this becomes
\begin{align*}
\sum_{i=1}^N p_i\cdot (x_{i+1}-x_i) \leq 0,
\end{align*}
which is precisely the classical cyclical monotonicity inequality.
[/step]
[step:Conclude the equivalence of the two notions]
If $\Gamma$ is $c$-cyclically monotone, then the cost inequality holds for every $N \in \mathbb{N}$ and every finite family $(x_i,y_i)_{i=1}^N \subset \Gamma$. By the preceding steps, for every such family the classical inequality
\begin{align*}
\sum_{i=1}^N y_i\cdot (x_{i+1}-x_i) \leq 0
\end{align*}
holds. Thus $\Gamma$, with second coordinate denoted by $p=y$, is cyclically monotone in the classical sense.
Conversely, if $\Gamma$ is classically cyclically monotone, then the same chain of equivalences in reverse gives
\begin{align*}
\sum_{i=1}^N c(x_i,y_i) \leq \sum_{i=1}^N c(x_i,y_{i+1})
\end{align*}
for every $N \in \mathbb{N}$ and every finite family $(x_i,y_i)_{i=1}^N \subset \Gamma$. Therefore $\Gamma$ is $c$-cyclically monotone. This proves the equivalence.
[/step]
