Properties of 3-vector dot product I've been playing around with an extension of a dot product to three vectors, as set forth in this question.  Basically, if you have three vectors A, B, and C, then you could compute the following
$$TD(A,B,C) = \sum_{i=1}^N a_i b_i c_i$$
where TD means "triple dot." I realize this isn't an accepted notation but it is useful for this question.
I'm curious if anyone has shown that
$$TD(A,B,C)\leqslant \lVert A \rVert \cdot\lVert B \rVert\cdot \lVert C \rVert.$$
I suppose it might involve an extension of the Rearrangement inequality to three vectors.
 A: Upon Jeff's request here the solution I provided. The validity of the proof as it is right now requires that:  $$\sum_{1\leqslant 1<i<j\leqslant n}b_ic_j\geqslant 0.$$

Using Lagrange's identity, for $(a_i)$ and $(b_ic_i)$, one has: 
$$\textrm{TD}(A,B,C)^2=\|A\|^2\cdot\sum_{i=1}^n(b_ic_i)^2-\underbrace{\sum_{1\leqslant i,j\leqslant n}^n(a_ib_jc_j-a_jc_ib_i)^2}_{\geqslant 0}.$$
Hence, one derives:
$$\textrm{TD}(A,B,C)^2\leqslant\|A\|^2\cdot\sum_{i=1}^n(b_ic_i)^2.\tag{1}$$
Besides, one has: $$\langle B,C\rangle^2=\sum_{i=1}^n(b_ic_i)^2+\underbrace{2\sum_{1\leqslant i<j\leqslant n}b_ic_j}_{\geqslant 0}.$$
Therefore, one derives: $$\sum_{i=1}^n(b_ic_i)^2\leqslant\langle B,C\rangle^2.\tag{2}$$
Using $(1)$ and $(2)$, one has: $$|\textrm{TD}(A,B,C)|\leqslant \|A\|\cdot|\langle B,C\rangle|.\tag{3}$$
Finally, using Cauchy-Schwarz inequality, one gets: $$|\textrm{TD}(A,B,C)|\leqslant\|A\|\cdot\|B\|\cdot\|C\|.$$
Remark. Inequality $(3)$ is stronger than the one you are interested in.
A: A short proof can be given using the Frobenius norm: As already pointed out in the question you linked to, we have
$$
 \mathrm{TD}(a,b,c) = a^T \cdot \mathrm{diag}(b_1, \dotsc, b_n) \cdot c
$$
where $\mathrm{diag}(b_1, \dotsc, b_n)$ denotes the diagonal matrix with diagonal entries $b_1, \dotsc, b_n$. Using the submultiplicativity of the Frobenius norm we find that
\begin{align*}
 \mathrm{TD}(a,b,c)
 &\leq |\mathrm{TD}(a,b,c)|
 = \|\mathrm{TD}(a,b,c)\|_F
 = \|a^T \cdot \mathrm{diag}(b_1, \dotsc, b_n) \cdot c\|_F \\
 &\leq \|a^T\|_F \cdot \|\mathrm{diag}(b_1, \dotsc, b_n)\|_F \cdot \|c\|_F
 = \|a\| \cdot \|b\| \cdot \|c\|,
\end{align*}
where $\|\cdot\|_F$ denotes the Frobenius norm.
Another proof can be given by using the Cauchy-Schwarz inequality: We can assume w.l.o.g. that $a_i, b_i, c_i \geq 0$ for every $1 \leq i \leq n$. Because $b_i \geq 0$ for all $1 \leq i \leq n$ we find that the bilinear form $\langle \cdot, \cdot \rangle_b$ defined via
$$
 \langle x,y \rangle_b
 = x^T \cdot \mathrm{diag}(b_1, \dotsc, b_n) \cdot y
 = \mathrm{TD}(x,b,y)
$$
is symmetric and positive semidefinite with
$$
 \|x\|_b
 = \sqrt{\langle x,x \rangle_b}
 = \sqrt{ \sum_{i=1}^n b_i x_i^2 }.
$$
Notice that for all $1 \leq i,k \leq n$ we have $b_i b_k \leq \sum_{j=1}^n b_j^2$: If $i = k$ this is clear and if $i \neq k$ then
$$
 b_i b_k \leq 2 b_i b_k \leq b_i^2 + b_k^2 \leq \sum_{j=1}^n b_j^2.
$$
So from the Cauchy-Schwarz inequality it follows that
\begin{align*}
 \mathrm{TD}(a,b,c)
 &= \langle a, c \rangle_b
 \leq \|a\|_b \cdot \|c\|_b
 = \sqrt{ \sum_{i=1}^n b_i a_i^2} \cdot \sqrt{\sum_{k=1}^n b_k c_k^2 } \\
 &= \sqrt{ \sum_{i,k=1}^n a_i^2 b_i b_k c_k^2}
 \leq \sqrt{ \sum_{i,j,k=1}^n a_i^2 b_j^2 c_k^2 } \\
 &= \sqrt{\sum_{i=1}^n a_i^2} \cdot \sqrt{\sum_{j=1}^n b_j^2}
    \cdot \sqrt{\sum_{k=1}^n c_k^2}
 = \|a\| \cdot \|b\| \cdot \|c\|.
\end{align*}
It is worth noticing that using the approach via the Frobenius norm we can also directly generalize our results to arbitrary $x^1, \dotsc, x^m \in \mathbb{C}^n$, in the sense that
\begin{align*}
 \left|\sum_{i=1}^n x^1_i \dotsm x^m_i\right|
 &= \left\|
    (x^1)^T
    \cdot \mathrm{diag}(x^2_1, \dotsc, x^2_n)
    \dotsm \mathrm{diag}(x^{m-1}_1, \dotsc, x^{m-1}_n)
    \cdot x^m
   \right\|_F \\
 &\leq
    \|(x^1)^T\|_F
    \cdot \|\mathrm{diag}(x^2_1, \dotsc, x^2_n)\|_F
    \dotsm \|\mathrm{diag}(x^{m-1}_1, \dotsc, x^{m-1}_n)\|_F
    \cdot \|x^m\|_F \\
 &= \|x^1\| \dotsm \|x^m\|.
\end{align*}
