Prove some properties of an alternate definition of the dot product

Instead of the usual definition of the dot product, suppose we define the dot product of $A = (a_1, \ldots, a_n)$ and $B = (b_1, \ldots, b_n)$ by the formula

$A \cdot B = \sum_{k=1}^n | a_k b_k |.$

Which of the properties of Theorem 12.2 (on page 451 of Apostol) are still valid under this new definition? Is the Cauchy-Schwarz inequality still valid?

Property (a) is still valid since
$A \cdot B = \sum_{k=1}^n | a_k b_k | = \sum_{k=1}^n | b_k a_k| = B \cdot A.$
Property (b) fails since
$A \cdot (B+C) = \sum_{k=1}^n | a_k (b_k +c_k) | = \sum_{k=1}^n | a_k b_k + a_k c_k |$

while

$A \cdot B + A \cdot C = \sum_{k=1}^n |a_k b_k| + \sum_{k=1}^n |a_k c_k| = \sum_{k=1}^n ( a_k b_k + a_k c_k).$

But, $| a_k b_k + a_k c_k|$ is not necessarily equal to $|a_k b_k| + |a_k c_k|$ .
Property (c) is still valid since
$\begin{align*} c(A \cdot B) &= c \sum_{k=1}^n |a_k b_k| \\ &= \sum_{k=1}^n c|a_k| |b_k| = (cA) \cdot B \\ &= \sum_{k=1}^n |a_k c|b_k| = A \cdot (cB). \end{align*}$
Property (d) is still valid since
$A \cdot A = \sum_{k=1}^n |a_k^2| = \sum_{k=1}^n a_k^2 > 0$

if $A \neq O$ .
Property (e) is still valid since
$A \cdot A = \sum_{k=1}^n |a_k^2| = \sum_{k=1}^n 0 = 0$

if $A = O$ .
The Cauchy-Schwartz inequality still holds since the formula is identical after squaring.