Inequality of norms $\| \frac{a}{d}-b\|$ < $\| \frac{a}{d}-c\| \Leftrightarrow \| a-bd\|$ < $\| a-cd\|$ How would one show or disproof the following statement
$\forall a,b,c,d \in \mathbb{C}\setminus \{0\} :$
$\left\| \frac{a}{d}-b\right\|_p$ < $\left\| \frac{a}{d}-c \right\|_p \Leftrightarrow \left\| a-bd\right\|_p$ < $\left\| a-cd \right\|_p$
 A: Hint: Note that $$a-bd=d\left(\frac{a-bd}{d}\right)=d \left(\frac{a}{d}{-b}\right).$$
A: p-norm is defined as
$$
||{\mathbf x}||_p  \equiv \left( {\sum_{n \mathop = 0}^d {|x_n|}^p} \right)^{1/p}
$$
For $x_n \in \mathbb{C}$, $|x_n|$ is usually defined by complex module
$$
|x_n| = |A + Bi| \equiv \sqrt{A^2 + B^2}
$$
Because $a,b,c,d \in \mathbb{C}$ in the question, I assume all the p-norm values in the inequation are the p-norms of the one-dimension vector which are scalers. First, we have to prove $||ad||_p = ||a||_p ||d||_p$.
For $A, B, C, D \in \mathbb{R}$
$$\begin{align*}
||(A + Bi)(C + Di)||_p
&= ||(AC - BD) + (AD + BC)i||_p \\
&= \left( |(AC - BD) + (AD + BC)i|^p \right)^{1/p} \\
&= |(AC - BD) + (AD + BC)i|
  && (|(AC - BD) + (AD + BC)i| \in \mathbb{R}) \\
&= \sqrt{(AC - BD)^2 + (AD + BC)^2} \\
&= \sqrt{A^2 C^2 + B^2 D^2 - 2ABCD + A^2 D^2 + B^2 C^2 + 2ABCD} \\
&= \sqrt{A^2 C^2 + B^2 D^2 + A^2 D^2 + B^2 C^2} \\
&= \sqrt{(A^2 + B^2)(C^2 + D^2)} \\
&= \sqrt{(A^2 + B^2)} \sqrt{(C^2 + D^2)} \\
&= |A + Bi| |C + Di| \\
&= ||A + Bi||_p ||C + Di||_p
\end{align*}$$
Using the identity above, we can prove the inequality.
$$\begin{align*}
\left\| a-bd\right\|_p \lt \left\| a-cd\right\|_p
&\Rightarrow \left\| \frac{a}{d}-b\right\|_p \left\|d\right\|_p \lt \left\| \frac{a}{d}-c\right\|_p \left\|d\right\|_p \\
&\Rightarrow \left\| \frac{a}{d}-b\right\|_p \lt \left\| \frac{a}{d}-c\right\|_p
  && \left( \left\|d\right\|_p \gt 0 \right)
\end{align*}$$
You can use the same method to prove the inversed part.
