Useful identity, $|ac - bd| \leq \max(a,b) | c-d| + \max(c,d)|a-b|$, how to prove?

Question

For $a,b,c,d \geq 0$, the following identity showed up in a paper I was reading.

$$ |ac - bd| \leq \max(a,b) | c-d| + \max(c,d)|a-b| $$

Well, to be precise, it's simply that they cited a step in a proof which seemed to imply this identity (the form they used actually used $L^\infty$ norms over the functions in this identity).

The statement seems somewhat intuitive, and it seems to be correct, but I was struggling to actually formulate a proof. Does anyone have any pointers?

Answer 1

It is enough to use triangular inequality: $$ \begin{align} |ac - bd| &\leq |ac-ad| + |ad-bd| \\ &\leq a|c-d| + d|a-b| \\ &\leq \max(a,b) | c-d| + \max(c,d)|a-b| \end{align} $$

Answer 2

$$\begin{align} |ac-bd| &= |a(c-d) + d(a-b)| \qquad (-ad + da = 0) \\ &\leq a|c-d| + d|a-b| \qquad (\textrm{by triangle inequality}) \\ &\leq \max(a,b)|c-d| + \max(c,d)|a-b|. \end{align}$$