Given $a \le b < c < d$, show that $|a - d| > |b - c|$ This is for a real analysis class.
For real numbers $ a,b,c,d$, is it sufficient to state the following?
$$|a-d| = |a-b| + |b-c| + |c-d| > |b-c|$$
since $|c-d| > 0$.
How would I provide more detail to demonstrate that the equality holds?

$a\leq b$
$\implies a-b\leq 0$
$\implies |a-b|\geq 0$
And, $c\lt d$
$\implies c-d\lt 0$
$\implies |c-d|\gt 0$
$\begin{align}\therefore &|a-b|+|c-d|\gt 0\\ & |b-c|+|a-b|+|c-d|\gt |b-c|\\ & |a-d|\gt |b-c|\end{align}$. QED
