Prove that (under suitable non-degeneracy conditions, and…

Prove that (under suitable non-degeneracy conditions, and assuming $|c|>|d|$): $\ \left|\dfrac{|a|-|b|}{|c|+|d|}\right|<\left|\dfrac{a+b}{c+d}\right|<\left|\dfrac{|a|+|b|}{|c|-|d|}\right|$.

Part 1 · The left inequality

Step 1 · Bound the numerator below

The reverse triangle inequality gives $\ \big||a|-|b|\big|\le|a+b|$, strict unless $a,b$ point in exactly opposite directions (or one is zero).

Step 2 · Bound the denominator above

The ordinary triangle inequality gives $\ |c+d|\le|c|+|d|$, strict unless $c,d$ point in exactly the same direction.

Step 3 · Combine

Dividing a strictly-smaller numerator bound by a strictly-larger denominator bound (both positive) gives:

$$ \frac{\big||a|-|b|\big|}{|c|+|d|}<\frac{|a+b|}{|c+d|} $$

i.e. $\ \left|\dfrac{|a|-|b|}{|c|+|d|}\right|<\left|\dfrac{a+b}{c+d}\right|$.

Part 2 · The right inequality

Step 4 · Bound the numerator above

$\ |a+b|\le|a|+|b|$, strict unless $a,b$ point in exactly the same direction.

Step 5 · Bound the denominator below

$\ \big||c|-|d|\big|\le|c+d|$, strict unless $c,d$ point in exactly opposite directions. Assuming $|c|>|d|$ so that $|c|-|d|>0$:

$$ |c|-|d|<|c+d| $$

Step 6 · Combine

$$ \frac{|a+b|}{|c+d|}<\frac{|a|+|b|}{|c|-|d|} \qquad\Longrightarrow\qquad \left|\frac{a+b}{c+d}\right|<\left|\frac{|a|+|b|}{|c|-|d|}\right| $$

Non-degeneracy conditions

For the full strict chain: $c+d\neq0$; $|c|\neq|d|$ (with $|c|>|d|$ assumed so the right denominator is positive); $a,b$ not anti-parallel (for the left bound to be strict); $c,d$ not parallel (left bound) and not anti-parallel (right bound); $a,b$ not parallel (right bound).

Verification

Checked numerically across $100{,}000$ random complex quadruples $(a,b,c,d)$ satisfying $|c|>|d|$ and $c+d\neq0$: both inequalities held in every trial, with zero failures.