Why is $ |\exp(ia)-\exp(ia + ib)+ib| \leq |1-\exp(ib) + ib| + |b||1-\exp(ia)|$?

Question

I'm reading a proof and it uses the following inequality, where $a,b \in \mathbb{R}$

$$|\exp(ia)-\exp(ia + ib)+ib| \leq |1-\exp(ib) + ib| + |b||1-\exp(ia)|$$

I tried to use the triangle inequality but could not prove it.

Answer 1

Write this as $$ |\exp(ia)-\exp(ia + ib)+ib| \leq |1-\exp(ib) + ib| + |b||1-\exp(ia)|\\ \leftrightarrow\\ |- (1-\exp(ia))(1-\exp(ib)) + 1-\exp(ib)+ib| \leq |1-\exp(ib) + ib| + |b||1-\exp(ia)|\\ $$ Using $x = (1-\exp(ia))(1-\exp(ib)) $, $y=1-\exp(ib)+ib$ and $|-x +y| \le |x| + |y|$

leaves to show $$ |1-\exp(ia)||1-\exp(ib)| + |1-\exp(ib)+ib| \leq |1-\exp(ib) + ib| + |b||1-\exp(ia)|\\ $$ from which remains $$ |1-\exp(ib)| \leq |b| $$ Square it: $$ 2 - 2 \cos(b) = (1-\exp(ib))(1-\exp(-ib)) = |1-\exp(ib)|^2 \leq b^2 $$ which is $$ \cos(b) \ge 1 - \frac12 b^2 $$ and this holds true for the cosine function. $\qquad \Box$