Prove that for $a, b \in \mathbb{R}$ $|a + b -a| \geq |a| - |b-a|$

Question

Prove that for $a, b \in \mathbb{R}$ $|a + b -a| \geq |a| - |b-a|$

I'm not sure how to go about proving this, I can't seem to find any reasonable way to use the triangle inequality and this doesn't seem to follow immediately from the other properties of the absolute value function.

Answer 1

$|a+b-a|+|b-a|=|a+b-a|+|a-b|\ge|(a+b-a)+(a-b)|=|a|$

Answer 2

Proof: Observe that $|b| = |a +b - a|$, and by the triangle inequality we have $|a| = |a-0| \leq |a-b| + |b-0| = |a-b| + |b|$.

Hence $|a| \leq |a-b| +|b|$, but $|a-b| = |b-a|$ so we have $|a| \leq |b-a| + |b|$ which implies $|a| - |b-a| \leq |b| = |a +b - a|$ as desired. $\square$

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Thanks @Bram28 for the hint in the comments section.

Answer 3

Alternatively: $$|a|=|a-b+b|\le |a-b|+|b|=|b-a|+|a+b-a|.$$