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

Question

Here is my proof can anyone check if it needs more detail or if it's good?

Since a = a - b + b, then by the triangle inequality, |a| $\geq$ ||a| - |b|| + |b| so that |a| - |b| $\geq$ ||a| - |b|| since ||a| - |b|| = |a| - |b| $\therefore$ |a - b| $\geq$ ||a| - |b||

Does not look good. You say "the triangle inequality", but what you apply is not it. Take $x=a-b$ and $y=b$, then you have $a=x+y$. How does the triangle inequality for $|x+y|$ look like? — A.Γ., Sep 18 '15 at 05:30
BTW you can find several posts about this inequality on this site. Like this post and other posts linked there. — Martin Sleziak, Sep 18 '15 at 08:45

score 1 · Answer 1 · answered Sep 18 '15 at 05:38

1

Since both sides are non-negative, the inequality holds if and only if $$(a-b)^2\ge (|a|-|b|)^2,$$ that is, if and only if $$a^2-2ab+b^2\ge a^2-2|ab|+b^2.$$ But this inequality is obvious.

answered Sep 18 '15 at 05:38

André Nicolas

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score 0 · Answer 2 · answered Sep 18 '15 at 05:54

0

If $a,b \in \mathbb{R}$, then $a = tb$ for some $t \in \mathbb{R}$. Then $|a-b| = |b||t - 1|$ and $||a| - |b|| = |b|||t|-1|$. If $t \geq 0$ then $|a-b| = ||a| - |b||$; if $t < 0$ then $||t| - 1| = |-t-1|$, so $|a-b| > ||a| - |b||$.

answered Sep 18 '15 at 05:54

Yes

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

2 Answers2