$|a-b| \geq | |a| - |b| | \Longleftrightarrow |a-b| \geq |a| - |b| $?

Question

Does the following always hold?

$|a-b| \geq | |a| - |b| | \Longleftrightarrow |a-b| \geq |a| - |b| $

Certainly, the second inequality implies the first, but the converse isn't so clear to me.

If it does hold, then why do we define the 'reverse triangle inequality' as the first inequality in the above?

Answer 1

Both inequalities are always true, hence the bi-conditional is valid.

Explicitly, $$|a| = |(a-b) + b| \le |a-b| + |b|$$ hence $|a-b| \ge |a|-|b|$.

Thus, the RHS is always true.

Similarly, $$|b| = |(b-a) + a| \le |b-a| + |a| = |a-b| + |a|$$ hence $|a-b| \ge |b|-|a|$.

Since we have both $$|a-b| \ge |a|-|b|$$ $$|a-b| \ge |b|-|a|$$ it follows that $|a-b| \ge ||a|- |b||$.

Thus, the LHS is always true.

Answer 2

The if and only if holds if you clarify "for all $a,b$".

The first inequality, for fixed $a,b$, implies the second inequality. The second inequality, on the other hand, does not imply the first inequality if you consider it for fixed $a,b$; you need it to hold for $b,a$.

In summary, the first inequality contains more information.

Answer 3

The RHS is always true by the triangle inequality... the RHS implies the LHS... Therefore the LHS is always true...

So they're equivalent...

Answer 4

I think the question that you're trying to ask is "how do we use the triangle inequality (which can be rearranged to $|a-b| \geq |a| - |b|$) to prove the "reverse triangle inequality, namely $|a-b| \geq | |a| - |b| |$.

Here's one way to do it: note that if we exchange the roles of $a$ and $b$ in the first inequality, we find that $$ |b-a| \geq |b| - |a| $$ However, we know that $|b - a| = |-(a - b)| = |a - b|$. So, the above implies that $|a - b| \geq |b| - |a|$. That is, we know that $$ |a - b| \geq |a| - |b| \quad \text{AND} \quad |a - b| \geq |b| - |a| $$ To put it more succinctly, we have $$ |a - b| \geq \max\{|a| - |b|, |b| - |a|\} = \max\{|a| - |b|, -(|a| - |b|)\} = ||a| - |b|| $$ which is the desired result.

Answer 5

For fun:

$\Rightarrow:$

$|a-b|\ge ||a|-|b|| =$

$\max((|a|-|b|), (|b|-|a|)) \ge $

$|a|-|b|.$

$\Leftarrow:$

1) $|a-b| \ge |a|-|b|;$

Interchange $a$ and $b$:

2) $|b-a| = |a-b| \ge |b| -|a|.$

Hence:

$|a-b| \ge $

$\max((|a|-|b|), (|b| -|a|))=$

$||a|-|b||$.