Prove Triangle Inequality for Two Special Cases

Question

I am reading "Understanding Analysis" - it is a great book. However, I lack a solutions manual and some of these answers are difficult to find. It has been a while since I've written proofs and I'm wondering if these proofs are sufficient for the problem, or I can do better:

The problem states:

Use the triangle inequality to establish the inequalities:

(a) $|a - b| \le |a| + |b|$

(b) $||a| - |b|| \le |a - b|$

Starting with (a):

Note that $|a - b| = |a + (-b)|$

so $|a + (-b)| \le |a| + |-b|$ by the definition of the Triangle Inequality.

So $|a + (-b)| \le a + b$ by applying the rules of the absolute value.

By definition $|a + b| \le a + b$

which surely means $|a + b| \le |a| + |b|$ by the rules of the absolute value.

and since we have shown $|a + (-b)| \le a + b$, $|a + b| \le a + b$ and, $|a + b| \le |a| + |b|$

Then it must the be case where $|a - b| \le |a| + |b|$

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For (b) it seems we can proceed directly:

$||a| - |b|| = |a - b|$ by the definition of the absolute value.

Then it must be the case that $|a - b| \le |a - b|$

and so $||a| - |b|| \le |a - b|$

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I hope I did these correctly. It would be very helpful if I could be provided the correct answer if I am wrong - and methods to write cleaner proofs. Thank you!

Answer 1

(a) Prove $|a - b| \le |a| + |b|$

We know that $|x|+|y| \ge |x+y|$

$$\text{Let $x=a$ and let $y=-b$. Substituting we get}$$

\begin{align} |x|+|y| &\ge |x+y| \\ |a| + |-b| &\ge |a + (-b)| \\ |a| + |b| &\ge |a-b| \\ |a-b| &\le |a| + |b| \end{align}

(b) Prove $||a| - |b|| \le |a - b|$

$$\text{Let $x=a$ and $y=b-a$.}$$ \begin{align} |x|+|y| &\ge |x+y| \\ |a| + |b-a| &\ge |b| \\ |a| + |a-b| &\ge |b| \\ |b| &\le |a| + |a-b| \\ -|a-b| &\le |a| - |b| \tag{1.} \end{align}

$$\text{Next, let $x=b$ and $y=a-b$.}$$ \begin{align} |x|+|y| &\ge |x+y| \\ |b| + |a-b| &\ge |a| \\ |a| &\le |b| + |a-b| \\ |a| - |b| &\le |a-b| \tag{2.} \end{align}

$$\text{From $(1.)$ and $(2.)$, we get}$$ \begin{align} -|a-b| \le |a| - |b| \le |a-b| \\ ||a| - |b|| \le |a - b| \end{align}

Answer 2

If you have the already proven the standard form of the triangle inequality.

$|a+b| \le |a|+|b|$

then you can say.

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

But you didn't stop there. And, what you do say after that is incorrect. You say "$|a+b| \le a+b$" when, in fact, $|a+b| \ge a+b$

As for the second one, I am not following your logic. Here is what I might say.

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

If you don't have the the proof of the standard form in your back pocket, then square both sides, or brute force the cases:

$a\ge 0$ and $b\ge 0, a<0$ and $b\ge 0, a\ge 0$ and $b< 0, a< 0$ and $b< 0.$

Answer 3

$ \lvert a+b \rvert \leq a+b$ is untrue. A counter example is $a=b=-1$. In fact, the reverse inequality is true by the triangle inequality: $ \lvert a+b \rvert \geq a+b$.

You need to use two facts for proving the first inequality:

(1) $a+(-b)=a-b$

(2) $ \lvert -b \rvert= \lvert b \rvert$

For (b), $\lvert \lvert a \rvert -\lvert b\rvert \rvert=\lvert a-b\rvert$ is untrue. A counter examples is a=1 and b=-1.

A correct reasoning would be noting $\lvert a-b \rvert=\lvert b-a \rvert$. And $\lvert a-b \rvert +\lvert b \rvert \geq \lvert (a-b)+b \rvert=\lvert a \rvert$ by the trinagle inequality. Similarly $\lvert b-a \rvert +\lvert a \rvert \geq \lvert (b-a)+a \rvert=\lvert b \rvert$ Therefore we have $\lvert a-b \rvert \geq max(\lvert a \rvert-\lvert b \rvert,\lvert b \rvert-\lvert a \rvert)=\lvert \lvert a \rvert-\lvert b \rvert \rvert$ by the definition of the absolute value.