Exercise 1.2.6

Question

\begin{enumerate}[label=(\alph*)] 
  \item Verify the triangle inequality in the special case where $a$ and $b$ have the same sign.
  \item Find an efficient proof for all the cases at once by first demonstrating $(a+b)^{2} \leq(|a|+|b|)^{2}$
  \item Prove $|a-b| \leq|a-c|+|c-d|+|d-b|$ for all $a, b, c$, and $d$.
  \item Prove $\| a|-| b|| \leq|a-b|$. (The unremarkable identity $a=a-b+b$ may be useful.)
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item We have equality $|a + b| = |a| + |b|$ meaning $|a + b| \le |a| + |b|$ also holds.
  \item $(a+b)^2 \le (|a| + |b|)^2$ reduces to $2ab \le 2|a||b|$ which is true as the left side can be negative but the right side can't.
    and since squaring preserves inequality this implies $|a + b| \le |a| + |b|$.
  \item I would like to do this using the triangle inequality, I notice that $(a-c)+(c-d)+(d-b) = a-b$. Meaning I can use the triangle inequality for multiple terms
    $$|a - b| = |(a-c)+(c-d)+(d-b)| \le |a-c| + |c-d| + |d-b|$$
    The general triangle inequality is proved by repeated application of the two variable inequality
    $$|(a + b) + c| \le |a + b| + |c| \le |a| + |b| + |c|$$
  \item Since $||a|-|b|| = ||b|-|a||$ we can assume $|a| > |b|$ without loss of generality. Then
    $$
    ||a|-|b|| = |a|-|b| = |(a - b) + b| - |b| \le |a-b| + |b| - |b| = |a-b|
    $$
   \end{enumerate}

Exercise 1.2.6

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Exercise 1.2.6

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