Prove Triangle Inequality

[nycmath]

Precalculus
David Cohen
Edition 3
Chapter 1, Section 1.2

71. Prove the triangle inequality.

[Math Tutor]

Claim: [tex]|a+b| \le |a| + |b|[/tex] for all real [tex]a, b[/tex].

Start from the fact that every real number lies between the negative and positive of its own absolute value:

[tex]-|a| \le a \le |a| \qquad \text{and} \qquad -|b| \le b \le |b|[/tex]

Add the two chains:

[tex]-\left(|a|+|b|\right) \le a+b \le |a|+|b|[/tex]

Now use the rule [tex]-c \le x \le c \iff |x| \le c[/tex] (with [tex]c = |a|+|b| \ge 0[/tex]):

[tex]|a+b| \le |a| + |b| \quad \blacksquare[/tex]

Equality holds exactly when [tex]a[/tex] and [tex]b[/tex] have the same sign (or one of them is 0).