Example 20: Triangle inequality $|\vec a+\vec b|\le|\vec a|+|\vec b|$

Basics you need

Solution

1. The case of a zero vector

   If $\vec a = \vec 0$⁠, then $\vec a + \vec b = \vec b$ and $|\vec a| = 0$⁠, so $|\vec a + \vec b| = |\vec b| = |\vec a| + |\vec b|$⁠. The case $\vec b = \vec 0$ is the same. The inequality holds, with equality.

   This answers the book's "(How?)". The zero vector is set aside because the argument below uses the angle between the vectors, which is not defined when one of them is $\vec 0$⁠.
   $\vec a + \vec 0 = \vec a$ (Section 10.4); $|\vec 0| = 0$ (Section 10.3). Lesson: adding arrows, head to tail

2. Expand the square of the length

   $$\begin{aligned} &|\vec a + \vec b|^2 = (\vec a + \vec b)\cdot(\vec a + \vec b) \\ &= \vec a\cdot\vec a + \vec a\cdot\vec b + \vec b\cdot\vec a + \vec b\cdot\vec b \\ &= |\vec a|^2 + 2\,\vec a\cdot\vec b + |\vec b|^2. \end{aligned}$$
   The square of a length is the scalar product of the vector with itself. Distributivity multiplies out the brackets, and commutativity ($\vec b\cdot\vec a = \vec a\cdot\vec b$⁠) joins the two middle terms, exactly like $(x + y)^2 = x^2 + 2xy + y^2$⁠. The book writes $(\vec a + \vec b)^2$ as a short form for $(\vec a + \vec b)\cdot(\vec a + \vec b)$⁠.
   Property 1, Observations 3 and 7 (Section 10.6.1). Basics: dot products obey the same rule

3. Bound the middle term

   $$\begin{aligned} &|\vec a + \vec b|^2 \\ &\le |\vec a|^2 + 2|\vec a\cdot\vec b| + |\vec b|^2 \\ &\le |\vec a|^2 + 2|\vec a||\vec b| + |\vec b|^2. \end{aligned}$$
   First inequality: every real number is at most its absolute value, $x \le |x|$⁠, and here $x = \vec a\cdot\vec b$⁠, which may be negative. Second inequality: the Cauchy-Schwarz inequality $|\vec a\cdot\vec b| \le |\vec a||\vec b|$⁠, proved in Example 19 from $|\vec a\cdot\vec b| = |\vec a||\vec b||\cos\theta|$ and $|\cos\theta| \le 1$⁠. Doubling both sides of an inequality and adding the same numbers keeps its direction.
   $x \le |x|$ for all real $x$⁠; Cauchy-Schwarz inequality. Example 19

4. A perfect square, then square roots

   $$|\vec a|^2 + 2|\vec a||\vec b| + |\vec b|^2 = (|\vec a| + |\vec b|)^2,$$

   so $|\vec a + \vec b|^2 \le (|\vec a| + |\vec b|)^2$⁠. Both $|\vec a + \vec b|$ and $|\vec a| + |\vec b|$ are non-negative, so $|\vec a + \vec b| \le |\vec a| + |\vec b|$.

   Squaring is increasing for non-negative numbers, so $p^2 \le q^2$ with $p, q \ge 0$ gives $p \le q$⁠. The book goes straight to "Hence"; this sentence is the reason. Without the sign condition the step can fail: $2^2 \le (-3)^2$ is true, but $2 \le -3$ is false.
   $(x + y)^2 = x^2 + 2xy + y^2$ with $x = |\vec a|$⁠, $y = |\vec b|$⁠. Basics: $(x + y)^2 = x^2 + 2xy + y^2$

5. When does equality hold?

   Equality needs equality in both steps of Step 3: $\vec a\cdot\vec b = |\vec a\cdot\vec b|$ (so $\vec a\cdot\vec b \ge 0$⁠) and $|\vec a\cdot\vec b| = |\vec a||\vec b|$ (so $\theta = 0$ or $\pi$⁠). Together, $\theta = 0$⁠: for nonzero vectors, $|\vec a + \vec b| = |\vec a| + |\vec b|$ exactly when $\vec a$ and $\vec b$ point the same way.

   This is the book's Remark after Example 20: if $|\vec a + \vec b| = |\vec a| + |\vec b|$⁠, then in the triangle law $|\overrightarrow{AC}| = |\overrightarrow{AB}| + |\overrightarrow{BC}|$⁠, and A, B, C are collinear, with B between A and C.
   Remark after Example 20 (Section 10.6.2); $\theta = 0$ gives $\vec a\cdot\vec b = |\vec a||\vec b|$ (Section 10.6.1, Observation 3). Lesson: the dot product

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