Prove that for any two vectors $A$ and $B$, the length of the vector sum is smaller than the sum of the lengths of $A$ and $B$ taken separately.

Question

I know that the sum of two sides of a triangle is greater than the third side. But how do I prove the above question?

Answer 1

We need to prove that $$||A+B||\leq ||A||+||B||$$

You can reach here easily by the definition of norm and linearity and other some property of inner product. So,

$$||A+B||^2=||A||^2+2Re(A,B)+||B||^2(\because z+z̅=2Re)$$

Since $Re(z)\leq |z|$ where $z$ is a complex number.

Therefore,

Now using Schwarz's inequality we have, $$||A+B||^2\leq ||A||^2+2||A||•||B||+||B||^2$$ $$||A+B||^2\leq (||A||+||B||)^2$$ $$||A+B||\leq (||A||+||B||)$$

Answer 2

The property you are asking about is actually a defining property of any normed vector space (see property 4 on the linked Wikipedia page). As such, what I believe you are really asking is why this property (aka the Triangle Inequality) holds in $\mathbb{R}^2$, $\mathbb{R}^3$, . . . , $\mathbb{R}^n$.

The short answer is because it holds in $\mathbb{R}$. And this, in turn, implies that it also holds in $\mathbb{R}^2$, $\mathbb{R}^3$, . . . , $\mathbb{R}^n$.

To see this, first one needs to convince themselves that for real numbers $a$ and $b$ $$|a+b|\leq |a|+|b|.$$

Once we believe this is true for all real numbers, we pick three points in $\mathbb{R}^n$. (For concreteness, in what follows we choose $n=3$ so we are dealing with three points in $\mathbb{R}^3$.) Label the three points: $A=(a_1, a_2, a_3)$, $B=(b_1, b_2, b_3)$, and $C=(c_1, c_2, c_3)$. We want to show

$$||\overrightarrow{AC}||\leq ||\overrightarrow{AB}||+||\overrightarrow{BC}||.$$

To see this consider that

$\begin{align*}||\overrightarrow{AC}||^2 &= |c_1-a_1|^2 + |c_2-a_2|^2 + |c_3-a_3|^2 \\ &= |c_1-b_1+b_1-a_1|^2 + |c_2-b_2+b_2-a_2|^2 + |c_3-b_3+b_3-a_3|^2 \\ &\leq(|c_1-b_1|+|b_1-a_1|)^2 + (|c_2-b_2|+|b_2-a_2|)^2 + (|c_3-b_3|+|b_3-a_3|)^2 \\ & \leq |c_1-b_1|^2+|b_1-a_1|^2 + |c_2-b_2|^2+|b_2-a_2|^2 + |c_3-b_3|^2+|b_3-a_3|^2 \\ &= \underset{||\overrightarrow{BC}||^2}{\underbrace{ |c_1-b_1|^2 + |c_2-b_2|^2 + |c_3-b_3|^2 }} + \underset{||\overrightarrow{AB}||^2}{\underbrace{ |b_1-a_1|^2 + |b_2-a_2|^2 + |b_3-a_3|^2 }} \\ \end{align*}$

where the first inequality is due to the fact that the Triangle Inequality holds in $\mathbb{R}$. Thus, $$||\overrightarrow{AC}||^2\leq ||\overrightarrow{AB}||^2 + ||\overrightarrow{BC}||^2.$$

Taking the square root of both sides yields: $$ ||\overrightarrow{AC}||\leq \sqrt{||\overrightarrow{AB}||^2 + ||\overrightarrow{BC}||^2} \leq \sqrt{||\overrightarrow{AB}||^2} + \sqrt{||\overrightarrow{BC}||^2} = ||\overrightarrow{AB}|| + ||\overrightarrow{BC}||$$ since the square root is subadditive over positive real numbers (i.e. for positive $x$ and $y$, $\sqrt{x+y} \leq \sqrt{x} +\sqrt{y}$ ).

[Reason: For positive $x$ and $y$ , we have $x+y \leq x+2\sqrt{x}\sqrt{y}+y = (\sqrt{x}+\sqrt{y})^2$ and so (by taking the square root of both ends) we get $\sqrt{x+y} \leq \sqrt{(\sqrt{x}+\sqrt{y})^2} = |\sqrt{x}+\sqrt{y}| = \sqrt{x}+\sqrt{y}$.]

Answer 3

Apply the cosine rule to the triangle formed by the two vectors

\begin{align} |AB|^2 & = |A|^2+|B|^2 -2|A||B|\cos\theta\\ & = (|A|+|B|)^2- 2|A||B|(1+\cos\theta) \\ & \le (|A|+|B|)^2 \end{align}

Thus, $|AB| \le |A|+|B|$.