Algebraic proof that $ ||\vec{a} +\vec{b}|| \le ||\vec{a}|| + ||\vec{b}|| $

Question

That is, that the sum of two vectors is never longer than the sum of the lengths of the vectors.

It's easy to see this is true geometrically: if you attach the vectors head to tail, the longest they can be is if they are in the same direction. Alternatively, you can see the two vectors and their sum form a triangle, and the length of a side can never be longer than the sum of the other two sides (the longest it can be is a degenerate triangle, with angles 0-0-180).

How can we prove this with high-school algebra (i.e. [algebra-precalculus]), using vectors in component form, $\vec{a} = (r,s), \vec{b} = (p,q)$?

Note: the "duplicate" question does not request high-school algebra, and accepted an answer using the Cauchy–Schwarz inequality and vector dot product, which are not usually taught in high school at all.

Answer 1

Here it is an approach using Cauchy-Schwarz inequality

\begin{align*} \lVert\textbf{a}+\textbf{b}\rVert^{2} & = \langle\textbf{a}+\textbf{b},\textbf{a}+\textbf{b}\rangle = \langle\textbf{a},\textbf{a}\rangle + 2\langle\textbf{a},\textbf{b}\rangle + \langle\textbf{b},\textbf{b}\rangle = \lVert\textbf{a}\rVert^{2} + 2\langle\textbf{a},\textbf{b}\rangle + \lVert\textbf{b}\rVert^{2}\\ & \leq \lVert\textbf{a}\rVert^{2} + 2\lVert\textbf{a}\rVert\lVert\textbf{b}\rVert + \lVert\textbf{b}\rVert^{2} = (\lVert\textbf{a}\rVert + \lVert\textbf{b}\rVert)^{2} \Rightarrow \lVert\textbf{a}+\textbf{b}\rVert\leq\lVert\textbf{a}\rVert + \lVert\textbf{b}\rVert \end{align*}

Answer 2

With $\vec{a} = (r,s), \vec{b} = (p,q), \vec{a}+\vec{b} = (r+p,s+q)$, the inequality becomes:

$$ \sqrt{ (r+p)^2 + (s+q)^2 } \le \sqrt{r^2+s^2} + \sqrt{p^2+q^2} $$

Square both sides, to eliminate the LHS root.

$$ (r+p)^2 + (s+q)^2 \le r^2+s^2 + 2\sqrt{ (r^2+s^2)(p^2+q^2)}+ p^2+q^2 $$

Expand the LHS and cancel equal terms:

$$ 2rp + 2sq \le 2\sqrt{ (r^2+s^2)(p^2+q^2)} $$

Remove the common $2$ and square again to remove the RHS root (this squaring introduces an extraneous solution, by a negative LHS being folded into a positive value, but fortunately both solutions obey the inequality - it doesn't matter that we also prove the extraneous solution.)

$$ r^2p^2 + 2rpsq + s^2q^2 \le (r^2+s^2)(p^2+q^2) $$

Expand the RHS and cancel equal terms:

$$ 2rpsq \le r^2q^2+s^2p^2 $$

Move $2rpsq$ to the RHS side, and (luckily!) it's a perfect square: $$ 0 \le (rq)^2 -2(rq)(sp) +(sp)^2 $$

This algebraically reduces to:

$$ 0 \le (rq-sp)^2 $$

which is true because squared real numbers are always $\ge 0$.

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BTW: this is zero when $rq=sp$. Dividing by $rp$ gives $\frac{q}{p}=\frac{s}{r}$, which says that the slopes are equal. Slopes are an imperfect representation of vector direction, because they can't represent vertical vectors, and they don't distinguish between exactly opposite directions. e.g. $\frac{q}{p} = \frac{-q}{-p}$.

That is, slopes are equal for vectors with the same direction - but also for vectors in exactly opposite directions.

The extraneous solution occurs when $rp+sq$ is negated. I'm not sure how to show this, but I think this means that one of $r,p$ and one of $s,q$ must be negative. When both are from the same vector, it is the opposite vector with the same slope.