Proving the triangle inequality for the euclidean distance in the plane

Question

I'm looking to introduce my students to the triangle inequality in the plane with the regular euclidean distance. They have no knowledge of functions or vectors (and therefore norms) so the proof should contain no mention of those concepts. I'm finding it rather difficult to prove with such basic tools.

Let $A=(x_A,y_A)$, $B=(x_B,y_B)$ and $C=(x_C,y_C)$ be points in $\mathbb{R}^2$. I want to prove that

$$ \sqrt{(x_A-x_B)^2+(y_A-y_B)^2} \leq \sqrt{(x_A-x_C)^2+(y_A-y_C)^2} + \sqrt{(x_C-x_B)^2+(y_C-y_B)^2} $$

Any ideas?

@Rohan Not a duplicate, since the linked question uses norms and vectors. — Lundborg, Jan 31 '17 at 12:10
What about the original proof in Euclid's Elements? http://farside.ph.utexas.edu/Books/Euclid/Elements.pdf , Book 1, Proposition 20. — johnnycrab, Jan 31 '17 at 12:13
@johnnycrab that is an option, I have considered it but I was really interested in an algebraic/analytic proof even though I am aware that I'm quite limited. — Lundborg, Jan 31 '17 at 12:14
Are your students familiar with the statement that the shortest path between two points in the Euclidian plane is the straight line? That's not a proof in itself, but a pretty good motivation for one. — dxiv, Feb 01 '17 at 07:38
Align three rods, of lengths $a$, $b$, and $a+b$ so that the endpoints coincide. If you want to form a triangle with basis $a+b$, you will have to lengthen one of $a$ or $b$. — , Feb 01 '17 at 08:02
By definition, the Euclidean distance is the length of the shortest path between two points. So wherever $C$, $AC+CB$ cannot be shorter than $AB$. — , Feb 01 '17 at 08:08
See also: Proof for triangle inequality for vectors and Prove that Euclidean distance in $\mathbb{R}^n$ is a distance — Martin Sleziak, Aug 28 '19 at 07:16

Anatoly · Accepted Answer · 2017-02-01T08:52:55.657

I would suggest you to set the point $C $ at the origin, so that the inequality becomes

$$ \sqrt{(x_A-x_B)^2+(y_A-y_B)^2} \leq \sqrt{x_A^2+y_A^2} + \sqrt{x_B^2+y_B^2} $$

Expanding the LHS we obtain

$$ \sqrt{x_A^2+x_B^2 -2x_Ax_B +y_A^2 +y_B^2 -2y_Ay_B } \leq \sqrt{x_A^2+y_A^2} + \sqrt{x_B^2+y_B^2} $$

Squaring both sides (there are no sign problems because both quantities are positive) we get

$$x_A^2+x_B^2 -2x_Ax_B +y_A^2 +y_B^2 -2y_Ay_B \leq x_A^2+y_A^2 + x_B^2+y_B^2 +2 \sqrt{x_A^2+y_A^2} \sqrt{x_B^2+y_B^2}$$

and simplifying

$$-(x_Ax_B+y_Ay_B) \leq \sqrt{x_A^2+y_A^2} \sqrt{x_B^2+y_B^2}$$

Now we have to consider two cases, i.e. $x_Ax_B+y_Ay_B\,\, \,\,$ is positive or negative. In the first case, the inequality holds. In the second one, we can continue by squaring both sides:

$$x_A^2x_B^2+y_A^2y_B^2+2x_Ax_By_Ay_B \\ \leq x_A^2x_B^2+y_A^2y_B^2 +x_A^2y_B^2 + y_A^2x_B^2$$

$$2x_Ax_By_Ay_B \leq +x_A^2y_B^2+ y_A^2x_B^2$$

$$0 \leq (x_Ay_B- y_Ax_B)^2 $$

which is trivial. Note that the only possibilities where the equality holds are:

$x_A=x_B=0\,\,\,$ and only one between $y_A$ and $y_B=0$ is negative (this is because if $y_Ay_B $ is positive we are in the first case described before and the inequality holds). This is the case where both the points $A $ and $B $ are on the $y $-axis, on opposite sides with respect to the origin, and the triangle reduces to a vertical line);
$y_A=y_B=0 \,\,\, $ and only one between $x_A$ and $x_B=0$ is negative (this is again because if $x_Ax_B $ is positive we are in the first case described before and the inequality holds). This is the case where the points $A $ and $B $ are on the $x$-axis, on opposite sides with respect to the origin, and the triangle reduces to a horizontal line);
at least three among $x_A,x_B,y_A,y_B\,\,$ are equal to zero. This corresponds to the limit case where two vertices of the triangles coincide with the origin.

Thank you for your note. I edited my answer accordingly. – Anatoly Feb 01 '17 at 08:53 — Anatoly, Feb 01 '17 at 08:53

Proving the triangle inequality for the euclidean distance in the plane

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