Proving the triangle inequality for the euclidean distance in the plane 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?
 A: 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.
