A question on Triangle Inequality in $\mathbb{R}^n$

Question

I'm reading a textbook on Topology. We know that $(\rho,\mathbb{R}^n)$ is a metric space, where
$$\rho(x,y)=\sqrt{\sum_{i=1}^n(x_i-y_i)^2}$$for any $x=(x_1,x_2,\ldots,x_n),y=(y_1,y_2,\ldots,y_n)\in\mathbb{R}^n$. When proving that $\rho(x,z)\le \rho(x,y)+\rho(y,z)$, the author uses Schwarz Inequality.

I can understand the method, but I wonder if we can do it directly. We know that three non-collinear points can determine a plane. If those three points $x,y,z$ are on a single line, then of course we can apply the Triangle Inequality on $\mathbb{R}$; if they are not, then they are on a same plane, still we can apply the Triangle Inequality. Isn't it just a question on $\mathbb{R}^2$ essentially?

Maybe I'm missing something, but I can't find it myself. Is my reasoning correct? Thank you!

Answer 1

Good point! I can’t explain why the author would go through a full algebraic proof rather than using what you noticed to simplify the problem considerably. However, I can give some advice about formalizing your idea into a full proof of the triangle inequality in $\mathbb R^n$.

Your observation that

Any three points $\vec x, \vec y, \vec z\in\mathbb R^n$ either lie on the same line or the same plane.

is equivalent to the statement

For any $\vec x, \vec y, \vec z\in\mathbb R^n$, there exists an isometry (distance-preserving transformation) mapping all but the first three coordinates of $\vec x, \vec y,$ and $\vec z$ to zero.

This statement shouldn’t be difficult to prove using a bit of linear algebra to show that a certain system of equations has at least one solution. Once you’ve done that, the triangle inequality in $\mathbb R^n$ follows, because any triplet of points $\vec x, \vec y, \vec z$ can be reduced by isometry to points of the form $(a,b,c,0,...,0)$, at which point the triangle inequality on $\mathbb R$ or $\mathbb R^2$ can be applied.