Prove the Euclidean Metric by Induction I'm working on "Metric Spaces" by Michael O Searcoid. I have done an introductory Metric Spaces course before but want to review and hopefully improve. However on just the first problem set I'm stumped by the following question,
" Use induction to verify that the function $ \mu_2 $ defined on $ \mathbb R^n \mathrm x \mathbb R^n$ by $ (a,b)\mapsto \sqrt{\sum_{i=1}^n (b_i -a_i)^2}$ satisfies the triangle inequality and is therefore a metric"
I've only ever seen this done by inducing a norm using the metric and using Cauchy-Shwarz Inequality. Struggling to make an inductive proof without that and I feel like that is not expected to be used as not mentioned in the book. 
Case $ n=1$ is just the absolute value I think and I claimed it holds there due to fact absolute value is itself a metric but establishing the inductive step is proving impossible for me. Maybe I'm overthinking it (or not enough)?
I've looked online and in other texts and no proofs of this by induction. All either use Cauchy-Swarz on the norm defined by the metric or use induction to prove general $\mu_p$ metric.
I'm really stumped. Maybe I'm reading too much into the question or missing something obvious. The Cauchy-Swarz inequality has not been introduced so I really don't want to use that standard proof using it. 
Can anyone tell me how to establish the inductive step? I it even possible without Cauchy-Schwarz?
I'll add this is not homework. I'm studying this book of my own accord in order to sharpen my metric spaces and topology understanding. I've spent few hours at it and reading other texts, no where left to turn.
 A: It seems to me that it is smooth sailing after the case $n=2$ has been established. Let me take the liberty of working with the norm rather than the metric, so the thing to be proved is
$$
\begin{align*} 
\|x+y\|_n \leq \|x\|_n+\|y\|_n && && n=1,2,3, \ldots && x,y \in \mathbb{R}^n
\end{align*}
$$
where, for $x = (x_1,\ldots,x_n) = \mathbb{R}^n$, we have
$$ \| x\|_n = \sqrt{ x_1^2 + \ldots x_n^2}.$$
Suppose the base case $n=2$ has been established and that $n>2$. Let $P : \mathbb{R}^n \to \mathbb{R}^{n-1}$ denote projection onto the first $n-1$ factors. That is, send $x = (x_1,\ldots,x_n)$ to $Px = (x_1,\ldots,x_{n-1})$. Note that $P$ is linear.  Now, observe that, for any $x \in \mathbb{R}^n$, we have
$$\|x\|_n = \sqrt{ \|Px\|_{n-1}^2 + x_n^2} = \|(\|Px\|_{n-1},x_n) \|_2.$$
From here, it is easy to use the triangle inequality of $\|\cdot\|_2$ and $\|\cdot \|_{n-1}$, together with the fact that $P(x+y) = Px+Py$ to show the triangle inequality for $\|\cdot\|_n$. 
Edit: actually I think I am using one more property, albeit one which is pretty easy to see holds. Namely, that $\|(a,b)\|_2 \leq \|(a',b)\|_2$ when $|a| \leq |a'|$. 
