I want to prove that the following is a metric space Let $p \geq 1, a = (a_{1},a_{2}) \in \mathbb{R}^2, b = (b_{1},b_{2}) \in \mathbb{R}^{2}$. Denote $d_{p}(a,b) = (|a_{1} - b_{1}|^{p} + |a_{2} - b_{b}|^p)^{\frac{1}{p}}$.
Prove that $(\mathbb{R}^2, d_{p})$ is a metric space.
Attempt:
Intuitively, $d_{p}(a,b) = d_{p}(b,a)$ and $d_{p}(a,b) = 0 \Rightarrow a = b$ make sense because if you think of $p = 2$, it is just the usual distance formula that we all know about in algebra. Therefore, those conditions seem to make sense to me. As for the triangle inequality, I am not sure how to prove it is true. 
 A: the metric you defined is induced by a norm, the so called $p$-Norm. especially this defines a metric on any $\mathbb{R}^n$ for any $p\in \mathbb{N}\cup \{ \infty \} $. the triangle inequality is called "Minkowski inequality" wich is proofed by using the "Hölder inequality".
I found a pdf file online where you can read the proof for $\mathbb{R}^n$. It's not that trivial at all. So a good exercise would be to rewrite the proof for $\mathbb{R^2}$.
here is the link: http://www.cs.umb.edu/~dsim/cs724/norms.pdf
It's chapter 5.2 from the beginning up to Corollary 5.11
EDIT (on your comment):
Sure you can write down it's clear, but if this is a homework for a basic lecture: i would like to see an argument why it's clear to you (if i were supposed to correct your homework).
so $$d_p (a,b) = (|a_1-b_1|^p+|a_2-b_2|^p)^{\frac{1}{p}}=0 \\
\Leftrightarrow (|a_1-b_1|^p+|a_2-b_2|^p)=0^p=0\\
\Leftrightarrow |a_1-b_1|=0\text{ and }|a_2-b_2|=0
\\ \Leftrightarrow a=b$$
and the symmetry is really straightforward:
$$d_p (a,b) = (|a_1-b_1|^p+|a_2-b_2|^p)^{\frac{1}{p}} \\
= (|b_1-a_1|^p+|b_2-a_2|^p)^{\frac{1}{p}} =d_p(b,a)$$
So you see the tricky part is really the triangle inequality.
In homework problems, you have to convince the reader that you understand what you're saying and "it's trivial" is not very convincing ;)
