Prove that Euclidean distance in $\mathbb{R}^n$ is a distance I'm trying to show that:
$$\forall x,y\in\mathbb{R}^n, d(x,y)=\left(\sum_{i=1}^n(x_i-y_i)^2\right)^{1/2}$$
is a distance. However I have not proved Cauchy-Schwarz yet and I'm pretty sure I wouldn't know how to use it in this case. Can someone give me a hint on how to solve this?
Thanks a lot for helping!!!
 A: $$ \sum_i (x_i + \lambda y_i)^2\geqslant 0 $$
$$ \sum_i x_i^2+\lambda x_iy_i +\lambda x_iy_i+\lambda^2y_i^2 \geqslant0 $$
Let $\lambda=-(\sum_jx_jy_j)/(\sum _jy_jy_j)$ then:
$$\sum_i x_i^2-\frac{(\sum_ix_iy_i)(\sum_jx_jy_j)}{(\sum _jy_jy_j)}-\frac{(\sum_ix_iy_i)(\sum_jx_jy_j)}{(\sum _jy_jy_j)}+(\frac{\sum_jx_jy_j}{\sum _jy_j^2})^2 \sum_iy_i^2 \geqslant0$$
Simplifying and adjusting the terms on both sides of the inequality yields:
$$(\sum_i x_iy_i)^2 \leqslant  \sum_i x_i^2 \sum_iy_i^2  \tag{cqfd!}$$ 
EDIT
Now to  the triangle inequality:
$$\sum_i(x_i+y_i)^2=\sum_i x_i^2 +2\sum_i x_iy_i + \sum_iy_i^2 \leqslant \sum_ix_i^2 +2\sqrt{\sum_ix_i^2 \sum_iy_i^2} +\sum_iy_i^2$$
$$=(\sqrt{\sum_ix_i^2} + \sqrt{\sum_iy_i^2} )^2 \tag{cqfd!}$$
A: Put $\mathbf{v} = \mathbf{x} - \mathbf{y}$
Then
$d(x,y) = \|\mathbf{x}-\mathbf{y}\| = \|\mathbf{v}\|$
and
$\langle{\mathbf{v},\mathbf{v}}\rangle = \langle{\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}}\rangle = \displaystyle{\sum\limits_i^n(x_i-y_i)^2}$
and so $\|v\| = \sqrt{\langle\mathbf{v},\mathbf{v}\rangle}$ implies the desired result.
