Triangle Inequality for $\ell^2$ Metric

Question

I've been staring at this simple question for the better part of a day now but can't seem to figure it out. How would one prove the Triangle Inequality for the $\ell^2$ metric given by
$$d_2(x,y):=\sqrt{\sum_{n=1}^\infty(x_n-y_n)^2},$$
where $x=(x_1,x_2,…),y=(y_1,y_2,…)$? I know how to prove it for the respective $\ell^2$ norm (that is, $\|x+y\|_2\leq\|x\|_2+\|y\|_2$), but I'd like to prove it in the form $d_2(y,z)\leq d_2(x,y)+d_2(x,z)$.

Answer 1

Let's denote $y_k-x_k=a_k$, $z_k-y_k=b_k$. Then we have $z_k-x_k=a_k+b_k$. Inequality

$$\sqrt{\sum\limits_{k=1}^{n}(z_k-x_k)^2}\leqslant \sqrt{\sum\limits_{k=1}^{n}(x_k-y_k)^2}+\sqrt{\sum\limits_{k=1}^{n}(z_k-y_k)^2}\quad (1)$$

becomes

$$\sqrt{\sum\limits_{k=1}^{n}(a_k+b_k)^2}\leqslant \sqrt{\sum\limits_{k=1}^{n}a_k^2} + \sqrt{\sum\limits_{k=1}^{n}b_k^2}$$

which easily comes from mentioned in comment Cauchy–Schwarz inequality.

Now, from inequality $(x_k \pm y_k)^2\leqslant 2 (x_k^2+y_k^2)$ we have that series $\sum\limits_{k=1}^{\infty}(x_k-y_k)^2$ converge, when converges $\sum\limits_{k=1}^{\infty}x_k^2$ and $\sum\limits_{k=1}^{\infty}y_k^2$ i.e. metric in $\ell^2$ is well defined. Taking limit in $(1)$ gives Triangle Inequality.