$X$ be set consisting of all sequences $(x_1,x_2, \dots)$ s.t $x_i \in \mathbb R$ and $\sum x_i^2$ converges
I need to prove triangle inequality for the metric on $X$ given by,
$d(x,y) = [ \sum_{i=1}^{\infty} (x_i - y_i)^2]^{\frac12}$
How to go about it? A hint will suffice.
My try after seeing the hint:
$\sum_{i=1}^{n} (x_i - y_i)^2 = \sum_{i=1}^{n}x_i^2 + \sum_{i=1}^{n} y_i^2 -2\sum_{i=1}^{n} x_i y_i \le | \sum_{i=1}^{n}x_i^2| + | \sum_{i=1}^{n} y_i^2| + 2|\sum_{i=1}^{n} x_i y_i|$
Using Cauchy Schwarz:
$| \sum_{i=1}^{n}x_i^2| + | \sum_{i=1}^{n} y_i^2| + 2|\sum_{i=1}^{n} x_i y_i| \le | \sum_{i=1}^{n}x_i^2| + | \sum_{i=1}^{n} y_i^2| + 2 (\sum_{i=1}^{n}x_i^2)^\frac12 (\sum_{i=1}^{n}y_i^2)^\frac12 $
So, $[\sum_{i=1}^{n} (x_i - y_i)^2]^\frac12 \le [ \sum_{i=1}^{n}x_i^2 + \sum_{i=1}^{n} y_i^2 + 2 (\sum_{i=1}^{n}x_i^2)^\frac12 (\sum_{i=1}^{n}y_i^2)^\frac12]^\frac12 $
How to proceed further.
Prove that $[\sum_{i=1}^{n} (x_i - y_i)^2]^{\frac12} \leq [\sum_{i=1}^{n} (x_i)^2]^{\frac12} + [\sum_{i=1}^{n} (y_i)^2]^{\frac12}$ holds for all $x_1, ..., x_n, y_1, ..., y_n \in \mathbb{R}$ and all positive integers $n$, then you can substitute $x_i$ with $x_i - z_i$ and $y_i$ with $y_i - z_i$ because they are real numbers too. Then take the limit as n goes to infinity
Remark: The partial sums $\{[\sum_{i=1}^{n} (x_i - y_i)^2]^{\frac12}\}_{n \in \mathbb{Z}_+}$ can be considered as a sequence and $\{[\sum_{i=1}^{n} (x_i)^2]^{\frac12} + [\sum_{i=1}^{n} (y_i)^2]^{\frac12}\}_{n \in \mathbb{Z}_+}$ can be considered a sequence too.