Proving that the sequence space is a metric space in $\mathbb{R}$ I need to show that $(X, d)$ is a metric space:
$X$ is the sequence space where its elements are sequences of real numbers, and $d : X \times X \rightarrow \mathbb{R}$ where d is defined as:
$$d(x,y) = \sum_{j=1}^\infty\frac{1}{2^j} \frac{|x_j - y_j|}{1 + |x_j - y_j|}$$
I think that I need to prove the right $\frac{|x_j - y_j|}{1 + |x_j - y_j|}$ is montone increasing but I have no clue how to start with proving the triangle inequality for this expression.  Any help is appreciated.

Let $f(x) = \frac{x}{1 + x}$. We wish to show that, for $x, y \ge 0$, we have $f(x + y) \le f(x) + f(y)$ (i.e. the function is subadditive over $[0, \infty)$). I hope you can see how this would prove triangle inequality for the proposed metric.
Suppose $x, y \ge 0$. Consider
\begin{align*}
f(x + y) \le f(x) + f(y) &\iff \frac{x + y}{1 + x + y} \le \frac{x}{1 + x} + \frac{y}{1 + y} \\
&\iff 1 - \frac{1}{1 + x + y} \le 2 - \frac{1}{1 + x} - \frac{1}{1 + y} \\
&\iff \frac{1}{1 + x} + \frac{1}{1 + y} \le 1 + \frac{1}{1 + x + y} \\
&\iff (1 + x + y)(1 + y) + (1 + x + y)(1 + x) \le (2 + x + y)(1 + x)(1 + y) \\
&\iff 1 + x + y \le (1 + x)(1 + y) \\
&\iff xy \ge 0,
\end{align*}
which is true.
You can also use the representation $f(x) = 1 - \frac{1}{1 + x}$ to quickly conclude that $f$ increases over $[0, \infty)$.
