triangle inequality to show metric

Question

$d(x,y)= \begin{cases} 0 &\mbox{if } x=y \\ 1+\frac{1}{x+y} & \mbox{if } x\neq y \end{cases} $.

Show that $(\mathbb{Z}^+,d)$ is a metric space.

I'm stuck in proving triangle inequality.

Answer 1

The triangle inequality is $d(x,y)\leq d(x,z) + d(z,y)$. If any two of $x$, $y$, and $z$ are equal then the triangle inequality holds. Suppose then that $x$, $y$, and $z$ are distinct, in which case the triangle inequality reads $$ 1 + \frac{1}{x+y} \leq 1 + \frac{1}{x+z} + 1 + \frac{1}{z+y}. $$ Subtracting the $1$ gives $$ \frac{1}{x+y} \leq 1 + \frac{1}{x+z} + \frac{1}{z+y}, $$ and that inequality is what we would like to prove. I think you're overthinking it: in fact, the weaker inequality $$ \frac{1}{x+y} \leq 1 $$ is true when $x$ and $y$ are in $\mathbb Z^+$.

Answer 2

HINT: Notice that $\frac{1}{x+y}\leq \frac{1}{2}$.

Possible solution:

! Let $z\in\mathbb{Z^+}$, where $z\neq x$ and $z\neq y$. Using the hint, we obtain \begin{align*} d(x,y)=1+\frac{1}{x+y} < 2 < 2+\frac{1}{x+z}+\frac{1}{z+y}=d(x,z)+d(z,y) \end{align*}