Is this the correct way to show this function is a metric?

Question

Is this the correct way to prove that this is metric function ?

let $d_1:\Bbb Z \times \Bbb Z \rightarrow \Bbb R^+_0$

$d(m,n):=|m-n|^3$

i. $d(x,y)\geq 0$ with equality iff x=y

$|m-n|\geq0$ because it is an absolute value

if $m \neq n$ $|m-n|>0$ if m=n $|m-n|=0$

ii.$ d(m,n)=|m-n|^3=|n-m|^3=d(n,m)$

iii.

$d(m,p)=|m-p|^3$

$d(m,n)=|m-n|^3$

$d(n,p)=|n-p|^3$

$|m-p|^3\leq |m|^3+|p|^3$

$|m-n|^3\leq |m|^3+|n|^3$

$|n-p|^3\leq |n|^3+|p|^3$

so

$|m-n|^3+|n-p|^3 =|m-n+n-p|^3=d(m,p)$

and so $d(m,p)\leq d(m,n)+d(n,p)$

Answer 1

The last part goes completely wrong as Daniels comment suggests.
You need to show $$ |a-b|^3 \leq |a-c|^3 +|c-b|^3 $$ which is not true. Note that if $x$ close to $0$, then $x^3$ is much and much closer to $0$. An easy counter example would thus be to take $a=0$, $b=2$ and $c=1$ to get $$ 2^3 > 1^3 + 1^3 $$ which is thus exactly the other way around.