What values of $p$ make $d$ a metric? I'm trying to determine what values of p make the following a metric:
$$d(x,y)=|x-y|^p$$ for x,y∈R.
Obviously, it's not difficult to show that this satisfies the first two conditions for most values of $p$. I'm just not seeing how to prove the triangle inequality one way or the other for some values of $p$, specifically values in $(-1,1)$ and for positive non-integers. Any guidance would be appreciated.
HINT: The triangle inequality applied to the points $0,1$, and $2$ already imposes a significant condition on $p$. Notice also that 

$$|x-z|^p\le|x-y|^p+|y-z|^p\quad\text{for all }x,y,z\in\Bbb R$$

if and only if

$$|u+v|^p\le|u|^p+|v|^p\quad\text{for all }u,v\in\Bbb R\;,$$

and you may find the latter version easier to think about and work with.
Try using minkowski's inequality. You then can use it to prove that $(a+b)^p&lt;a^p+b^p$ with $0&lt;p&lt;1$. Finally work with this to get the triangle inequality, dunno if this helps, Im a first timer in analysis.