Distance in metric space, triangle inequality problem Let $(X, d)$ be a metric space. Let $t\in (0,1]$. Show that $d^t: X\times X\to\mathbb{R}$
$$d^t (x,y) := d(x,y)^t, \forall x,y\in X$$
is also a distance function.
Problematic bit is the triangle inequality, when $0<t<1$
$$d (x,y)^t\leq d (x,z)^t+d (z,y)^t$$
Not sure how to tackle this one: we know that if $0<x<1, y\geq 1$ and $0<t<1$, then $x<x^t<1$ and $1\leq y^t\leq y$. So, if $d(x,y)<1$ and at least one of the right hand distances is $\geq 1$, then everything is fine.  
In general, I think I am overlooking something about such problems, I always want to systematically work through every possible case. Doesn't seem to be too efficient. Please, give hints on how to solve the problem.

I think I got it now (hope I'm not confusing things again). Setting $a := d(x,z)/d(x,y), b:= d(z,y)/d(x,y)$ the problem reduces to showing
$$ a^t + b^t \geq 1$$
If $0 < x < 1$, for $0 < t < 1$ we have $ x < x^t$, which gives
$$ \left(\frac{a}{a+b}\right)^t + \left(\frac{b}{a+b}\right)^t \geq \left(\frac{a}{a+b}\right) + \left(\frac{b}{a+b}\right) = 1$$
which shows
$$ a^t + b^t \geq (a+b)^t \geq 1$$ as $a+b\geq 1$.
