Let $p>0$ and consider a metric space $(X,d)$. I have recently come across a problem where the space $(X,d^q)$ provides is natural; where $q>1$. However, the triangle inquality break (i.e. it is a "semimetric space").
I'm wondering, in this case, does $(X,d^p)$ satisfy some kind of "generalized triangle inequality"?
More generally, let $p>1$ and $a,b>0$, and set $\tilde{d}:X\times X\rightarrow [0,\infty)$ defined by $$ \tilde{d}(x,z)\mapsto a \,d(x,z) + b\,d(x,z)^p \qquad (\text{ for all }x,z\in X) $$ Then, for every $x,y,z\in X$ we have that $$ \begin{aligned} \tilde{d}(x,z) = &\, a \,d(x,z) + b\,d(x,z)^p \\ \le & a(d(x,y)+d(y,z)) + b(d(x,y)+d(y,z))^p\\ \le & a(d(x,y)+d(y,z)) + 2^{p-1}\,b(d(x,y)^p+d(y,z)^p)\\ \le & 2^{p-1}\,a(d(x,y)+d(y,z)) + 2^{p-1}\,b(d(x,y)^p+d(y,z)^p)\\ = & 2^{-1}\big(\tilde{d}(x,y)+\tilde{d}(x,z)\big). \end{aligned} $$
For more details, see Qinglan Xia - The Geodesic Problem in Quasimetric Space, J. Geom. Anal. 2009.
