Let $(X,\varrho)$ be a metric space, and define $d:X\times X\to \mathbb{R}$ by
$$d(x,y)=\inf\lbrace\varrho(x,z)+\varrho(z,y):z\in X\rbrace$$
Show that $d$ is a metric on $X$.
I have all the axioms but the triangle inequality, and just need to find the way to actually leverege that $\varrho$ obeys the triangle inequality in order to finish, but that infimum is stopping me.
On
Another approach (and generating an insight that may not be obvious) is to show that $d(x,y) = \rho (x,y)$ holds.
We have
$$d(x,y) = \inf\,\{\rho(x,z)+\rho(z,y): z \in X\} \le \rho(x,y)+\rho(y,y) = \rho(x,y), \tag{1}\label{eq1}$$
by simply choosing $z=y$ and using that $\rho(y,y)=0$.
On the other hand, due to triangualr inequality, we have
$$\forall z \in X: \rho(x,z)+\rho(y,z) \ge \rho(x,y). \tag{2}\label{eq2}$$
So $\rho(x,y)$ is a lower bound of the set $\{\rho(x,z)+\rho(z,y): z \in X\}$. With the infimum of a set being the largest lower bound, we get
$$d(x,y) = \inf\,\{\rho(x,z)+\rho(z,y): z \in X\} \ge \rho(x,y).$$
Now \eqref{eq1} and \eqref{eq2} together show $d(x,y) = \rho (x,y).$
$$d(a,c)\le\rho(a,b)+\rho(b,c)$$$$\le d(a,b)+d(b,c),$$ where both $\le$ are by definition of $d,$ and the second one ($\rho\le d$) also uses that $\rho$ satisfies the triangular inequality.