For $A,B,C\subset X$where $X$ is a metric space under some $d$, check if the triangle inequality holds for $d_m(A,B)=\min_{x\in A,y\in B}\{d(x,y)\} $ $$d_m(A,B)=\min_{x\in A,y\in B}\{d(x,y)\} $$
Is it the case that $$d_m(A,C)\leq d_m(A,B)+d_m(B,C)$$ based on the definition of $d_m$ and the fact that $d$ is already some arbitrary metric on $X$?
I am rather unsure how to start this particular problem. I already know $$d(x,z)\leq d(x,y) + d(y,z)$$ from the fact that $d$ itself is a metric on $X$, which I imagine will be useful.
However, I need a hint to help me towards proving (or contradicting) the statement in the question. Any starting points will be appreciated.
 A: Hint: 
Suppose A and B have a point in common, and B and C have a point in common, and A and C have no points in common.
Spoiler:

If $A\cap B\neq\emptyset$ then $d(A,B)=0$ and similarly for $B\cap C$. Now if $A\cap C=\emptyset$, $d(A,C)>0$, so it doesn't hold.

A: Try to create some extreme situations where the claim fails. Take $\mathbb R$ with the Euclidean metic, and consider $A$ and $C$ to be some really far away subsets, say even just a singleton each, but far away. Clearly, $d(A,C)>0$. Now for $B$ you can choose all of $\mathbb R$. What happens then?
Remark: the minimum in the definition of distance between subsets need not be attained. One thus uses the infimum. 
A: You need to use infimum instead of minimum in the definition.
Let $A\subseteq X$ and define $d(x,A)=\inf\lbrace d(x,a) |a\in A\rbrace$, you can prove that


*

*$d(x,A)\leq d(x,y)+d(y,A)$


Then for $A,B, C\subseteq X$, let $c\in C$, now
$$\begin{align}
d(A,B)=&\inf\lbrace d(a,b)| a\in A\;,b\in B\rbrace\\
&=\inf \lbrace d(a,B) |a\in A\rbrace\\
&\leq \inf\lbrace d(a,c)+d(c,B) |a\in A\rbrace\\
&\leq \inf \lbrace d(a,c) |a\in A\rbrace +d(c,B)\\
&=d(A,c)+d(c,B)\quad\quad \forall\; c\in C
\end{align}
$$
and that is the best you can say regarding a "triangle inequality".
