TITLE: infimum in inequalities
QUESTION [5 upvotes]: I am just starting on analysis and I have a question about the concept of infimum:
Consider this scenario:
Theres is metric space (X,d) and a set $A \subseteq X$, hence for any $x,y \in X$ and $a\in A$
$ d(x,a) \leq d(x,y) +d(y,a) \; \forall a \in A$ .... this is true since d() is a valid metric. 
Now can I justify the following statement: 
$\inf_{a\in A} d(x,a) \leq d(x,y) + \inf_{a \in A} d(y,a)$ 
Basically $ d(x,A) \leq d(x,y) + d(y,A) $?
Is is it OK to take the infimum on both sides of the inequality? How to argue that doing so is right? the '$a\in A$' which minimises (weakly speaking here) d(x,a) need not be the same '$a\in A$' which minimises d(y,a)?
P.S: This question is in the backdrop of proving that the function d(x,A) is continuous

REPLY [4 votes]: Let $x,y$ be fixed members of the metric space and suppose $A$ is a subset of the metric space. Let $D = \{ d(x,a) \in \mathbb{R} : a \in A \}$ and let $E = \{ d(x,y) + d(y,a) \in \mathbb{R} : a \in A\}.$ 
Let arbitrary $e \in E$ be given. Then $e = d(x,y) + d(y,a)$ for some $a \in A$. Thus, there is some $d \in D$ (namely, $d(x,a)$) such that $d(x,a) \leq e.$ By definition of $\inf,$ $\inf D \leq d$ and therefore $\inf D \leq e.$ Since $e$ was arbitrary, this is true $\forall e \in E$. This shows that $\inf D$ is a lower bound for $E.$ Again, by definition of $\inf,$ this must mean that $\inf D \leq \inf E,$ as required. Yes, your assertion is ok.
Apologies for my earlier comment which was misleading because these two propositions are different:

$\forall b \in B, \exists a \in A : a \leq b$
$a \in A, b \in B \Rightarrow a \leq b$