Please if someone can help, show that $$diam(A\cup B)\le diam(A)+diam(B)+d(A,B)$$ Where $diam(A)=sup\{d(x,y)|x,y\in A\}$ and $d(A,B)=inf\{d(a,b)|a\in A,b\in B\}$
Where $A,B\subseteq X$ and $(X,d)$ is a metric space.
I hope I've been clear enough, Thanks.
Let $x,y\in A \cup B$. Then either
$$d(x,y) \le \operatorname{diam}(A) \le \operatorname{diam}(A) + \operatorname{diam}(B) + d(A,B).$$
$$d(x,y) \le d(x,a) + d(a,b) + d(b,y) < \operatorname{diam}(A) + (d(A,B) + \epsilon) + \operatorname{diam}(B) .$$
Since $\epsilon$ was arbitrary, $$d(x,y) \le \operatorname{diam}(A) + \operatorname{diam}(B) + d(A,B).$$
$x\in B, y\in A$. Same reasoning as in $2$.
$x, y\in B$. Same reasoning as in $1$.
So in all cases, $d(x,y)$ is bounded above by $\operatorname{diam}(A) + \operatorname{diam}(B) + d(A,B)$. Since $x$ and $y$ were arbitrary, $\operatorname{diam}(A) + \operatorname{diam}(B) + d(A,B)$ is an upper bound for the set $\{d(x,y): x,y\in A\cup B\}$. Therefore
$$\operatorname{diam}(A\cup B) \le \operatorname{diam}(A) + \operatorname{diam}(B) + d(A,B).$$
Correct me if I'm wrong (please do), but I think that kobe's answer is only partially correct, as in case no. 2, they made the incorrect assumption that for any y from B and that for any $\epsilon \in \mathbb{R}$ there is an $a \in A$ such that $d(a, y) \leq d(A, B) + \epsilon$. You can draw a pretty simple Venn diagram-like picture to see that this is false. If you want another solution, see Show that $\operatorname{diam}(A\cup B)\le \operatorname{diam}(A)+\operatorname{diam}(B)+d(A,B)$.
