Consider $f(x) = \sin x$ , then $f''(x) = -\sin x$ . Since $A,B$ and $C$ are angles of a triangle, then it follows that $0 <A,B,C<\pi$ and also $0 < \sin \dfrac A2 , \sin B , \sin C \leq 1$ , this tells us that $-1 \leq -\sin \dfrac A2 ,- \sin B , -\sin C < 0$ , so $f(x)$ is concave for all values $\dfrac A2 , B, C$ .

Applying Jensen's inequality , we get

$\dfrac{2\sin (A/2)}4 + \dfrac{\sin B}4 + \dfrac {\sin C}4 \leq \sin \left ( \dfrac{2 (A/2) + B+C}4 \right) = \sin \left ( \dfrac{A+B+C}4 \right) = \sin \left(\dfrac\pi4 \right) = \dfrac{\sqrt2}2$

Simplifying this gives

$2\sin \dfrac A2 + \sin B + \sin C \leq 2\sqrt2 .$

Equality holds when $\dfrac A2 =B=C=\dfrac \pi4$ .

So our answer is $\lfloor 10^4 \cdot 2\sqrt2 \rfloor = \boxed{28284}$ .