TITLE: Is [mD] very ample if D is ample?
QUESTION [10 upvotes]: Let D be an ample R-divisor, is the round down [mD] very ample for any sufficiently divisible number m?
I think it's true. But I do not know how to arrange an argument.

REPLY [5 votes]: I am not sure if this is the shortest proof, but I think that it is a proof. 
Let $A=$ very ample line bundle. After replacing D by a multiple, you may assume that 
$$C=D - K_X - (n+1) A$$ is ample where $n=\dim X$. 
By Angehrn and Siu, we know that $$K_X+(n+1)A + \text{(ample line bundle)}$$ is very ample. 
Now 
$$[mD] = K_X+ (n+1) A  + C + [mD] - D$$ 
and you just need to make sure that $[mD]+C-D$ is ample if $m\gg 0$. This is easy to check by Diophantine approximation.