Let D be an ample Rdivisor, 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.
Let D be an ample Rdivisor, 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. 


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 $[mD]+CD$ is ample if $m\gg 0$. This is easy to check by Diophantine approximation. 


Edit 3: This is my 3rd attempt to give an elementary proof. It is essentially the same proof as in Edits 1 and 2, but with some corrections, and hopefully will be clearer. I hope you see that the idea is very simple and geometrically almost obvious. If it seems complicated, then the fault is in my exposition. Set Up: Let $D_1, \ldots, D_k$ be ample divisors and $D := \sum_j r_jD_j$ for positive real numbers $r_1, \ldots, r_k$. Also, let $D_j = \sum_{i=1}^N a_{ji} C_i$, for irreducible divisors $C_i$ and integers $a_{ji}$. We want to show that $[mD]$ is very ample for large $m$. In the proof we will use the following fact about finite sums of integral points in a lattice: Lemma: Let $v_1, \ldots, v_k \in \mathbb{Z}^N$ such that $\mathbb{Z}$span of $v_j$'s equals $\mathbb{Z}^N$. Let $P$ be the convex hull (over $\mathbb{R}$) of $\lbrace 0, v_1, \ldots, v_k \rbrace$. Then there exists a positive real number $c$ such that for all $n \geq 1$, if $v \in nP \cap \mathbb{Z}^N$ such that the (Euclidean) distance of $v$ from both the origin and the boundary of $nP$ is greater than $c$, then $v$ is in fact an nonnegative integral linear combination of $v_1, \ldots, v_k$. The above statement (actually a more precise formulation of it) is due to Khovanskii. The proof is very elementary and beautiful, and is in Proposition 2 of this article. Here starts the proof: Step 1: Without loss of generality we may assume that $\mathbb{Z}$span of $D_j$'s equals the $\mathbb{Z}$span of $C_i$'s. Indeed, it follows from Kleiman's criterion, and finite dimensionality of $N_1(X)$ that for every $m \gg 1$ and $\epsilon := (\epsilon_1, \ldots, \epsilon_N) \in \lbrace 1, 0, 1 \rbrace^N$, $D_{m,\epsilon} := mD_1 + \sum_{i=1}^N\epsilon_i C_i$ is ample. Choosing different values of $\epsilon$ and $m$ and adding $D_{m,\epsilon}$'s to the collection of $D_j$'s, we may ensure that $\mathbb{Z}$span of $D_j$'s equals the $\mathbb{Z}$span of $C_i$'s. Moreover, and this is essential, choosing $D_{m,\epsilon}$'s to be sufficiently close to the ray generated by $D_1$, we may ensure that $D$ still lies in the interior of the cone generated by $D_j$'s, i.e. $D = \sum_{j=1}^k r_jD_j$ with each $r_j$ being a positive real number. Step 2: For each $j$, $1 \leq j \leq k$, let $v_j := (a_{j1}, \ldots, a_{jN}) \in \mathbb{R}^N$, i.e. $v_j$ is the "coordinate" vector of $D_j$ for each $j$ (and therefore $v_j \in \mathbb{Z}^N$ for each $j$). Adding some big multiples of $D_j$'s to the existing collection of $D_j$'s if necessary, we may assume that $v := \sum r_j v_j$ is in the interior of the convex hull $P$ of $0, v_1, \ldots, v_k$. Step 3: For each $j$, $1 \leq j \leq k$, there exists a positive integer $m_j$ such that $mD_j$ is very ample for all $m \geq m_j$. Indeed, there is $l_j, n_j$ such that $n_jD_j$ is very ample and $mD_j$ is globally generated for all $m \geq l_j$. Setting $m_j := l_j + n_j$ does the job (due to Exercise II.7.5(d) of Hartshorne). Step 4: There exists a positive integer $m_0$ such that $m_0(D_1 + \cdots +D_k) + \sum s_jD_j$ is very ample for all collections of nonnegative integers $s_1, \ldots, s_k$. Indeed, set $m_0 := \max \lbrace m_1, \ldots, m_k \rbrace$ and apply the same exercise of Hartshorne. Step 5: Let $v, v_1, \ldots, v_k$ and $P$ be as in Step 2. Let $c$ be the constant we get from applying Khovanskii's lemma to $v_1, \ldots, v_k$. Let $v_0 := m_0(v_1 + \cdots + v_k)$, where $m_0$ is as in Step 4. Since $v$ is in the interior of $P$, it follows that if $m$ is sufficiently large, then $[mv]  v_0$ is in the interior of $mP$ and the distance of $[mv]  v_0$ from the origin and the boundary of $mP$ is bigger than $c$. Therefore, Khovanskii's lemma implies that $[mv]  v_0 = \sum a_j v_j$ for nonnegative integers $a_j$. Consequently, if $m$ is sufficiently large, then $$[mD] = m_0(D_1 + \cdots + D_0) + \sum a_j D_j$$ for nonnegative integers $a_1, \ldots, a_k$. Step 4 then tells that $[mD]$ is very ample. 

