Let $\pi:X\to\mathbb{P}^3$ be the blowing up at single point with $E$ be the exceptional divisor. Let $H=\pi^\ast\mathcal{O}\_{\mathbb{P}^3}(1)$.


In [Ample divisors on the blow up of $\mathbb{P}^3$ at points](http://www.ams.org/journals/proc/2002-130-09/S0002-9939-02-06488-2/S0002-9939-02-06488-2.pdf) , we know that the divisor on $X$ of the form $$dH-E$$ with $d\geq5$ is ample. But in this case I am very much willing to know a easy and straightforward proof.



Ample divisors on $\mathbb{P}^3$ blow-up along single point
Let $[x:y:z:w]$ be the homogeneous coordinates on $\mathbb{P}^3$ and suppose that the point blown up is given by the vanishing of the first three coordinates. Introduce a projective plane with coordinates $[X:Y:Z]$ then the blowup is a closed subset in $\mathbb{P}^3 \times \mathbb{P}^2$ given by the equations
$$
xY = yX; xZ = zX; yZ = zY.
$$
The bundle $\mathcal{O}(1) \boxtimes \mathcal{O}(1)$ on $\mathbb{P}^3 \times \mathbb{P}^2$gives the Segre Embedding, so it is very ample on the product of projective spaces. Its restriction to the blowup is also very ample and it must be of the form $dH - k E$. We can computed $d$ and $k$ by intersecting with (a) a line in $\mathbb{P}^3$ that avoids the point blown up, which gives $d = 2$; and (b) with a line in the exceptional fiber $[0:0:0:1] \times \mathbb{P}^2$, which gives $k=1$. So $2H - E$ is very ample. 


