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Suppose $L$ is an effective divisor and $H$ is ample (On a smooth 3-fold) such that $L+H$ is nef.

Then show that $L+H$ is big ( $(L+H)^3 > 0$) ?

This was claimed in a paper, without proof. So I assume it should be well-known. I am not sure if the restriction on dimension is necessary or not.

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up vote 5 down vote accepted

This is indeed true. See for example Lemma 2.60 in Kollar-Mori Birational Geometry of algebraic varieties. In particular, it is shown that a Cartier divisor $D$ is big if and only if $mD \sim A + E$ for some ample divisor $A$ and effective divisor $E$. This is also proven in Corollary 2.2.7 in Lazarsfeld's Positivity of in Algebraic Geometry I.

Anyway, they make no assumptions on the dimension or singularities of the ambient variety, you also don't need the nef assumption.

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2 basic comments:

1) $L+H$ big is not equivalent to $(L+H)^3>0$

example: $\mathbb P^2$ blown up at a point, $H=$ any ample and $L=mE$ where $E=$exceptional curve and $m\gg 0$.

2) this is the proof (as mentioned above, it can be found in books and it doesn't require $H+L$ nef):

$h^0(X,a(L+H))\ge h^0(X,aH)\ge \frac { a^n}{n!} H^n + \mathcal O (a^{n-1})$

where $n=\dim X$ and $a\gg 0$ (simple consequence of Riemann-Roch and Serre vanishing).

$\Rightarrow L+H $ is big

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But the OP assumes that $D=L+H$ is nef, and in that case bigness is equivalent to $D^3>0$. – J.C. Ottem Jun 9 '11 at 12:01
Good comment... – Mohammad F. Tehrani Jun 9 '11 at 14:05

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