Let $X$ be a complex projective manifold of complex dimension $n$ and $A\to X$ an ample line bundle. Let $L\to X$ be a line bundle such that
$$
c\_1(L)^k\cdot c\_1(A)^{n-k}>0,\quad k=1,\dots,n.
$$
Is it true that then $L$ is big?


The answer is yes if $n\le 2$: for $n=1$ there is nothing to prove, and for $n=2$ the positivity of the top self intersection $c\_1(L)^2>0$ says that $L$ or its dual is big. But then $c\_1(L)\cdot c\_1(A)>0$ implies that in fact $L$ is big.


The answer is again yes in all dimensions if $X$ is an abelian variety: in this case $L$ is moreover ample. This is because one can represent $c\_1(L)$ and $c\_1(A)$ by "constant" hermitian forms, the second being positive definite, and thus the intersection conditions simply tell that the elementary symmetric polynomials in the eigenvalues of the hermitian form representing $c\_1(L)$ are all positive. Thus, $L$ is positively curved and hence ample.


I strongly suspect anyway that the result is false in general.


Could you give for instance a counterexample in dimension three?


Thanks in advance.



A line bundle not big but with good intersection numbers
Here is a very straightforward contre-example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$, and take the following bundle:


$$L\_n=\pi^\*(O(n))\otimes O(E),$$
where $n$ satisfies two conditions: $$c\_1(O(n))^2\cdot c\_1(A)>-c\_1(O(E))^2\cdot c\_1(A),\;\;\;\;
c\_1(O(n))\cdot c\_1(A)^2>-c\_1(O(E))\cdot c\_1(A)^2,$$ 
it is obvious that such $n$ exists.


To that this bundle is what you want we just need the following two simple facts: $c\_1(\pi^\*(O(n)))\cdot c\_1(O(E))=0$ and $c\_1(O(E))^3=1$. $L\_{n}$ is not big because the $H^0(kL\_n)$ grows quadratically with $k$. 


Idea of this example works in dimensions $2m+1$. We take a semi-ample line bundle $L\_{sa}$ with Itaka dimension $2m$ <http://en.wikipedia.org/wiki/Iitaka_dimension> (in particular it is not big), and tensor it with a line bundle corresponding to an exceptional divisor $E$. We chose them so that $c\_1(L\_{sa})\cdot c\_1(O(E))=0$, i.e., these bundles "don't interact". For large $n$ the class $(c\_1(nL\_{sa}))^k$ (provided
$k\le 2m$) is represented by a cycle of a high degree (with respect to $A$), so it "beats" $(c\_1(O(E)))^k$. Finally
 $E^{2m+1}=1$.


