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If X is a normal projective variety with an ample line bundle $L$, and $\pi:Y\to X$ a resolution of $X$ and $E$ be the exceptional divisor, then is it true that $A\pi^{\star}L-[E]$ is always ample for $A$ sufficiently large?

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You should probably state what you want a little more precisely. As it is currently stated, it allows for the possibility that $\pi$ is not projective in which case there is no chance. The statement also allows a small resolution in which case $E$ is empty and $a\pi^*L$ is not ample for any $a$.

Unfortunately, even if you assume that $\pi$ is projective and $E$ is the entire exceptional set(!), it is possible that this fails. The condition you need is that $-E$ has to be relatively ample for $\pi$. This can fail already for surfaces. For instance, assume that $E$ has two components, both with negative self-intersection, say $-n$ and $-m$, and let's say that the intersection number of the two components is $n+e$ for some positive number $e$. The intersection matrix has to be negative definite, which means that we need $nm> (n+e)^2$, but this is easy to satisfy by making $m$ really big. In this case, $E$ restricted to the component with self-intersection $-n$ has positive degree, so $-E$ cannot be ample.

This suggests that you cannot allow $E$ to have more than one component. On the other hand, in that case you are OK. Alternatively, if you allow different coefficients for the components of $E$ then a similar statement holds.

Addendum: In response to @freidtchy's comment-question below here is an explanation of the last sentence above. Yes, I meant exactly that there exist $a_i>0$ such that $A\pi^*L-\sum a_iE_i$ is ample. Since $\pi$ is ample, there exists a $\pi$-ample Cartier divisor on $Y$. With a little bit of work one may assume that there is one which is entirely supported on $E$ (the possible components that are a priori not can be exchanged to something pulled back from $X$ plus something supported on $E$ and the pull-back of anything is $\pi$-trivial). In other words, one has a $\pi$-ample Cartier divisor $\sum b_iE_i$. Then the Negativity Lemma [Kollár-Mori-98, 3.39] tells us that all the $b_i$ are negative and then choosing a large enough $A$ gives the claimed statement.

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  • $\begingroup$ Can you clarify what you mean by if you allow different coefficient for the components of E then a similar statement holds? Do you mean something like if $E_i$ are the irreducible components of $E$, then there always exists $a_i>0$ such that $A\pi^{\star}L-\sum a_iE_i $ is ample? Can you point me to some reference? $\endgroup$ – freidtchy Nov 13 '19 at 1:49
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    $\begingroup$ Yes. I added some explanation and a reference in the body of my answer. $\endgroup$ – Sándor Kovács Nov 13 '19 at 18:50

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