Choose $0<\epsilon\ll1$ as in (c). Then
$$
D=(D-\epsilon A)+\epsilon A
$$
is the sum of a <nef divisor> and an <ample real divisor>. The <nef-plus-ample ampleness lemma> gives
$$
\boxed{D\text{ is ample}.}
$$
For clarity, this last lemma follows from <Kleiman's criterion> and the <convex cone> property: if $x$ is an interior point of the <nef cone> and $y$ lies in that cone, translating a small neighbourhood of $x$ by $y$ stays in the cone. Thus $x+y$ remains in its interior, which is the <ample cone> on a <projective scheme>.

The complete argument proves the <real Nakai–Moishezon criterion> rather than assuming it: curve positivity gives nefness, rational approximation and a proved section-count inequality give bigness, induction and the finite-support argument give a uniform <ample> subtraction, and the nef-plus-ample lemma concludes <ampleness>. The zero-dimensional case is automatic, and <ampleness on reduced components> handles reducibility and nilpotents.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 134 / 3 / iii / d / Solution

{title2=$D^{\dim V}\cdot[V]>0\quad\text{for all }\dim V>0$}

A real Cartier class on a <projective scheme> is <ample> if and only if its top self-intersection on every positive-dimensional integral <subvariety> is positive. For the converse curve tests give nefness. Small <ample> perturbations and rational approximation give <ample> rational $B,C$ with $D-A=B-C$ satisfying the <algebraic Morse inequality for ample divisors>, hence bigness. Induction gives <ample> restrictions on codimension-one <subvarieties>; <uniform ample subtraction from a big divisor with ample exceptional restrictions> makes $D-\epsilon A$ <nef>. The <nef-plus-ample ampleness lemma> concludes. For reducible schemes perform the finite component tests simultaneously.

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 Incoming links: Real Nakai–Moishezon criterion