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Choose an ample Cartier divisor $H$. Put initially $A=\delta H$, $A^{\prime }=tH$ for small positive real $\delta ,t$. Since $D$ is nef, the nef-plus-ample ampleness lemma makes $D+tH$ ample. The polynomialhas $P(0,0)=D^n\cdot [X]>0$. Thus the desired strict inequality holds when $\delta ,t$ are sufficiently small and positive.

We must also arrange rationality of the two specified classes; $D$ itself need not be rational. Choose rational ample classes $B$ and $C$ sufficiently near $D+tH$ and $(\delta +t)H$, and defineThen $A^{\prime }$ is close to $tH$ and $A$ is close to $\delta H$, so both are ample real divisors by openness of the ample cone. Also $D+A^{\prime }=B$ and $A+A^{\prime }=C$ are rational and ample. Continuity preserves the strict inequality, giving

Here is the needed algebraic Morse inequality for ample divisors, with its section-count proof. Choose rational Cartier divisor representatives of $B,C$ and a common positive integer $q$ making $qB,qC$ very ample integral Cartier divisors. For the section-count argument rename these scaled representatives $B,C$; undoing this scaling restricts section indices to sufficiently divisible multiples and leaves bigness unchanged. Choose an effective Cartier divisor $G\in |C|$ by taking a defining section that avoids the associated points of ${\mathcal{O}}_X$. Repeated divisor restriction exact sequences giveBecause $C$ is very ample, for each $j$ a section of $jC$ avoiding the finitely many associated points of ${\mathcal{O}}_G$ gives an injection into ${\mathcal{O}}_G(mB)$. Thus every summand is at most $h^0(G,{\mathcal{O}}_G(mB))$. By Serre vanishing and asymptotic Riemann–Roch for the ample $B$,For $n=1$, the restriction term is the constant length of $G$, giving the same formula directly. The positive coefficient proves that $B-C$, hence $D-A$, is big. Scaling back preserves bigness, so

For a general projective scheme, enforce the same strict inequality separately on each positive-dimensional reduced irreducible component $V_j$, using its own dimension $n_j$. At $(\delta ,t)=(0,0)$ every such expression equals the positive number $D^{n_j}\cdot [V_j]$. Finitely many conditions are preserved by one sufficiently small choice and one sufficiently close rational approximation on $N^1(X{)}_{\mathbb{R}}$. The top-dimensional inequalities imply the printed inequality for $[X]$ with its positive generic multiplicities; the section proof on each component makes $D-A$ componentwise big. This avoids inferring bigness on every component from just a positive sum. The displayed inequality is used for $n\ge 1$. For $n=0$, its literal $n-1$ intersection power is undefined; handle this vacuous positivity case separately. Every line bundle is ample, all numerical classes are zero, and the componentwise bigness convention makes the conclusions automatic.

Choose $0<ϵ\ll 1$ as in (c). Thenis the sum of a nef divisor and an ample real divisor. The nef-plus-ample ampleness lemma givesFor 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.

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 $D-ϵA$ nef. The nef-plus-ample ampleness lemma concludes. For reducible schemes perform the finite component tests simultaneously.