Lemma 8.2.4.94C4E018DD25 If $\alpha ,\gamma \in \Gamma _ m(\omega )$, then
\[ \frac{\alpha ^{m-1}\wedge \gamma \wedge \omega ^{n-m}}{\omega ^ n} \frac{\alpha \wedge \omega ^{n-1}}{\omega ^ n} \geq \frac1{m!} \frac{\alpha ^ m\wedge \omega ^{n-m}}{\omega ^ n} \frac{\gamma \wedge \omega ^{n-1}}{\omega ^ n}. \]
The assertion extends to $\gamma \in \overline{\Gamma _ m(\omega )}$ by continuity.
Proof.
Write $\sigma _ k=\alpha ^ k\wedge \omega ^{n-k}/\omega ^ n$, and $M_ k=\alpha ^{k-1}\wedge \gamma \wedge \omega ^{n-k}/\omega ^ n$. For $1\leq k<m$, the cone shift with index $k+1$ and Gårding positivity give
\[ M_{k+1}\geq tM_ k \quad \left(0\leq t<\frac{\sigma _{k+1}}{(k+1)\sigma _ k}\right). \]
Let $t$ increase to the displayed endpoint. Iterating, starting with $M_1=\gamma \wedge \omega ^{n-1}/\omega ^ n$, yields
\[ M_ m\geq \frac{\sigma _ m}{m!\sigma _1}M_1. \]
All denominators are positive. Approximating a form in the closed cone by $\gamma +\delta \omega $, $\delta \downarrow 0$, gives the last assertion.
$\square$
Pome
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