Let $X$ be a compact Kähler manifold of complex dimension $n$, and let
$\omega\_1, \omega\_2$ be Kähler classes on $X$. Denote the Lefschetz
operator of a Kähler class $\omega$ by $\Lambda\_{\omega}$. Then
$$
\Lambda\_{\omega\_1} \omega\_2 + \Lambda\_{\omega\_2} \omega\_1 \geq 2n
$$
with equality if and only if $\omega\_1 = \omega\_2$.


Has anyone seen this inequality? I checked in the usual places like
Lazarfeld's Positivity, but didn't find it. It seems simple enough that
someone must have come across it before.


(Aside one: We can rewrite this as an inequality on intersection numbers if
we want, since $\Lambda\_{\omega\_1} \omega\_2 = n \int\_X \omega\_2 \cup
\omega\_1^{n-1} / \int\_X \omega\_1^n$.)


(Aside two: On a complex torus, everything reduces to linear algebra. There
one can whittle the statement of the inequality down to that
$\operatorname{tr}(H) + \operatorname{tr}(H^{-1}) \geq 2n$ for any
Hermitian positive-definite matrix $H$, which follows from the
arithmetric-geometric inequality.)


The proof I found is a little amusing. The Kähler cone of $X$ has a
natural Riemannian metric $g$, which at a point $\omega$ is given by the
inner product that $\omega$ induces on real $(1,1)$-classes. If $u,v$
are such classes, then
$$
g(u,v)\_{\omega} = \Lambda(u) \Lambda(v) - \tfrac12 \Lambda^2(u \cup v).
$$
Given two points $\omega\_1, \omega\_2$ in the Kähler cone, we can
estimate the Riemannian distance between them by calculating the length
of any path between them. Doing so for the segment of the Euclidean line
that runs through the two gives the estimate
$$
d(\omega\_1,\omega\_2)^2 
\leq \Lambda\_{\omega\_1} \omega\_2 + \Lambda\_{\omega\_2} \omega\_1 - 2n,
$$
from which the inequality follows.



Inequality on Kähler classes
Here is a simple proof using Theorem 1.6.1 in Lazarsfeld book, which is the following:



> 
> **Theorem** (Demailly)
> 
> 
> If $H\_1,\ldots,H\_n$ are Kähler classes in a compact Kähler manifold of dimension $n$, then the following inequality holds:
> 
> 
> $$(H\_1 \cdots H\_n)^n \ge (H\_1^n)\cdots(H\_n^n).$$
> 
> 
> 


Let $H\_1$ and $H\_2$ be Kähler classes. By the AM–GM inequality,


$$\left[\frac{1}{2}\left( \frac{{H\_1^{n-1}H\_2}}{H\_1^{n}} + \frac{H\_2^{n-1}H\_1}{H\_2^{n}} \right)\right]^2 \ge\frac{{H\_1^{n-1}H\_2}}{H\_1^{n}} \cdot \frac{H\_2^{n-1}H\_1}{H\_2^{n}}. $$


Using the above theorem, it is easy to see that


$$({H\_1^{n-1}H\_2}) \cdot ({H\_2^{n-1}H\_1}) \ge {H\_1^{n}}{H\_2^{n}},$$


which finishes the proof.


