Inequality on Kähler classes

It is a consequence of the Khovanskii-Teissier inequality for Kähler classes (which was proved by Gromov and Demailly on Kähler manifolds, the algebraic case is also in Lazarsfeld's book):

$$\int_X \omega_1^{n-1}\wedge\omega_2 \geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{1}{n}},$$

which implies

$$\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)\geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n+1}{n}}.$$

Take this inequality and the same inequality with the roles of $\omega_1$ and $\omega_2$ exchanged, and sum them up:

$$\begin{split}&\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)+\left(\int_X \omega_2^{n-1}\wedge\omega_1 \right)\left(\int_X\omega_1^n\right)\\ &\geq \left(\int_X \omega_1^n\right)^{\frac{n-1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n+1}{n}}+\left(\int_X \omega_1^n\right)^{\frac{n+1}{n}}\left(\int_X\omega_2^n\right)^{\frac{n-1}{n}}.\end{split}$$

Now use the Young inequality $$2xy \leq x^{\frac{n-1}{n}}y^{\frac{n+1}{n}}+x^{\frac{n+1}{n}}y^{\frac{n-1}{n}},$$

and get

$$\left(\int_X \omega_1^{n-1}\wedge\omega_2 \right)\left(\int_X\omega_2^n\right)+\left(\int_X \omega_2^{n-1}\wedge\omega_1 \right)\left(\int_X\omega_1^n\right)\geq 2\left(\int_X\omega_1^n\right)\left(\int_X\omega_2^n\right),$$

which is equivalent to your inequality. If equality holds, then equality must hold in Khovanskii-Teissier, hence $\omega_1$ and $\omega_2$ are proportional by Boucksom-Favre-Jonsson, but equality also holds in Young, hence $\omega_1=\omega_2$.

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.