We now compare the mixed non-pluripolar mass of $n$ closed positive currents, which may lie in different classes, with their individual masses. The idea is to replace each current by a solution of a Monge–Ampère equation with one and the same smooth right-hand side. Since singularity envelopes preserve mixed masses (proposition 16.21.2(3)), such a replacement does not change the mixed mass, and for the solutions the pointwise mixed lower bound of proposition 13.5.8 integrates to the desired inequality. The equations can be solved because, by theorem 17.5.37, every current of positive mass lies in a big class.
Theorem 17.5.41.0A392CD9486E Let $X$ be a connected compact Kähler manifold of dimension $n\geq 1$, and let $T_1,\ldots ,T_ n$ be closed positive $(1,1)$-currents on $X$. Then
AC065DF59BDD
17.5.42
\begin{equation} \int _ X\langle T_1\wedge \cdots \wedge T_ n\rangle \geq \prod _{i=1}^ n\left(\int _ X\langle T_ i^ n\rangle \right)^{1/n}. \end{equation}
Consequently the functions
\[ T\longmapsto \left(\int _ X\langle T^ n\rangle \right)^{1/n}, \qquad T\longmapsto \log \left(\int _ X\langle T^ n\rangle \right) \]
are concave on the convex cone of closed positive $(1,1)$-currents on $X$, with the convention $\log 0=-\infty $ for the second function.
Proof.
All the masses involved are finite by theorem 13.5.4. If one of the masses $\int _ X\langle T_ i^ n\rangle $ vanishes, the right side of (17.5.42) is zero and there is nothing to prove. Otherwise put $m_ i=\int _ X\langle T_ i^ n\rangle >0$. Both sides of (17.5.42) are positively homogeneous of degree one in each $T_ i$, by the multilinearity of non-pluripolar products (proposition 13.2.6), so after replacing $T_ i$ by $m_ i^{-1/n}T_ i$ we may assume $m_ i=1$ for all $i$.
By theorem 17.5.37, the class of each $T_ i$ is big. Let $\theta _ i$ be a smooth closed real $(1,1)$-form in the class of $T_ i$ (definition 10.2.17). By corollary 10.2.19 we can write $T_ i=\theta _ i+\mathrm{dd}^{\mathrm{c}}u_ i$ with $u_ i\in \mathrm{PSH}(X,\theta _ i)$. Set $\phi _ i=P_{\theta _ i}[u_ i]$. By the mass preservation in proposition 16.21.2(3) and by theorem 16.22.11, $\phi _ i$ is a model potential of mass one.
Let $\nu =\omega ^ n/\int _ X\omega ^ n$. This smooth probability measure is non-pluripolar, as noted after definition 17.5.2. In each big class $\{ \theta _ i\} $, theorem 17.5.15 gives
\[ w_ i\in \mathcal E(X,\theta _ i,\phi _ i),\qquad S_ i=\theta _ i+\mathrm{dd}^{\mathrm{c}}w_ i\geq 0,\qquad \langle S_ i^ n\rangle =\nu . \]
By proposition 16.22.14, $P_{\theta _ i}[w_ i]=P_{\theta _ i}[\phi _ i]=\phi _ i$. Applying the mixed mass preservation of proposition 16.21.2(3) first to the tuple $(w_1,\ldots ,w_ n)$ and then to the tuple $(u_1,\ldots ,u_ n)$, we obtain
\[ \int _ X\langle S_1\wedge \cdots \wedge S_ n\rangle =\int _ X\left\langle \bigwedge _{i=1}^ n (\theta _ i+\mathrm{dd}^{\mathrm{c}}\phi _ i)\right\rangle =\int _ X\langle T_1\wedge \cdots \wedge T_ n\rangle . \]
On the other hand, proposition 13.5.8, with all densities equal to one, gives $\langle S_1\wedge \cdots \wedge S_ n\rangle \geq \nu $. Hence the common value of these mixed masses is at least $\nu (X)=1$, which is the normalized inequality. Undoing the scalings proves (17.5.42).
For the concavity, let $T,S$ be closed positive $(1,1)$-currents and $0<t<1$; for $t=0,1$ there is nothing to prove. Put $a=\int _ X\langle T^ n\rangle $ and $b=\int _ X\langle S^ n\rangle $. By multilinearity,
\[ \int _ X\langle ((1-t)T+tS)^ n\rangle =\sum _{k=0}^ n\binom nk(1-t)^ kt^{n-k} \int _ X\langle T^ k\wedge S^{n-k}\rangle . \]
For $1\leq k\leq n-1$, (17.5.42) with $k$ factors $T$ and $n-k$ factors $S$ gives $\int _ X\langle T^ k\wedge S^{n-k}\rangle \geq a^{k/n}b^{(n-k)/n}$; the terms with $k=n$ and $k=0$ are $a$ and $b$, also when one of them vanishes. The binomial formula therefore yields
\[ \int _ X\langle ((1-t)T+tS)^ n\rangle \geq \big ((1-t)a^{1/n}+tb^{1/n}\big )^ n. \]
Taking $n$th roots proves that the function $T\mapsto (\int _ X\langle T^ n\rangle )^{1/n}$ is concave. If $a$ and $b$ are both positive, then, since the logarithm is increasing and concave,
\[ \begin{aligned} \log \Big(\int _ X\langle ((1-t)T+tS)^ n\rangle \Big) & \geq n\log \big ((1-t)a^{1/n}+tb^{1/n}\big )\\ & \geq (1-t)\log a+t\log b. \end{aligned} \]
If one of them is zero, the right side is $-\infty $ for $0<t<1$, so there is nothing to prove.
$\square$
The connectedness of $X$ cannot be omitted when $n\geq 2$. Let $X$ be the disjoint union of two compact Kähler manifolds $X',X''$ of dimension $n\geq 2$, with Kähler forms $\omega '$ and $\omega ''$. Let $T_1$ equal $\omega '$ on $X'$ and zero on $X''$, let $T_2$ equal zero on $X'$ and $\omega ''$ on $X''$, and let $T_3,\ldots ,T_ n$ equal $\omega '$ on $X'$ and $\omega ''$ on $X''$. Every mass $\int _ X\langle T_ i^ n\rangle $ is positive, but $\langle T_1\wedge \cdots \wedge T_ n\rangle =0$. The theorem applies separately on each connected component.
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