After regularizing we may assume that all functions involved are model functions. Write, symbolically, $T=(\omega+dd^{c}\varphi_{2})\wedge\dots\wedge(\omega+dd^{c}\varphi_{n})$. Then

$$\int(\psi-\varphi)\MA(\varphi_{1},\dots,\varphi_{n})=\int(\psi-\varphi)\omega\wedge T+\int(\psi-\varphi)\,dd^{c}\varphi_{1}\wedge T.$$

Since $0\leq\int(\psi-\varphi)\omega\wedge T\leq M$, the first term in the right-hand side satisfies

$$\int(\psi-\varphi)\omega\wedge T\leq M^{\frac{1}{2}}\left(\int(\psi-\varphi)\omega\wedge T\right)^{\frac{1}{2}}.$$

By the Cauchy-Schwarz inequality (Corollary 3.3), the second term is bounded by

$$\left(\int(\psi-\varphi)\,dd^{c}(\varphi-\psi)\wedge T\right)^{\frac{1}{2}}\left(\int(-\varphi_{1})\,dd^{c}\varphi_{1}\wedge T\right)^{\frac{1}{2}}$$

By the assumption that $-M\leq u_{1}\leq 0$ and $\int\omega^{n}=1$ we have

$$0\leq\int(-\varphi_{1})\,dd^{c}\varphi_{1}\wedge T=\int\varphi_{1}\omega\wedge T-\int\varphi_{1}(\omega+dd^{c}\varphi_{1})\wedge T\leq M.$$

Similarly,

	$\displaystyle 0\leq\int(\psi-\varphi)\,dd^{c}(\varphi-\psi)\wedge T$	$\displaystyle=\int(\psi-\varphi)(\omega+dd^{c}\varphi)\wedge T-\int(\psi-\varphi)(\omega+dd^{c}\psi)\wedge T$
		$\displaystyle\leq\int(\psi-\varphi)(\omega+dd^{c}\varphi)\wedge T.$

Putting this together, and using the concavity of the square root, we get

$$\int(\psi-\varphi)\MA(\varphi_{1},\dots,\varphi_{n})\leq\\ M^{\frac{1}{2}}\left(\left(\int(\psi-\varphi)\omega\wedge T\right)^{\frac{1}{2}}+\left(\int(\psi-\varphi)(\omega+dd^{c}\varphi)\wedge T\right)^{\frac{1}{2}}\right)\\ \leq 2M^{\frac{1}{2}}\left(\int(\psi-\varphi)(\omega+dd^{c}\frac{\varphi}{2})\wedge T\right)^{\frac{1}{2}}.$$

The lemma follows (with the constant $4M/(2M)^{\frac{1}{2^{n}}}<4M$) by repeating this argument $n-1$ times, successively replacing $\varphi_{2},\dots,\varphi_{n}$ by $\varphi/2$. ∎