Given a real smooth function $u$ on $\mathbb{C}^n$. Let the complex Hessian $$\left\{u_{j\bar{k}} := \frac{\partial}{\partial z^j}\frac{\partial}{\partial \bar{z^k}} u\right\}_{j,k\in\{1,..,n\}}$$ be a semi-positive Hermitian matrix with unitary eigenvectors $w_1, .., w_n$. Denote that $z^j = x^{2j-1} + i x^{2j}$. Then, $$u_{j\bar{k}} = \frac{1}{4}\left[(u_{2j-1, 2k-1} + u_{2j, 2k}) + i(u_{2j-1, 2k} - u_{2j, 2k-1})\right],$$ where $u_{\alpha, \beta} := \frac{\partial}{\partial x^\alpha} \frac{\partial}{\partial x^\beta} u.$ Denote $D^2u := (u_{j,k})_{j,k\in\{1,..,2n\}}$ be the real Hessian and $w_l = \xi_l+i\eta_l$, where $\xi_l$ and $\eta_l$ are real vectors. Now, we define vectors $$v_l := [\xi_l^1, \eta_l^1, \xi_l^2, \eta_l^2, ..., \xi_l^n, \eta_l^n]^T$$ and $$\tilde{v}_l := [-\eta_l^1, \xi_l^1, -\eta_l^2, \xi_l^2, ... , -\eta_l^n, \xi_l^n]^T$$.
The eigenvalues $\lambda_l$ of the complex Hessian $\{u_{j\bar{k}}\}$ can be expressed as $$\lambda_l =\sum_{j,k} w_l^j u_{j\bar{k}} \overline{w_l^k} = \frac{1}{4}\sum_{j,k} (\xi_l^j + i \eta_l^j)\left[(u_{2j-1, 2k-1} + u_{2j, 2k}) + i(u_{2j-1, 2k} - u_{2j, 2k-1})\right](\xi_l^k - i \eta_l^k) = \frac{1}{4}\left( v_l^T (D^2u) v_l + \tilde{v}_l^T (D^2u) \tilde{v}_l \right). $$ Hence, the determine of the complex Hessian $$\det(u_{j\bar{k}}) = \prod_{l=1}^n w_l^j (u_{j\bar k}) \overline{w_l^k} = \frac{1}{4^n} \prod_{l=1}^{n} \left( v_l^T (D^2u) v_l + \tilde{v}_l^T (D^2u) \tilde{v}_l \right) \geq \frac{1}{2^n} \sqrt{\prod_{l=1}^{n} \left( v_l^T (D^2u) v_l \right) \left( \tilde{v}_l^T (D^2u) \tilde{v}_l \right)}. $$
I want to prove an estimate of the real Hessian. Therefore, I want to show that whether the $$ \prod_{l=1}^{n} \left( v_l^T (D^2u) v_l \right) \left( \tilde{v}_l^T (D^2u) \tilde{v}_l \right) $$ can control the product of the diagonal part of the real Hessian $$\prod_{p=1}^{2n} (D^2 u)_{pp}.$$