The largest element in magnitude of a Hermitian positive definite matrix is on the diagonal

Anyone can help provide a proof for

The largest element in magnitude of a Hermitian positive definite matrix is on the diagonal.

I found one related question Why is the largest element of symmetric, positive semidefinite matrix on the diagonal? whose answer proves a similar claim for symmetric positive definite matrix. However, I think the prove fails for a matrix with complex numbers, as the imaginary part also contributes to the magnitude.

Suppose the complex matrix $M$ is positive definite. The diagonal entries $m_{ii}$ are real and non-negative. As a consequence the trace, $tr(M) ≥ 0$. Furthermore, since every principal sub matrix (in particular, $2\times 2$) is positive definite, $${\displaystyle |m_{ij}|\leq {\sqrt {m_{ii}m_{jj}}}\leq {\frac {m_{ii}+m_{jj}}{2}}}$$ and thus $${\displaystyle \max |m_{ij}|\leq \max |m_{ii}|}$$