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I've found two versions of Hadamard's inequality :

(1) If $P$ is a $n\times n$ positive-semidefinite matrix, then : $$\det(P)\le\prod_{i=1}^n p_{ii} .$$ (2) For any $M$ is a $n\times n$ matrix, then : $$|\det(M)|\le\prod_{i=1}^n ||m_i || =\prod_{i=1}^n \left(\sum_{j=1}^n |a_{ij}|^2\right)^{1/2}.$$ It's easy to prove that $(2)\Rightarrow (1)$, but do we have $(1)\Rightarrow (2)$ ?

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    $\begingroup$ Yes it does. Let $M$ be any matrix. Then $MM^t$ is a positive semi-definite matrix. $\endgroup$ – mouthetics Jan 3 at 16:02
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Using comment of @mouthetics,

From (1) we know that $\sqrt{\det(MM^T)}\le\prod_{i=1}^n\{MM^T\}_{ii}^{1/2}=\prod_{i=1}^n \left(\sum_{j=1}^n |m_{ij}|^2\right)^{1/2}=\prod_{i=1}^n ||m_i ||$

Since $\det(MM^T)=\det(M)\det(M^T)=\det(M)^2$

we have $|\det(M)|=\sqrt{\det(MM^T)}\le\prod_{i=1}^n ||m_i ||$.

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