I am currently working on this problem;
Let $A=[a_{ij}] \in M_n$ be positive definite. Partition $$ A=\begin{bmatrix} A_{11}&x\\x^* &a_{nn}\end{bmatrix}$$ in which $A_{11}\in M_{n-1}$. Show that $$\det(A) = (a_{nn}-x^*A_{11}^{-1}x) \det (A_{11}) \leq a_{nn} \det (A_{11})$$ with equality iff $x=0$.
Your assistance will be highly appreciated.
Let $$ Y=\begin{bmatrix}I&0\\-x^*A_{11}^{-1}&1\end{bmatrix},\ \ Z=\begin{bmatrix}I&-xA_{11}^{-1}\\0&1\end{bmatrix}. $$ These are triangular matrices with diagonal equal $1$, so they have determinant equal to $1$. Then $$ \det A=\det YAZ=\det\begin{bmatrix}A_{11}&0\\0&a_{nn}-x^*A_{11}^{-1}x\end{bmatrix}=(a_{nn}-x^*A_{11}^{-1}x)\,\det A_{11}. $$ From $\det A>0$, $\det A_{11}>0$, we deduce that $a_{nn}-x^*A_{11}^{-1}x>0$. For $y=A_{11}^{-1}x$, $$ x^*A_{11}^{-1}x=y^*A_{11}y>0, $$ so $0<a_{nn}-x^*A_{11}^{-1}x\leq a_{nn}$. Equality implies $x^*A_{11}^{-1}x=0$, which occurs precisely when $x=0$.