Given, Toeplitz matrix $T \in R^{n \times n}$: $$ T= \begin{bmatrix} \tau_0 & \tau_1 & \cdots & \tau_{n-1} \\ \tau_1 & \tau_0 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \tau_1 \\ \tau_{n-1} & \cdots & \tau_1 & \tau_0 \\ \end{bmatrix} $$ and denoted the $k$-th order leading principal submatrix of $T$ by $T_k$.
If $T$ is a positive definite matrix, how to prove the following inequality: $$ \det T_{k+1} \le \frac{(\det T_k)^2}{\det T_{k-1}} $$ ,where $\forall k \in \{1, \cdots, n\}$.
And, when the equality is attained?
The inequality can be rewritten as $\dfrac{\det{T_{k-1}}}{\det{T_k}}\le\dfrac{\det{T_k}}{\det{T_{k+1}}}$ or equivalently (since $X^{-1}=\frac{\operatorname{adj} X}{\det X}$), $$ (T_k^{-1})_{11}\le(T_{k+1}^{-1})_{11}. $$ In other words, it means the $(1,1)$-th element of $T_k^{-1}$ is weakly increasing.
To prove it, write $T_{k+1}=\pmatrix{T_k&v\\ v^T&\tau_0}$. Using Schur complement, we have $$ T_{k+1}^{-1} =\pmatrix{T_k&v\\ v^T&\tau_0}^{-1} =\pmatrix{\left(T_k-\frac{1}{\tau_0}vv^T\right)^{-1}&\ast\\ \ast&\ast}. $$ Since $\left(T_k-\frac{1}{\tau_0}vv^T\right)^{-1}\succeq T_k^{-1}$, we have $(T_{k+1}^{-1})_{11}=\left(\left(T_k-\frac{1}{\tau_0}vv^T\right)^{-1}\right)_{11}\ge(T_k^{-1})_{11}$.