TITLE: Why does $A \circ {B^{ - 1}} + {A^{ - 1}} \circ B \ge 2{I_{n \times n}}$?
QUESTION [6 upvotes]: Let $A, B \in M_n$ be positive definite and $A \circ B = \left[ {{a_{ij}}{b_{ij}}} \right]$.
Why does $A \circ {B^{ - 1}} + {A^{ - 1}} \circ B \ge 2{I_{n \times n}}$ ?

REPLY [11 votes]: First observe that for any two elements $X,Y \in M_{n}$, we have $X \circ Y$ is a one of the principal minor of $X \otimes Y$. And it is also straightforward to observe that any principal minor of positive definite matrix is also positive definite.
So to prove $A\circ B^{-1} + A^{-1} \circ B \geq 2 I_n$, It is sufficient to prove $$A\otimes B^{-1} + A^{-1} \otimes B \geq 2 I_{n^2}.$$
Now $A,B$ are given positive definite. So both the element $A\otimes B^{-1}$ and its inverse $(A \otimes B^{-1})^{-1} = A^{-1}\otimes B$ is postive definte.
Now using spectral theorem we will prove that for any positive definite matrix $T \in M_k$, one will have $$T+T^{-1} \geq 2I_k.$$
Since $T$ is positive definite, there exist a orthonormal eigen basis with respect to which  $T$ will be of the diagonal form and $T +T^{-1}$ will look like 
\begin{align}
\begin{pmatrix}
t_1 + \frac{1}{t_1} &0  &\cdots &0\\
0 & t_2 + \frac{1}{t_2} &\cdots &0\\
0& 0 &\cdots &0\\
0& 0& \cdots & t_k + \frac{1}{t_k}
\end{pmatrix}
\end{align}
where $t_i$'s are positive eigen value of $T$. And since $x+\frac{1}{x} \geq 2$ for all $x>0$, we will have  $$T+T^{-1} \geq 2I_k.$$