Why does $A \circ {B^{ - 1}} + {A^{ - 1}} \circ B \ge 2{I_{n \times n}}$? Let $A, B \in M_n$ be positive definite and $A \circ B = \left[ {{a_{ij}}{b_{ij}}} \right]$.
Why does $A \circ {B^{ - 1}} + {A^{ - 1}} \circ B \ge 2{I_{n \times n}}$ ?
 A: 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  $$T+T^{-1} \geq 2I_k.$$
A: This answer was completed with the help of user timon who has already provided an excellent answer which is quite general. This is a different approach. 
For $x>0$, you have that $x+\frac{1}{x}\geq 2$.
Now, Let $A=\sum_{i}\alpha_i x_i x_i^H$ be its Eigen-decomposition so that $\alpha_i>0$ are the eigenvalues. Similarly $B=\sum_{j}\beta_i y_i y_i^H$ so that $\beta_i>0$ are the eigenvalues. Then, try to show that
$$C=A\circ B^{-1}+A^{-1}\circ B=\sum_{i,j}\left(\frac{\alpha_i}{\beta_j}+\frac{\beta_j}{\alpha_i}\right)(x_i\circ y_j)(x_i\circ y_j)^H$$
Now, note that co-efficients of all terms are greater than two. Define the vectors $r_{ij}=x_i\circ y_j$ for all $i,j$ and also constants $\gamma_{ij}=\frac{\alpha_i}{\beta_j}+\frac{\beta_j}{\alpha_i}$. Note that $\gamma_{ij}\geq 2$
Thus, we have 
\begin{align}
C&=\sum_{i,j}\left(\frac{\alpha_i}{\beta_j}+\frac{\beta_j}{\alpha_i}\right)(x_i\circ y_j)(x_i\circ y_j)^H \\ &\geq 2\sum_{i,j}(x_i\circ y_j)(x_i\circ y_j)^H \\ &=2\left(\left(\sum_{i}x_ix_i^H\right)\circ \left(\sum_{j}y_jy_j^H\right)\right) \\ &= 2I
\end{align}
This proves the needed inequality. Steps above follow from following facts
$$\sum_{i}x_ix_i^H=I$$
This follows since $x_i$'s are a set of orthonormal vectors by definition. Similar property holds for $y_i$'s. Also we have
$$(x_i\circ y_j)(x_i\circ y_j)^H = \left(x_ix_i^H\right)\circ \left(y_jy_j^H\right)$$
Try to prove this yourself. 
