Show that $A + A^{-1} \geq 2I$ for $A > 0$. For a positive matrix $A$.  Here, we assume that all positive matrices are self-adjoint.  Show that 
\begin{align}
A + A^{-1} \geq 2I.
\end{align}
Here, $A≥0$ means that A is self-adjoint and for all $x∈ℂn,⟨Ax,x⟩≥0.$
 A: $A \geq 0$ means $\langle Ax,x \rangle \geq 0$. $A \geq B$ means $A-B \geq 0$ which translates to $\langle Ax,x \rangle \geq \langle Bx,x \rangle$.
Now for your problem we need to show $A+A^{-1} \geq 2I$.
So, we need to show $\langle A+A^{-1}x,x \rangle \geq 2 \langle x,x\rangle =2 \lVert x\rVert^2$.
Now, $A$ ,$A^{-1}$ are self adjoint, and commute, and hence can be simultaneously diagonalised.
So, let $A=diag(\lambda_1,...\lambda_n)$.
Then, $A^{-1}=diag(1/\lambda_1,...1/\lambda_n)$. Now, $\lambda_i$ is postive, as $A$ is a positive matrix. For a postive number $a$ we always hve $a+1/a \geq 2$. Hence your result follows.
A: Consider a diagonalization of $A$.
This is also a diagonalization of $A + A^{-1}$.
 You are then reduced to show that
$$
a + \frac 1{a} \ge 2
$$for each (potential eigenvalue) $a>0$.
The function $$
f(a) =  a + \frac 1{a}
$$ is minimum when $a=1$ and then the minimum is $2$.
A: $$A+A^{-1} -2 I =A^{-1}(A^2+ I -2A)= A^{-1}(A-I)^2 $$
and both $A^{-1}$ and $(A-I)^2$ are positive-definite.
