Show that if $A$ is positive definite then $A + A^{-1} - 2I$ is positive semidefinite 
Let $A$ be a real symmetric positive definite matrix. Show that $$A + A^{-1} -2I$$ is positive semidefinite.

I found that $A^{-1}$ is a positive definite matrix, thus $A + A^{-1}$ is also a positive definite matrix, moreover I know the form of the $z^TIz$ is as follows $(a^2 + b^2 + ... )$, where $a,b, ...$ are the components of vector $z$. 
I don't know what to do next...
 A: Hint: diagonalize your matrix $A$ in an orthonormal basis and study the real function $f(t)=t+\frac{1}{t}-2$. Or just use functional calculus an spectral mapping if you know about that.
A: Hint: Using the fact that $A$ is diagonalizable and the eigenvalues are positive, write $A:=B^2$, where $B$ is symmetric. Then $A+A^{-1}-2I=(B-B^{-1})^2=(B-B^{-1})^t(B-B^{-1})$, which is semi-positive definite (not necessarily positive definite, as $A=I$ shows). 
A: Hint: $A+A^{-1}-2I = A^{-1}(A^2-2A+I)=A^{-1}(A-I)^2$
A: Purely for fun:
\begin{align}
\min_x \quad &x^T(A+A^{-1}-2I)x\\
f(x)&=x^T(A+A^{-1}-2I)x\\
\nabla f(x)&=2\times (A+A^{-1}-2I)x=0\\
(A+A^{-1}-2I)x&=0\\
\implies x^T(A+A^{-1}-2I)x&\geq0 \qquad \forall x\\
\end{align}
A: Since $A$ is symmetric, it has an eigendecomposition $A = Q \Lambda Q^T$. Hence,
$$\begin{array}{rl} A + A^{-1} - 2 I &=  Q \Lambda Q^T + Q \Lambda^{-1} Q^T - 2 Q Q^T\\\\ &= Q (\Lambda + \Lambda^{-1} - 2 I) Q^T\\\\ &= Q \Lambda^{-\frac 12} (\Lambda^2 - 2 \Lambda + I) \Lambda^{-\frac 12} Q^T\\\\ &= Q \left(\Lambda^{-\frac 12} (\Lambda - I)^2 \Lambda^{-\frac 12}\right) Q^T\end{array}$$
If $A \succ 0$, then $A + A^{-1} - 2 I$ is at least positive semidefinite. It may be positive definite.
