Computing $\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr)$ Suppose that $A \in \mathbb{R}^{n \times n}$ is a symmetric positive semi-definite matrix  such that $\operatorname{Tr}(A)\le n$. I want a lower bound on the following quantity 
$$\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr)$$
where $I$ denotes the identity matrix.
My intuition tells me it should be
$$
\operatorname{Tr} \bigl( \bigl( (A+I )^{-1} \bigr)^2\bigr) \ge 1/4
$$
However, I'm not sure how to show it. Also my intuition might be wrong?
 A: By spectral theorem, there is invertible matrix $P$ such that
$$
P^{-1}(I+A)P=\pmatrix{1+\mu_1 \\ & \ddots & \\ & & 1+\mu_n}
$$
where $\mu_i$ is eigenvalue of $A$ and $\mu_i\geqslant0$. So 
$$
(I+A)^{-1}=P\pmatrix{\dfrac1{1+\mu_1} \\ & \ddots & \\ & & \dfrac1{1+\mu_n}}P^{-1}
$$
And
$$
((I+A)^{-1})^2=P\pmatrix{\dfrac1{(1+\mu_1)^2} \\ & \ddots & \\ & & \dfrac1{(1+\mu_n)^2}}P^{-1}
$$
So
$$
Tr(((I+A)^{-1})^2)=\sum_{i=1}^n\dfrac1{(1+\mu_i)^2}\leqslant n
$$
Then by $\sum_{i=1}^n\mu_i\leqslant n$, we have $\mu_i\leqslant n$ for all $i$. So a lower bound is
$$
Tr(((I+A)^{-1})^2)\geqslant\frac{n}{(1+n)^2}
$$ 
A: Since $A$ is real symmetric, there exists a matrix $P$ such that 
$$PAP^{-1} = D =\verb/diag/(\lambda_1,\lambda_2,\ldots,\lambda_n)$$
is diagonal with eigenvalues $\lambda_1,\ldots,\lambda_n$. This leads to
$$P (I_n+A)^{-2}P^{-1} = (I_n + PAP^{-1})^{-2} = \verb/diag/\left(\frac{1}{(1+\lambda_1)^2},\ldots,\frac{1}{(1+\lambda_n)^2}\right)\\
\implies \verb/Tr/((I_n + A)^{-2}) = \sum_{i=1}^{n}\frac{1}{(1+\lambda_i)^2}$$
Consider the function $g(x) = \frac{1}{(1+x)^2}$. It is easy to check
for $x \in (0,\infty)$,
$$g(x) > 0,
\quad
g'(x) = -\frac{2}{(1+x)^3} < 0
\quad\text{ and }\quad
g''(x) = \frac{6}{(1+x)^4} > 0$$
This means $g(x)$ is a positive, monotonic decreasing and convex function over $(0,\infty)$.
By Jensen's inequality, 
we obtain following lower bound:
$$\verb/Tr/((I_n + A)^{-2})
= \sum_{i=1}^{n} g(\lambda_i) 
\underbrace{\ge}_{\text{Jensen}}
n g\left(\frac1n\sum_{i=1}^n \lambda_i\right)
\underbrace{\ge}_{g\text{ decreasing}}  
n g(1) = \frac{n}{4}$$
Since $\verb/Tr/(I_n) \le n$ and $\verb/Tr/((I_n + I_n)^{-2}) = \frac{n}{4}$, above lower bound $\frac{n}{4}$ is achievable and hence the optimal one.  
As a consequence,
your intution $\verb/Tr/((I_n + A)^{-1}) \ge \frac14$ is true (but far from the optimal).
