My question is the following:
Suppose that $A, B$ are symmetric positive definite real matrices.
Is it true that
$$
\mathrm{trace}\big((A+B)^{-1}\big)\geq \frac{1}{2} \min\Big\{\mathrm{trace}(A^{-1}), \mathrm{trace}(B^{-1})\Big\}
$$
Some comments:
-
This is clearly true in the scalar case:
$$
\frac{1}{a + b} \geq \frac{1}{2 \max\{a, b\}} = \frac{1}{2} \min\{a^{-1}, b^{-1}\}.
$$ -
What I could show so far is that, using the inequality above,
$$
\mathrm{trace}((A+B)^{-1}) \geq \frac{1}{2} \sum_{i=1}^d \min\{\lambda_i(A^{-1}), \lambda_i(B^{-1})\},
$$
Above, $d$ is the dimension of the matrices and $\lambda_i$ denotes the $i$th largest eigenvalue. This isn't quite strong enough, though.
Best Answer
The answer to this is no in general. As a counterexample, take $$ A = \pmatrix{1 & 0\\0 & \epsilon}, \quad B = \pmatrix{\epsilon & 0\\0 & 1}, \quad \epsilon = 0.01. $$ We have $$ \operatorname{tr}((A + B)^{-1}) = \frac 2{1 + \epsilon} < 2 <\frac 12 \min \{\operatorname{tr}(A^{-1}),\operatorname{tr}(B^{-1}))\} = \frac {1 + \epsilon^{-1}}{2} = 50.5. $$