This is B.17 from Fundamentals of Convex Analysis by Jean-Baptiste Hiriart-Urruty, Claude Lemaréchal.
Let $f: S^{++}(\mathbb{R}^n) \to \mathbb{R}$ be
$$f(M) := \frac{-1}{tr(M^{-1})}$$
Then show $f$ is convex.
Is the trace of inverse matrix convex? shows that $tr(M^{-1})$ is convex in $M$, but this doesn't help with this problem.
If we let $A\in S^{++}(\mathbb{R}^n), B \in S(\mathbb{R}^n)$, and let $\phi(t) = f(A + tB)$, then try to show $\phi''(0) \ge 0$, we get
$$tr(A^{-1})tr(A^{-1}BA^{-1}BA^{-1})\ge tr(A^{-1}BA^{-1})^2$$
Then let $C = BA^{-1}$, and change to an eigenvector basis for $A^{-1}$, so that $A^{-1} = diag(\lambda_1, \cdots, \lambda_n), C = [c_{ij}]$, then it becomes
$$\sum_{ij}\lambda_i \lambda_j \sum_k c_{jk}^2 \ge \sum_{ij}\lambda_i \lambda_j c_{ii}c_{jj}$$
But I don't know how to prove this inequality.
Best Answer
The trace inequality can be shown as follows. By an orthonormal change of basis, we may assume that $A$ is a positive diagonal matrix. Let $H=A^{-1}BA^{-1}$ and $D$ be its diagonal part (i.e. $d_{ii}=h_{ii}$ for each $i$ and $d_{ij}=0$ when $i\ne j$). Then $(H^2)_{ii}=\sum_jh_{ij}^2\ge h_{ii}^2=(D^2)_{ii}$ and hence $$ \operatorname{tr}(AH^2)\ge\operatorname{tr}(AD^2).\tag{1} $$ Also, by Cauchy-Schwarz inequality, $$ \operatorname{tr}(A^{-1})\operatorname{tr}(AD^2) =\|A^{-1/2}\|_F^2\|A^{1/2}D\|_F^2 \ge\left(\operatorname{tr}(A^{-1/2}A^{1/2}D)\right)^2=(\operatorname{tr}(D))^2.\tag{2} $$ Combining $(1)$ and $(2)$, we get $$ \operatorname{tr}(A^{-1})\operatorname{tr}(HAH) =\operatorname{tr}(A^{-1})\operatorname{tr}(AH^2) \ge\operatorname{tr}(A^{-1})\operatorname{tr}(AD^2) \ge(\operatorname{tr}(D))^2 =(\operatorname{tr}(H))^2 $$ and this is just your trace inequality.