Evaluate product of trace of two matrices

Question

I have a question. Let A be a positive semi-definite matrix, H be a positive definite matrix. Is the following inequality right: $Tr(AH).Tr(AH^{-1}) \geq (Tr(A))^2$? I tried to take some concrete examples and always got the inequality above. If it's true, how can it be proved ? Thanks !

Answer 1

Since $H$ is positive definite, thus symmetric, it can be diagonalized, $H=QDQ^{-1}$ with $D$ diagonal matrix with positive entries. Then $$ tr(AH)tr(AH^{-1}) = tr(AQDQ^{-1})tr(AQD^{-1}Q^{-1})=tr(Q^{-1}AQD)tr(Q^{-1}AQD^{-1}). $$ Set $B:=Q^{-1}AQ$. Then $tr(B)=tr(A)$. Since $D$ is diagonal, we have $$ tr(BD) = \sum_{i=1}^n b_{ii}d_{ii}. $$ Defining $x:=sqrt(diag(B))=(\sqrt{b_{11}},\dots,\sqrt{b_{nn}})^T$ we can write $$ tr(BD)tr(BD^{-1}) =(x^TDx)(x^TD^{-1}x). $$ From Cauchy-Schwarz inequality you get for every positiv definite matrix $M$ (check Kantorovich inequality) $$ (x^Tx)^2 = (x^TM^{1/2}M^{-1/2}x)^2 \le (x^TMx)(x^TM^{-1}x)\quad \forall x. $$ Apply this with $M:=D$, to get $$ (x^TDx)(x^TD^{-1}x)\ge (x^Tx)^2 = tr(B)^2=tr(A)^2, $$ and the inequality is proven.

Answer 2

A slightly shorter proof (but essentially the same proof as the one by @daw): Apply Cauchy.-Schwarz on $$ \operatorname{tr}(A) = \operatorname{tr}((H^{-1/2} A^{1/2}) (A^{1/2} H^{1/2})). $$