Let $A$ and $B$ be positive semidefinite matrices of the same size. If the largest eigenvalues of $A$ and $B$ are less than equal to $1$. probe that: $AB+BA\geq- \frac{1}{4}I$
In the hint, it says use the fact: $0\leq (A+B-\frac{1}{2}I)^{2}$. In order for $ (A+B-\frac{1}{2}I)^{2}$ to be positive semidefinite, the matrix $ (A+B-\frac{1}{2}I)$ has to be positive semidefinite. The hint given above confused me because I know that $A+B$ is definitely positive semidefinite since it is the sum of two positive semidefinite matrices, but we can't be sure that $A+B-\frac{1}{2}I$ is positive semidefinite. Does anyone know how to prove that: $0\leq (A+B-\frac{1}{2}I)^{2}$?
Now assume the given hint is true, then: $0\leq A\leq I$ and $0\leq B\leq I$. It follows that:$0\leq (A+B-\frac{1}{2}I)^{2}=A^{2}+B^{2}+\frac{1}{4}I+A(B-I)+B(A-I)\leq A+B+\frac{1}{4}I+A(B-I)+B(A-I)$
I need to prove that:$A(B-I)+B(A-I)\leq 0$. This will be true if the product of a positive semidefinite matrix and a negative semidefinite matrix is a negative semidefinite matrix. Is this true?