$(A+B)^2\|B^{-1}\|-A$ is positive definite?

Question

If $A$ and $B$ are positive definite matrix, $(A+B)^2\|B^{-1}\|-A$ is also positive definite?

When $A$ and $B$ are real numbers, then $(A+B)^2\|B^{-1}\|-A$ is obviously positive definite. When $A$ and $B$ are matrix, I can use orthonormal matrix $O$ to transform $B$ an diagonal matrix $\operatorname{diag}\{\lambda_1,\cdots,\lambda_n\}$. Then I can transform the original problem to $(A+B)^2-\lambda_n A$ is it also positive definite?

Answer 1

If $A\ge 0$, $B\ge 0$ then $$A^{\frac12} B A^{\frac12} \ge 0$$

PS: this property does not require $A$ and $B$ to commute.

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Since $B \ge \lambda_n I$ then $A+B \ge \lambda_n I$ then,

\begin{align} \left(A+B\right)^2 &= \left(A+B\right)^{\frac12}\left(A+B\right)\left(A+B\right)^{\frac12}\\ &\ge \lambda_n \left(A+B\right)^{\frac12}\left(A+B\right)^{\frac12}\\ &= \lambda_n (A+B)\ge \lambda_n A \end{align}

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By the way, using the same idea you can prove the stronger result:

$$(A+B)^2\left\|\left(A+B\right)^{-1}\right\| - (A+B) \ge 0$$