Exercise 7(a) after $\S.$ 90 Minimax principle from Paul R. Halmos's "Finite-Dimensional Vector Spaces" (second edition) asks to prove or disprove the following assertion.
If $A$ is a positive linear transformation on a finite-dimensional inner product space, and if $AB$ is self-adjoint, then $$\vert(ABx, x)\vert \leq \Vert B \Vert \cdot (Ax, x)$$ for all vectors $x$.
The underlying field is not specified as real or complex. Also, for reference, $\S . 87$ and $\S. 88$ (from the book) define the norm $\Vert \cdot \Vert$ of a bounded linear operator as follows:
$\Vert A \Vert = \inf \big\{K: \Vert Ax \Vert \leq K \Vert x \Vert \text{ for all vectors } x \big\} = \sup \left\{\frac{\vert (Ax,y) \vert}{\Vert x \Vert \cdot \Vert y \Vert}: x \neq 0, y\neq 0 \right\}.$
In my attempt thus far, I have only been able to progress to the following conclusion, and would appreciate a pointer. Thanks.
The assertion obviously holds if $x = 0$. If $x \neq 0$, then the self-adjoint nature of $AB$ means that $$\vert(ABx, x)\vert \leq \Vert AB \Vert \cdot \Vert x \Vert^2 \leq \Vert A \Vert \cdot \Vert B \Vert \cdot \Vert x \Vert^2 .$$