Take symmetric and real matrices F, f and f' such that $F \geq f$ and $F>f'$. Here $F \geq f$ means that $F-f$ is positive semi-definite, and $F>f'$ means that $F-f'$ is positive definite. I want to show that if the inequality $F \geq f$ is sharp (defined below), then this means that $f>f'$.
$F \geq f$ is sharp means that for any matrix $A$ such that $F>A$, we must have $A \ngeq f$ (I'm not 100% sure of this definition). So if we take $A=f'$, we must have that $f'\ngeq f$.
So in effect I want to show that $F \geq f$, $F>f'$ and $f'\ngeq f$ together imply that $f>f'$.
Alternatively, it would be just as good to show that $F \geq f$, $F \geq f'$ and $f'\ngeq f$ together imply that $f \geq f'$.
As anther alternative, it would be almost as good to show that $F \geq f$, $F \geq f'$ and $f'\ngeq f$ together imply that $tr(f) \geq tr(f')$ where $tr$ is the trace.
Note that these are matrix inequalities, so $f'\ngeq f$ alone does not imply that $f'< f$.
Thanks!