I'm trying to show that $$\sup_{x}\{y^Tx - x^TAx\} = y^TA^{-1}y,$$ (where $A$ is positive definite) however I'm having a bit of trouble.
I know I can find the stationary point of the inside of the $\sup$, which gives me $$y^T - x^T(A^T + A) = 0 \implies x = (A^T + A)^{-1}y$$, however at this point I get a little stuck: plugging this back in for $x$ gives me $y^T(A^T + A)^{-1}y - y^T(A^T + A)^{-1}A(A^T + A)^{-1}y$. How do I proceed from here? As far as I know I can't do anything with the inverse-over-sum, but is there a known rule that I'm missing?.
Could someone suggest a good way forward?
As $A$ is (real) positive definite, $A=A^T$ and hence $(A^T+A)^{-1}=\frac12A^{-1}$.
And the answer you found is wrong. To find the answer using calculus is fine, but you can actually simply complete the square: $$ y^Tx-x^TAx = \frac14y^TA^{-1}y - \left(A^{1/2}x - \frac12A^{-1/2}y\right)^T \left(A^{1/2}x - \frac12A^{-1/2}y\right). $$ So the supremum is $\frac14y^TA^{-1}y$ (not $y^TA^{-1}y$) and it is attained at $x=\frac12A^{-1}y$.