<s>[INST] Duality by computing $P+Q$ Given $C$ is symmetric and positive definite, consider
\begin{align}
\min: &Q(y) = \frac{1}{2} y^T C^{-1} y - b^Ty\\
\text{ subject to :}&A^Ty = f
\end{align}
and 
$$\max: -P(x) = -\frac{1}{2}(Ax-b)^TC(Ax-b) + x^Tf.$$
I am asked to prove duality by computing $P(x)+Q(y)$. Here duality means that $-P(x)$ has a max and $Q(y)$ has a min, so there exists a saddle point $(x,y)$. 
I am really confused about this problem...
We showed when we defined the Langrange multiplier for the constraind problem, 
$$L(x,y) = Q(y) - x^T(A^Ty-f)$$
then $L(x,y)$ has a saddle point at where $Q(y)$ has a min (subject to the constrain) and $-P(x)$ has a max. 
 [/INST] We write out the sum and substitute $A^Ty= f$
\begin{align}
P+Q & = \left[\frac{1}{2}(Ax-b)^T C (Ax-b) - x^TA^T y\right] + \left[\frac{1}{2} y^T C^{-1} y - b^Ty\right]\\
& = \left[\frac{1}{2}(Ax-b)^T C (Ax-b) - x^TA^T y\right] + \left[\frac{1}{2} (C^{-1}y)^T C (C^{-1}y) - b^Ty\right]\\
& = \frac{1}{2}\bigg[(Ax-b)^T C (Ax-b) + (C^{-1}y)^T C (C^{-1}y)  \bigg] +( x^TA^t - b^T)y \\
& = \frac{1}{2}\bigg[(Ax-b + C^{-1}y)^T C (Ax-b + C^{-1}y)   - 2 (Ax-b)^T C C^{-1}y \bigg] +( x^TA^t - b^T)y \\
&= \frac{1}{2} (Ax-b + C^{-1}y)^T C (Ax-b + C^{-1}y) 
\end{align}
Since $C$ is positive definite, we know $P+Q \geq 0$, and it can achieve zero when we have 
$$Ax-b + C^{-1} y = 0 \text{ and }A^T y = f,$$
This condition implies that $Q(y)$ is a min and $-P(x)$ is a max. So we have 
$$\min_{constrain} Q(y) = \max -P(x).$$
</s>