Linear objective with quadratic constraints I have the problem
$$
\text{maximize } f= c^Tx \\
\text{subject to } x^T Q x \leq 1 \\
x,c \in \mathbb{R}^n \text{ , } Q \in \mathbb{R}^{n \times n}
$$
and $ Q $ is additionally symmetric positive definite. Using a Lagrange multiplier I get the answer
$$
f = +\sqrt{c^T Q^{-1} c}
$$
but I can't prove that it's correct. How would this be proven correct or incorrect?

With the Lagrangian
$$
L(x,\lambda) = -c^Tx + \lambda \frac12 (x^TQx -1)
$$
one finds the KKT conditions
$$
-c + \lambda Qx = 0, \ \lambda\ge0 , \ \lambda(x^TQx -1)=0, \ x^TQx\le 1.
$$
If $x^TQx<1$ then $\lambda=0$, and necessarily $c=0$. Hence $\lambda>0$, $x^TQx=1$,
$$
x = \frac1\lambda Q^{-1}c, \ x^TQx = \frac1{\lambda^2}c^TQ^{-1}c = 1,
$$
this implies $\lambda^2 = c^TQ^{-1}c$. And since $\lambda\ge0$, we get $\lambda = \sqrt{c^TQ^{-1}c}$, and 
$$
c^Tx = \lambda x^TQx = \sqrt{c^TQ^{-1}c}.
$$
Since the function to be maximized is concave and the feasible set is convex, the global maximum is realized at a KKT point. The only KKT point is $x= \frac1{\sqrt{c^TQ^{-1}c}} Q^{-1}c$ with $c^Tx=\sqrt{c^TQ^{-1}c}$. Hence this value is indeed the maximum.
