This problem is quite old and there should be similar problems. I know the following technique:
\begin{equation} \begin{aligned} q^TAq=q^TU\Lambda U^Tq=(U^Tq)^T\Lambda (U^Tq) \end{aligned} \end{equation}
- Since $A$ is symmetric, $U$ must be orthogonal.
- If picking $q=q_{\max}$ such that $Aq_{\max}=\lambda_{\max}q_{\max}$, so we have (also by the fact that $U$ is orthogonal)
$$ \begin{bmatrix} 0 & \cdots & 1 & 0 & \cdots \end{bmatrix} \Lambda \begin{bmatrix} 0 \\ \vdots \\ 1 \\ 0 \\ \vdots \end{bmatrix}=1\lambda_{\max}1=\lambda_{\max} $$
However, it does not really prove the title. If I can come up with $\bar{q}$ with $\|\bar{q}\|=1$, which is not the eigenvector of $A$ such that $\bar{q}A\bar{q}\geq q_{\max}Aq_{\max} $?
My work:
By contradiction, $$\bar{q}A\bar{q}- q_{\max}Aq_{\max}\geq 0 \Rightarrow [U^T(\bar{q}- q_{\max})]^T\Lambda [U^T(\bar{q}- q_{\max})]\geq 0$$
then how to do the next step?
$$\begin{array}{ll} \text{maximize} & x^T A x\\ \text{subject to} & x^T x = 1\end{array}$$
where $A \in \mathbb{R}^{n \times n}$ is symmetric and, thus, has real eigenvalues. We define the Lagrangian
$$\mathcal{L} (x,\lambda) := x^T A x - \lambda (x^T x - 1)$$
Taking the partial derivatives of $\mathcal{L}$ and finding where they vanish,
$$(A - \lambda I_n) \, x = 0_n \qquad \qquad \qquad x^T x = 1$$
Thus, we conclude that the maximum is attained at the intersection of one of the eigenspaces of $A$ with the unit Euclidean sphere.