I often find an inequality of the form
$ \lambda_{\min P} \|x\|^2 \leq x^T P x \leq \lambda_{\max P} \|x\|^2 $,
where
$P=P^T$, $\|x\|$ is the Euclidean norm of $x\in\mathbb R^n$, and $\lambda_{\min P}$ and $\lambda_{\max P} $ are the smallest and largest eigenvalues of $P$, respectively.
How can it proofed or from which "rule" can this result be derived?
When for instance $ Q(x) = \begin{bmatrix}x_1 \\ x_2 \end{bmatrix}^T \begin{bmatrix}q_{11} & 0 \\ 0 &q_{22} \end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \end{bmatrix}$ with $q_{11} \ll q_{22}$ then $Q(x) \leq q_{22} \|x\|^2$.
This seems conservative or isn't it? Is there an inequality where still both $q_{11}$ and $q_{22}$ appear on the right hand side of the $\leq$?
Hint: If there exists an orthonormal basis of eigenvectors $\{v_i\}_{i=1}^n$ of $P$ (e.g., when $P$ is symmetric) then given any $x \in \Bbb{R}^n$, you can write $x = \sum_{i=1}^n \alpha_i v_i$ for some scalars $\alpha_i$. Letting $\lambda_i$ be the eigenvalue corresponding to $v_i$ for each $i$, you have $$ x^T P x = \sum\limits_{i=1}^n \lambda_i \alpha_i^2 \|v_i\|^2 = \sum\limits_{i=1}^n \lambda_i \alpha_i^2. $$ From here, you have $$ \lambda_{\text{min}} \sum\limits_{i=1}^n \alpha_i^2 \leq \sum\limits_{i=1}^n \lambda_i \alpha_i^2 \leq \lambda_{\text{max}} \sum\limits_{i=1}^n \alpha_i^2 $$