Let $Q(x_1,\ldots,x_n)$ be a positive definite quadratic form.
I would like to show that there exists $C>0$ such that $Q(x_1,\ldots,x_n)\geq C(x_1^2+x_2^2+\ldots x_n^2)$.
Proof:
Let $\vec{x}=\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}$. The quadratic form can be expressed as $Q(x_1,\ldots,x_n)=\vec{x}^\top A\vec{x}$ for a symmetric matrix $A\in\mathbb{R}^{n\times n}$ with positive eigenvalues $\lambda_1,\ldots,\lambda_n$. Moreover, there exists orthogonal matrix $Q$ such that $$ Q(x_1,\ldots,x_n)=\vec{y}^\top D \vec{y}=\sum_{i=1}^n\lambda_i y_i^2\geq\lambda_\min\sum_{i=1}^n y_i^2\geq C\sum_{i=1}^ny_i^2=C\Vert y\Vert_2^2 $$ where
- $\vec{y}=Q^\top\vec{x}$,
- $D=Q^\top A Q=\textrm{diag}(\lambda_1,\ldots,\lambda_n)$,
- $\lambda_\min:=\min\{\lambda_1,\ldots,\lambda_n\}>0$,
- $0<C\leq\lambda_\min$.
Since $$ \Vert\vec{x}\Vert_2^2=\vec{x}^\top\vec{x}=(Q\vec{y})^\top(Q\vec{y})=\vec{y}^\top \underbrace{Q^\top Q}_{I_n}\vec{y}=\Vert\vec{y}\Vert_2^2, $$ the claim follows.
Do you agree?
The reasoning seems fine.
Here is a reorganization of the presentation of the solution.
As $A$ is a positive definite symmetric matrix, we can write $A=QDQ^T$ where $D$ is a diagonal matrix with positive diagonal entries and $Q$ is an orthogonal matrix.
We denote $y=Q^Tx$.
\begin{align} Q(x_1, \ldots, x_n) &= x^TAx\\&=x^T(QDQ^T)x\\&=(x^TQ)D(Q^Tx)\\&=(Q^Tx)^TD(Q^Tx) \\ &=y^TDy\\ &=\sum_{i=1}^n \lambda_i y_i^2\\ &\ge \lambda_{\min}\sum_{i=1}^n y_i^2\\ &=\lambda_{\min}\|y\|^2\\ &=\lambda_{\min}\|Q^Tx\|^2\\ &=\lambda_{\min}\|x\|^2 \end{align}
Hence we can pick $C$ to be $\lambda_{\min}$ (or any positive number smaller than $\lambda_{\min})$.