An inner product defined by a positive definite matrix

Suppose we have positive definite, real and symmetric matrix $D \in \mathbb{R}^{2x2}$. One might be interested in showing the following theorem.

Theorem 1: There exists a constant $C>0$ such that
$$x^T D x \geq C x^T x,$$ holds for every $x \in \mathbb{R}^2$.

This is not as trivial as it seems. To do the trick, one needs to know a few preliminary results. From now on we assume that we're dealing with vectors of $\mathbb{R}^2$ and the matrix $D$ is always $2\times 2$.

Theorem 2: Eigenvalues of a positive definite matrix are positive.

Proof: By definition, positive definiteness means
$$ x^T D x > 0,$$ for every $x$. Then for the eigenvectors $v$ and eigenvalues $\lambda$ of $D$ it holds that
$$ Dv = \lambda v,$$ which can be multiplied from left by $v^T$ to arrive at
$$ v^TDv = \lambda v^T v = \lambda \| v \|^2 > 0,$$ where we used the definition of positive definiteness. Thus, $\lambda > 0$. $\blacksquare$

Theorem 3: A real symmetric matrix $D$ can be decomposed as
$$D = Q \Lambda Q^T,$$ where $Q$ is orthogonal and $\Lambda$ is a diagonal matrix with eigenvalues of $D$ on the diagonal. This is known as the eigendecomposition of the matrix $D$.

Proof: Look up any book on matrix decompositions. $\blacksquare$

Then we may proceed to the original problem.

Proof of Theorem 1: Consider
$$ \frac{x^T D x}{x^T x} = \frac{x^T Q\Lambda Q^T x}{\| x \|^2} = y^T \lambda y,$$ where $y := Q^T \frac{x}{\|x\|}$. The length of $\frac{x}{\|x\|}$ is unity and so is the length of $y$ because $Q^T$ is orthogonal and such a mapping is either a rotation or a mirroring. Thus,
$$y^T \Lambda y = \lambda_1 y_1^2 + \lambda_2 y_2^2 \leq \lambda_{\text{min}} ( y_1^2 + y_2^2) = \lambda_{\text{min}},$$ where $\lambda_{\text{min}} = \min_j \lambda_j, j=1,2$. Multiplying with $x^T x$ gives the result. $\blacksquare$