Proving equivalence of norms in $\mathbb{R}^2$ 
Let $\lVert \cdot \rVert_*:\mathbb{R}^2 \to \mathbb{R}, (x,y) \rightarrow \sqrt{x^2+2xy+3y^2}$ be a norm.
How can I find to constants $k,K \in \mathbb{R}^{>0}$ so that the following equivalence is given:$$k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) \rVert_2.$$

My idea:
$$
k\lVert (x,y) \rVert_2 \leq \Vert (x,y) \rVert_* \leq K\Vert (x,y) \rVert_2
\\
\Leftrightarrow
k\sqrt{|x|^2+|y|^2} \leq \sqrt{x^2+2xy+3y^2} \leq K\sqrt{|x|^2+|y|^2}
$$
Now how can I find $k,K$? The problem is these constants must be minimal.
Can someone give me a tip?
Thx

Note that 
$$x^2 + 2xy + 3y^2 = \begin{pmatrix} x \\ y \end{pmatrix}^T \begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$ 
The eigenvalues of the matrix have sum 4 and product 2, so they are both stricly positive; let us call them $\lambda_1$ and $\lambda_2$. Since the matrix is symmetric, there exists $Q$ orthogonal such that:
$$\begin{pmatrix} 1 & 1 \\ 1 & 3 \end{pmatrix} = Q^T \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} Q.$$
Now let us note: $$\begin{pmatrix} q_1 \\ q_2 \end{pmatrix}= Q \begin{pmatrix} x \\ y \end{pmatrix}.$$ Then you can see that $\| \mathbf q \|_2 = \| \mathbf x\|_2$, and
$$x^2 + 2xy + 3y^2 = \begin{pmatrix} q_1 \\ q_2 \end{pmatrix}^T \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \begin{pmatrix} q_1 \\ q_2 \end{pmatrix} = \lambda_1 q_1^2 + \lambda_2 q_2^2,$$
from where it is easy to conclude.
Another way to proceed, adapted specifically to the example, is to notice that, on the one hand,
$$ x^2 + 2xy + 3y^2 \leq 2 x^2 + 4y^2 \leq 4(x^2 + y^2),$$
and on the other:
$$ x^2 + 2xy + 3y^2 = \left(\sqrt{a}x + \frac{y}{\sqrt{a}}\right)^2 + (1-a)x^2 + \left(3-\frac{1}{a}\right)y^2 \geq \min\left(1-a,3-\frac{1}{a}\right) (x^2 + y^2) \quad \forall a \in \mathbb R_+.$$
Any $a$ such that $1/3 < a < 1$ will guarantee that the $\min$ is strictly positive.
