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

Here's a version that's more explicitly geometric, but whose underlying mathematics resemble Roberto's matrix diagonalization.
Rewrite the norm in rotated coordinates $(x', y')$, where $x = x' \cos \theta - y' \sin \theta$ and $y = x' \sin \theta + y' \cos \theta$. We'll choose $\theta$ at our convenience—specifically, we'll choose it such that the coordinate axes of the rotated coordinate system align with the axes of the unit ellipse $||(x, y)||_* = 1$, thus making the $xy$ terms in the norm vanish. In this case, \begin{align*}|| (x, y)||_*^2 =& x'^2 \cos^2 \theta \tag{$x^2$} - 2x' y' \sin \theta \cos \theta + y'^2 \sin^2 \theta \\ &+ 2 (x'^2 - y'^2) \sin \theta \cos \theta + 2x' y' (\cos^2 \theta - \sin^2 \theta) \tag{$2xy$} \\
&+ 3x^2 \sin^2 \theta + 6x' y' \sin \theta \cos \theta + 3y^2 \cos^2 \theta \tag{$3y^2$} \end{align*}
The coefficient of $2x' y'$ is $4 \sin \theta \cos \theta + 2 \cos^2 \theta - 2 \sin^2 \theta = 2 \sin (2\theta) + 2 \cos (2\theta)$. So we can make this term disappear by choosing $\theta = -\pi/8$, for which $$\begin{align*} \sin^2 \theta &= \frac{2 - \sqrt{2}}{4} \\ \cos^2 \theta &= \frac{2 + \sqrt{2}}{4} \\ \sin \theta \cos \theta &= -\frac{\sqrt{2}}{4}\end{align*}$$
We thus have
$$ \begin{align*}
|| (x', y')||_*^2 &= x'^2  (\cos^2 \theta + 2 \sin \theta \cos \theta + 3 \sin^2 \theta) + y'^2 (\sin^2 \theta - 2 \sin \theta \cos \theta + 3 \cos^2 \theta) \\
&= (2 - \sqrt{2}) x'^2 + (2 + \sqrt{2}) y'^2\\ \end{align*}$$
Meanwhile, the ordinary Euclidean norm, invariant under rotation, is $$|| (x', y')||_2^2 = x'^2 + y'^2.$$ The path from here should be evident.
