Question: > 
> 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:$$k\lVert (x,y) \rVert\_2 \leq \Vert (x,y) \rVert\_\* \leq K\Vert (x,y) \rVert\_2.$$
> 
> 
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**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

Answer: 
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.
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.