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
 A: 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.
A: (How to make simple things difficult...)
By Young's inequality, for every $a>0$ you have that
$$
2|xy| = 2(a|x|)(|y|/a) \leq a^2 x^2 + \frac{y^2}{a^2}
$$
hence, for every $a, b > 0$,
$$
- a^2 x^2 - \frac{y^2}{a^2} \leq 2xy \leq b^2 x^2 + \frac{y^2}{b^2}\,.
$$
These inequalities give
$$
(1-a^2) x^2 + \left(3 - \frac{1}{a^2}\right) y^2
\leq x^2 + 2xy + 3y^2
\leq (1+b^2) x^2 + \left(3+\frac{1}{b^2}\right)y^2.
$$
In order to get multiples of the Euclidean norm, you can choose $a,b> 0$ such that
$$
1-a^2 = 3 - \frac{1}{a^2}\,,
\qquad
1+b^2 = 3 + \frac{1}{b^2}\,,
$$
hence
$$
a^2 = \sqrt{2}-1, \qquad
b^2 = \sqrt{2}+1.
$$
Substituting we finally get
$$
(2-\sqrt{2})(x^2+y^2) \leq x^2 + 2xy + 3y^2 \leq (2+\sqrt{2}) (x^2+y^2).
$$
A: 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.
