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

**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.
(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
$$
(2-\sqrt{2})(x^2+y^2) \leq x^2 + 2xy + 3y^2 \leq (2+\sqrt{2}) (x^2+y^2).
$$