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