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

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