I have to find A and C $ \in {R_{>0}}$ so that :
$$ A \space\sqrt{x^{2}+y^{2}}\leq \sqrt{x^{2}+2xy+3y^{2}} \leq C\sqrt{x^{2}+y^{2}}$$
I was only able to find C but i am not sure if it is correct
$\textbf{My try }$ $$ {x^{2}+2xy+3y^{2}} \leq 3x^{2}+6xy+3y^{2}= 3(x+y)^2 \leq 6(x^2+y^2)$$
$$\Rightarrow \sqrt{x^{2}+2xy+3y^{2}} \leq \sqrt{6}\sqrt{x^{2}+y^{2}}$$
Can you please check if i am right and give some hints to find A
On
So, I assumed both $x,y\in R$ are real valued.
For the left hand side of inequality we can write: $A^2 x^2+A^2 y^2\leq x^2+2xy + 3y^2$, or $(1-A^2)x^2+2xy+(3-A^2)y^2\geq 0$. Now, we can think of this as a quadratic function $f(x)$ and find its roots as follows:
$x = \dfrac{-2y \pm \sqrt{4y^2-4(1-A^2)(3-A^2)y^2}}{2(1-A^2)}$
We want $f(x)$, given any $y$ to be positive, for example make it a convex function of $x$, with no real roots (or only one). In order to be convex $f''(x)>=0$, or $A^2<1$, and since $A>0$, so $0<A<1$. For having no real roots, we need to have $4y^2-4(1-A^2)(3-A^2)y^2\leq 0$, or $(1-A^2)(3-A^2)\geq 1$ for $y\neq 0$. Then, we may conclude that $A\leq 2-\sqrt{2}$ or $A\geq 2+\sqrt{2}$. But, since $0<A<1$, so we can choose $A=2-\sqrt{2}$ as a valid candidate.
Choose spherical coordinates, let $x = r \cos \phi$, $y = r \sin \phi$, and $x^2 + y^2 = r^2$, then we are looking for
$$ A \leq \sqrt{1 + 2\cos \phi \sin \phi+ 2 \sin^{2} \phi} \leq C $$
Use the identity $2\cos \phi \sin \phi = \sin (2\phi)$ :
$$ A^2 \leq f(\phi) = {1 + \sin (2\phi) + 2 \sin^{2} \phi} \leq C^2 $$
The extrema of $f(\phi)$ are at $ 0 = f'(\phi) = 2 \cos (2 \phi) + 4 \cos\phi \sin \phi = 2 \cos (2 \phi) + 2 \sin (2 \phi)$, so
$ \tan (2 \phi) = -1 $ which gives $\phi = -\pi/8$ and $\phi = 3 \pi/8$. So the two extremes are
$f(\phi = - \pi/8) = 0.5858$ and $f(\phi = 3 \pi/8) = 3.4142$
So $A = 0.7654$ and $B = 1.8478$. Using known values for the trigonometric functions, this can be given as well as: $A = \sqrt{2 - \sqrt 2}$ and $B = \sqrt{2 + \sqrt 2}$.