Asymptotic Analysis for two variables i have a function  $ f(x,y)= \sqrt{x^2+2xy+3y^2}$ $\space$ , and $g(x,y)= \sqrt{x^2+y^2}$ 
I have to prove that $f$ is bounded both above and below by  $g$ asymptotically $\Rightarrow f(x,y)=\Theta(g(x,y)) \Rightarrow  \text{there is c and C $\in {R_{>0}}$ so that  $c g(x,y)\leq f(x,y) \leq C g(x,y)$}$
I was only able to find  the big C , but i am still not  sure how to find the  little  c 
$$ {x^{2}+2xy+3y^{2}} \leq 3x^{2}+6xy+3y^{2}= 3(x+y)^2 \leq 6(x^2+y^2)$$ 
$$\Rightarrow \sqrt{x^{2}+2xy+3y^{2}} \leq \sqrt{6}\sqrt{x^{2}+y^{2}}$$
which means that C =$\sqrt{6}$
Any hints how can i find c 
 A: $$c\sqrt{x^2+y^2}\leq \sqrt{x^2+2 x y+3 y^2}$$
square both sides
$$c^2 \left(x^2+y^2\right)\leq x^2+2 x y+3 y^2$$
expand and rearrange
$$\left(c^2-1\right) x^2-2 x y+\left(c^2-3\right) y^2\leq 0\quad(*)$$
which is true for any $x,y\in\mathbb{R}$ if
$$\left(1-\left(c^2-1\right) \left(c^2-3\right)\right) y^2<0\land c^2-1<0$$
that is
$$-c^4+4 c^2-2<0\land c^2-1<0$$
$c^2-1<0\to -1<c<1\quad(*)$
$-c^4+4 c^2-2<0\to c^4-4c^2+2>0$
set $c^2=d$ so the inequality becomes 
$d^2-4d+2>0\to d<2-\sqrt{2}\lor d>2+\sqrt{2}$
reset $d=c^2$
$c^2<2-\sqrt{2}\lor c^2>2+\sqrt{2}$
$c<-\sqrt{2+\sqrt{2}}\lor -\sqrt{2-\sqrt{2}}<c<\sqrt{2-\sqrt{2}}\lor c>\sqrt{2+\sqrt{2}}$
recalling $(*)$ and the constraint given that $c>0$ we get $0<c<\sqrt{2-\sqrt{2}}$
so we can take $\color{red}{c=\sqrt{2-\sqrt{2}}}$
hope this helps
$(*)$
Discriminant $\Delta=(2y)^2-4(c^2-1)(c^2-3)y^2=4y^2(1-(c^2-1)(c^2-3))$
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Edit
Using the same technique for the right bound I found 
$C=\sqrt{2+\sqrt{2}}$
which is a bit smaller than $\sqrt 6$
