Find the range of $xy$ under the conditions: $x^2-xy+y^2=9$ and $|x^2-y^2|<9$ Assume $x,y \in \mathbb{R}^+$, and satisfied the following expression:
$$x^2-xy+y^2=9$$
$$\left|x^2-y^2\right| < 9$$
find the range of $xy$

My approach:  $x^2-xy+y^2=9$ $\Rightarrow$ $xy+9=x^2+y^2 \geq 2xy$ $\Rightarrow$ $xy \leq 9$
But I don't know  how to find the lower bound. please help me..thanks very much. 

For the lower bound you can use the follwing facts:


*

*$x^2-xy+y^2 = (x-y)^2 +xy \Rightarrow 9-xy = (x-y)^2$

*$x^2-xy+y^2 = (x+y)^2 -3xy \Rightarrow 9+3xy = (x+y)^2$ 

*$\left|x^2-y^2\right| < 9 \Leftrightarrow (x+y)^2(x-y)^2 < 81$
Plugging 1. and 2. into 3. you get:
$$(x+y)^2(x-y)^2 < 81 \Leftrightarrow (9-xy)(9+3xy) < 81 \Leftrightarrow xy(6-xy) < 0 \stackrel{x,y >0}{\Leftrightarrow} \boxed{xy>6}$$
Together with your upper bound $\boxed{xy \leq 9}$ you get
$$\boxed{6 < xy \leq 9}$$
