How to show that this function $f(x,y,) <0$? I would like to show that the function $f(x,y) = -5 x^4 + 4 y^2 - 5 y^4 + x^2 (4 - 6 y^2)$ is less than $0$ for $1 < x^2 + y^2 <2$ (at least). Upon rearranging, I obtain
$f(x,y) = -5(x^2+y^2)^2 + 4(x^2 +y^2 + x^2y^2)$
But then I don't know how to proceed to get $f(x,y)<0?
Thanks!

Let us write this expression under the form:
$$f(x,y) = -5(x^2+y^2)^2 + 4(x^2 +y^2) + 4 x^2y^2$$
Let us convert this expression into polar coordinates. One gets:
$$F(r,\theta) = -5r^4 + 4r^2 + r^4 \sin ^2(2 \theta)$$
(using relationship $\sin (2 \theta) = 2 \sin \theta \cos \theta$).
Therefore:
$$F(r,\theta) = r^2 (r^2 (\sin ^2(2 \theta) - 5) + 4)$$
For $r>0$, the sign of $F$ is that of $g(r,\theta):=r^2 (\sin^2(2 \theta) - 5) + 4$
Let $1<r^2<2 \ \ \ (1) $. Taking into consideration that 
$0-5 \leq \sin^2(2 \theta) - 5 \leq -4 \ \ \ (2)$
Multiplication of (1) by (2) gives:
$$-5 \leq r^2 (\sin^2(2 \theta) - 5) \leq -4$$
Adding 4 to the three parts gives the result: $-1 \leq g(r,\theta) \leq 0$.
