How to show that $f(x,y)=x^4+y^4-3xy$ is coercive? How to show that $f(x,y)=x^4+y^4-3xy$ is coercive ?
This is my attempt :
$$f(x,y)=x^4+y^4-3xy$$
$$f(x,y)=x^4+y^4\left(1-\frac{3xy}{x^4+y^4}\right)$$
As $||(x,y)|| \to \infty $ , $\frac{3xy}{x^4+y^4} \to 0$
So $||(x,y)|| \to \infty $ , $f(x,y)=x^4+y^4-3xy \to \infty$.
Is this a valid method ?
I think it is not a rigorous proof. can anyone help me with a good proof ?
 A: I would go to polar coordinates, substituting $x=r\cos\theta$ and $y=r\sin\theta$. Then $\lVert(x,y)\rVert\rightarrow\infty$ becomes simply $r\rightarrow\infty$
$$f(r,\theta)=r^4(\cos^4\theta+\sin^4\theta)-3r^2\sin\theta\cos\theta$$
This is a polynomial in $r$, the coefficient of $r^4$ is positive (cannot have both $\sin\theta=0$ and $\cos\theta=0$ at the same time), so the limit when $r\rightarrow\infty$ is $\infty$ for fixed $\theta$
It is also possible to get a lower bound independent of $\theta$, in order to remove the "fixed $\theta$" condition:
$$f(r,\theta)=r^4[(\cos^2\theta+\sin^2\theta)^2-2\sin^2\theta\cos^2\theta]-3r^2\sin\theta\cos\theta\\
=r^4\left(1-\frac{1}{2}\sin^2(2\theta)\right)-\frac{3}{2}r^2\sin(2\theta)\ge \frac{r^4}{2}-\frac{3r^2}{2}\rightarrow\infty$$
A: Your answer looks right. 
Here's another way. Observe
\begin{align}
xy \le \frac{x^2+y^2}{2} \ \ \text{ and }\ \  x^4+y^4\geq \frac{1}{2}(x^2+y^2)^2
\end{align}
which means
\begin{align}
x^4+y^4-3xy \geq&\ x^4+y^4-\frac{3}{2}(x^2+y^2) \\
 \geq&\ \frac{1}{2}(x^2+y^2)^2-\frac{3}{2}(x^2+y^2)\\
=&\ \frac{1}{2}\left(x^2+y^2-\frac{3}{2}\right)^2-\frac{9}{8}.
\end{align}
Hence it follows as $x^2+y^2\rightarrow \infty$, we see that $x^4+y^4-3xy\rightarrow \infty$. 
Note: Using the above inequalities you could show
\begin{align}
1-\frac{3xy}{x^4+y^4} \geq 1- \frac{3(x^2+y^2)}{2(x^4+y^4)} \geq 1-\frac{3}{(x^2+y^2)}
\end{align}
which means for $x^2+y^2$ sufficiently big we have that
\begin{align}
1-\frac{3}{(x^2+y^2)} \ge \frac{1}{2}.
\end{align}
