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 ?

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:
$$f(r,\theta)=r^4[(\cos^2\theta+\sin^2\theta)^2-2\sin^2\theta\cos^2\theta]-3r^2\sin\theta\cos\theta\\
=r^4\left(1-\frac{1}{2}\sin^2(2\theta)\right)-\frac{3}{2}r^2\sin(2\theta)\ge \frac{r^4}{2}-\frac{3r^2}{2}\rightarrow\infty$$
