How to show that $f(x,y)=x^4+y^4-3xy$ is coercive?

Question

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 ?

Answer 1

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}

Answer 2

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:

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