How to prove the following function is coercive?
\begin{align*} f(x,y)= x^{2n}+y^{2n}-nx^2+2nxy-ny^2 \end{align*} I found the definition of a coercive function and tried some problems and figure out how to prove a function is coercive but in this case, there is the term $n$ is a bit confusing to me.
Definition: A continuous function $f(x)$ that is defined on $\mathbb{R}^n$ is coercive if, \begin{align*} \lim_{||x||\xrightarrow[]{}\infty} f(x)= +\infty \end{align*} that is for any $M>0$ there exists a constant $R_{M}>0$ such that $||f(x)||>M$ whenever $||x||>R_{M}$
Solution: \begin{align*} f(x,y)= x^{2n}+y^{2n}-nx^2+2nxy-ny^2 &= x^{2n}+y^{2n}-n(x-y)^2\\ x^{2n}+y^{2n}-n(x-y)^2 &\leq x^{2n}+y^{2n} \end{align*}
then how do we prove $x^{2n}+y^{2n} \xrightarrow[]{} \infty$ as $||x||\xrightarrow[]{} \infty$
Is my approach is correct if it is not could anyone please show me the right path? Thank you
hint 1): Since $f$ is a function on $\mathbb{R}^2$, if we translate the definition of coercivity in terms of both its arguments $x$ and $y$, the limit is actually $\|(x,y)\|_{\mathbb{R}^2}\to\infty$, not just $\|x\|_{\mathbb{R}}\to\infty$.
hint 2) The inequality you have is correct. However, we actually want to show directly that the limit on $f$ goes to $\infty$, or that a lower bound on $f$ goes to $\infty$. The upper bound $x^{2n}+y^{2n}$ that is currently written in your question will not prove what we need -- this is because, even if it is coercive, it will only imply $$\lim\limits_{\|(x,y)\|_{\mathbb{R}^2}\to\infty}f(x,y)\leq \lim\limits_{\|(x,y)\|_{\mathbb{R}^2}\to\infty}x^{2n}+y^{2n}=\infty,$$ which is not sufficient to show the limit on the left is equal to $\infty$. If, instead, we can show some lower bound $L(x,y)\leq f(x,y)$ is coercive, then we will have
$$\infty=\lim\limits_{\|(x,y)\|_{\mathbb{R}^2}\to\infty}L(x,y)\leq \lim\limits_{\|(x,y)\|_{\mathbb{R}^2}\to\infty}f(x,y)\leq\infty,$$ from which the desired equality follows.