How to show that $f(x_1,x_2)=e^{x_1^2}+e^{x_2^2}-x_1^2-x_2^2$ is coercive?

Question

I am trying to show $$f(x_1,x_2)=e^{x_1^2}+e^{x_2^2}-x_1^2-x_2^2$$ is a coercive function. I considered the inequality $e^a\geq 1+a \ \ \forall a\in\mathbb{R}$, so that \begin{align} e^{x_1^2}+e^{x_2^2}-x_1^2-x_2^2&\geq1+x_1^2+1+x_2^2-x_1^2-x_2^2 \\ &=2. \end{align}

However, from this, I am unable to show that $f$ is coercive by the below definition $$\lim_{\|x\|\rightarrow\infty} f(x_1,x_2)=+\infty.$$

Answer 1

Use $$e^{x^2}\geq1+x^2+\frac{x^4}{2}.$$ By C-S we obtain: $$e^{x_1^2}+e^{x_2^2}-x_1^2-x_2^2\geq2+\frac{x_1^4+x_2^4}{2}\geq2+\frac{(x_1^2+x_2^2)^2}{4}\rightarrow+\infty.$$

Answer 2

Writing $f(x_1,x_2)=e^{\|x\|^2}e^{-x_2^2}+e^{x_2^2}-\|x\|^2$, fix $\|x\|$. Then $$\frac\partial{\partial x_2}(e^{\|x\|^2}e^{-x_2^2}+e^{x_2^2}-\|x\|^2)=-2x_2e^{\|x\|^2}e^{-x_2^2}+2x_2e^{x_2^2}=0$$ for a minimum which yields $x_2^2=\|x\|^2/2$. Note $x_2=0$ immediately yields $\lim\limits_{x_1\to\infty}f(x_1,0)=+\infty$. Substituting back yields $$\lim_{\|x\|\to\infty} f(x_1,x_2)\ge\lim_{\|x\|\to\infty}(e^{\|x\|^2}e^{-\|x\|^2/2}+e^{\|x\|^2/2}-\|x\|^2)=+\infty.$$