Let $f(x_1, x_2) = 3 {{{x_1}}^{2}}+{{{x_2}}^{4}}-2 {{{x_2}}^{2}}-2{x_1} {x_2}+2 {x_1}-1 $
The above function is a coercive function, now prove that: Show that there exist constants $\lambda$ and $\mu$, where $\lambda > 0$ , such that for any $x_1,\,x_2$, $$ f(x_1, x_2) \geq \lambda (x_1^2 + x_2^2) + \mu $$ Determine the values of $\lambda$ and $\mu$. This implies that $f(x_1, x_2)$ is coervice.
Using the inequalities:
$$x^4\geq 4x^2-4~~,~~x_1^2+x_2^2\geq 2x_1x_2$$
we get that:
$$f(x_1,x_2)\geq 3x_1^2+2x_2^2-2x_1x_2+2x_1-5\\\geq 2x_1^2+x^2_2+2x_1-5\\=2(x_1+\frac{1}{2})^2+x_2^2-\frac{11}{2}$$
This inequality already suffices to show that $f$ is coercive, because if one performs the transformation to polar coordinates the limit $\rho \rightarrow \infty$ sends $f\rightarrow \infty$ for all $\theta.$ However, it is easy to show that
$$ 2x_1^2+x^2_2+2x_1-5\geq x_1^2+x_2^2-6$$ by rearranging the inequality and completing the square, so we obtain that such $\lambda, \mu$ exist and in particular the choice $\lambda=1, \mu=-6$ works.