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Consider a real polynomial \begin{equation} P(x)=\sum_{|\alpha| \leq N} c_{\alpha} x^{\alpha} \qquad (x \in \mathbb{R}^n), \end{equation} where the $c_{\alpha}$ are real numbers, and for every $\alpha=(\alpha_1,\dots,\alpha_n) \in \mathbb{N}^{n}$ and $x=(x_1,\dots,x_n) \in \mathbb{R}^n$, we set as usual $|\alpha|=\alpha_1+\dots+\alpha_n$ and $x^{\alpha}=x_1^{\alpha_1}\dots x_n^{\alpha_n}$. Assume that $P(x) > 0$ for all $x \in \mathbb{R}^n$.

Does there exist $C > 0$ such that $P(x) \geq C$ for all $x \in \mathbb{R}^n$?

Thank you very much in advance for your attention.

NOTE. The answer is clearly yes if $n=1$, since in this case we have \begin{equation} \lim_{|x| \rightarrow \infty} |P(x)| = \infty, \end{equation} so that if $r > 0$ is such that $|P(x)| \geq 1$ for all $x \in \mathbb{R}$, with $|x| \geq r$, and we set \begin{equation} m= \min_{\substack{x \in \mathbb{R} \\ |x| \leq r}} |P(x)|, \end{equation} we have $|P(x)| \geq \min \{m,1 \}$ for all $x \in \mathbb{R}$.

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$f(x,y)=(xy-1)^2+x^2>0$ for all $x,y$, but there is no $C>0$ such that $f(x,y)>C$ for all $x,y$. This was A1 in the 1969 Putnam exam. I have read that the people who put the exam together didn't see this possibility.

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    See also http://mathoverflow.net/questions/38019/zeros-of-gradient-of-positive-polynomials and http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/27201#27201 – 2017-01-16
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    Dear Gerry, thank you very very much for you help and for having shared your knowledge with me and with all users of the forum. Just a few minutes ago, I met the same example at p. 7 of the paper [Representations of non-negative polynomials](https://math.usask.ca/~marshall/bounds1.pdf) by Marshall. I have also found the wonderful book "Positive Polynomials and Sum of Squares" by the same author, who deeply investigates the subject of non-negative polynomials, which drew the attention of the young David Hilbert already in the last decades of the XIX century. – 2017-01-16