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Consider a complex polynomial \begin{equation} P(z)=\sum_{|\alpha| \leq N} c_{\alpha} z^{\alpha}, \end{equation} where as usual for every $\alpha=(\alpha_1,\dots,\alpha_n) \in \mathbb{N}^{n}$ we set $|\alpha|=\alpha_1+\dots+\alpha_n$, and $z^{\alpha}=z_1^{\alpha_1}\dots z_n^{\alpha_n}$. Assume that $P$ has no zeros on $\mathbb{R}^n$, and define $R:\mathbb{R}^n \rightarrow \mathbb{C}$ as \begin{equation} R(x)= \frac{1}{P(x)} \qquad (x \in \mathbb{R}^n). \end{equation} (i) Is it true that $R$ is a bounded function?

(ii) Is it true that all the partial derivatives $D_{i}R$, with $i=1,\dots,n$, are bounded functions?

Every help is welcome.

NOTE. The answers are trivially positive for the special case $n=1$, but for $n \geq 2$ things are more complicated. For example, it is not true that $|P(x)|$ must tend to infinity when $|x| \rightarrow \infty$, as the simple example in two variables $P(x,y)=1+x^2 y^2$ shows. Note that the same example shows that it is not true that all the derivatives $D^{\alpha}R$ must be bounded. Indeed, we have in this case \begin{equation} \frac{\partial^2 P}{\partial x^2}(x,y) = -\frac{2y^2}{(1+x^2 y^2)^2} + \frac{8x^2 y^4}{(1+x^2 y^2)^3}, \end{equation} so that \begin{equation} \frac{\partial^2 P}{\partial x^2}(0,y) = -2y^2. \end{equation}

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    I now realize that the answer to (ii) is negative, as the example in two variables $P(x,y)=1+ i xy$ shows. Even if we assume that the coefficients of $P$ are real, (ii) does not hold, as the example $P(x,y)=1+xy+ x^2 y^2$ shows. Anyhow I am quite convinced that (i) has a positive answer, even if I have no idea of how to prove it. – 2017-01-15

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The answers to my two questions are both negative for $n \geq 2$, even if we assume that $P$ is a real polynomial, as the following example in two variables shows \begin{equation} P(x,y)=x^2+(xy-1)^2. \end{equation}