Consider $x \in \Bbb{R}^n$ and suppose $f_m: \Bbb{R}^n \to \Bbb{R}$ which is defined as: $$f_m(x)=x^TAx+\alpha_{1}x_1^3+\alpha_{2}x_1^2x_2+\dots+\alpha_{r_1}x_n^3+\alpha_{r_1+1}x_1^4+\alpha_{r_1+2}x_1^3x_2+\dots+\alpha_{r_2}x_n^4+\dots+\alpha_{c}x_n^m$$
where subscript in $x_i$ refers to the $i$th element of $x$. If we know that $A$ is positive definite, what is the necessary and sufficient condition for $\alpha_1,...,\alpha_c$ to result in a positive function $f_m(x)$ for all $\Vert x\Vert < \rho$ and all $m\in\{3,4,...,M\}$? Is the following statement correct?
$f_m(x)$ is positive for $\Vert x\Vert < \rho$ if and only if $x=0$ is the only solution of $f_m(x)=0$ in $\Vert x\Vert < \rho$ and $A$ is positive definite.
If it is correct, how to show $x=0$ is the only solution? In other words, how to choose $\alpha_i$ so that $x=0$ becomes the unique solution?
For example in $x\in\Bbb{R}^2$, up to and including the 3rd order, $f_3(x)$ can be showed as follows:
$$f_3(x)=A_{1,1}x_1^2+A_{2,2}x_2^2+\alpha_{1}x_1^3+\alpha_{2}x_1^2x_2+\alpha_{3}x_2^2x_1+\alpha_{4}x_2^3$$ $$A_{1,1}>0,A_{2,2}>0$$ Choose $\alpha_1,\dots,\alpha_4$ on the basis of $A_{1,1}$ and $A_{2,2}$ so that $f_3(x)>0$ for $\Vert x\Vert < \rho$. Then choose $\alpha_5,\dots,\alpha_9$ on the basis of $A_{1,1}$ and $A_{2,2}$ so that $f_4(x)=f_3(x)+\alpha_{5}x_1^4+\alpha_{5}x_1^3x_2+\dots+\alpha_{9}x_2^4>0$ for $\Vert x\Vert < \rho$.
And so on.