Find the values of $c\neq 0$, such that $P(x,y)>0$ Consider the polynomial $P:\mathbb{R}^{2}\to\mathbb{R}$ given by, $P(x,y)=1+c(x^3+y^3)+(x^2+y^2)^2$. Find all values of $c\neq 0$, such that $P(x,y)>0$, for all $(x,y)\in\mathbb{R}^{2}$
My approach: Clearly $(x^2+y^2)^2$ is positive. However, how should I determine the values of c such that $P(x,y)$ is positive, because the cubic terms could be negative or positive?
 A: I'm nearly sure there must be more elegant solutions, but here is a "brute-force" approach:
It is easy to see that if $c$ is solution, then $-c$ is also solution, so it is sufficient to assume $c>0$
Convert to polar coordinates: $x=r\cos t$, $y=r\sin t$ so for $r\ge0$ and $0\le t\le 2\pi$:
$$P(r,t)=1+c r^3(\sin^3t+\cos^3t)+r^4>0$$
By continuity, periodicity and critical points, we see that $\sin^3t+\cos^3t$ takes all values in $[-1,1]$, so it is possible to reformulate the problem to seek $c>0$ such that:
$$f(r)=1+\alpha r^3+r^4>0$$
for all $r$ (not necessarily positive!) and all $\alpha$ with $\lvert\alpha\rvert\le c$
The minimum of $f$ is among the solutions of the equation $f'(r)=0$ which has solutions $r=0$ (double) and $r=-\frac{3\alpha}{4}$
As $f(0)$ is positive, it follows that $f$ is positive iff $f\left(-\frac{3\alpha}{4}\right)>0$ which has the solutions $\lvert\alpha\rvert<\frac{4\sqrt[4]{3}}{3}$
So we see that the positive solutions for $c$ are $0<c<\frac{4\sqrt[4]{3}}{3}$
Thus, the solution of the problem is $c\in (-\frac{4\sqrt[4]{3}}{3},\frac{4\sqrt[4]{3}}{3})\setminus \{0\}$
