Consider the polynomial $q(x, y) = a x y^3 + a x^3 y + b x^2 y^2$, where $a$ and $b$ are real constants. Suppose that
$$
q(z, \bar{w}) + q(w, \bar{z}) \le q(z, \bar{z}) + q(w, \bar{w})
\quad \text{for all } z, w \in \mathbb{C}.
$$
Is it true that $a = 0$?



Polynomials satisfying $q(z, \bar{w}) + q(w, \bar{z}) \le q(z, \bar{z}) + q(w, \bar{w})$ for all complex z and w
If $b= 0$, then $a$ must be as well. Consider $z = 1$ and $w = -1$. Then it's easy to show $LHS=-4a$ and $RHS = 4a$. This shows $a \ge 0$. Now if we insead use $z = i$ and $w = -z$, we get $LHS=4a$ and $RHS=-4a$, so $a \le 0$. 


For $b \neq 0$, we need to choose $z, w$ so that $|z| \gg |w|$ or $|w| \gg |z|$, so that one of the two $a$-terms would dominate the $b$ term, rendering the latter effectively infinitesimally small.


