By definition, a (multivariate) polynomial P is non-negative if for every real value we assign to the variables, the result is not negative. For instance, $P=a^2+b^2-2ab=(a-b)^2$ is non-negative.
Using this information, how can I prove that if P is not non-negative then there is a way we can assign its variables to rational numbers and get a negative answer?
P not being non-negative implies that there is a selection of real values that causes it to be negative, but how can I prove that there exists a choice of rational numbers that also results in a negative value?
Thanks!
Expanding on a suggestion in the comments, suppose that $P $ is negative when the variables $x_1, \cdots, x_i $ take the values $a_1, \cdots, a_i $. Then, because $P $ is continuous, there exists $\epsilon > 0$ such that the polynomial on $a_1 + \epsilon, \cdots, a_i + \epsilon $ is still negative (it has not changed signs). Then, because "there are a heck lot of rationals", there is, for each $j $, a rational between $a_j $ and $a_j + \epsilon $. Making the variables assume those values, the polynomial is still negative for those values.