Olimpiadi di Matematica

The answer is $\mathbf{(C)}$. By hypothesis, for $x$, $y$ such that $x y \neq 0$, we have
$$\frac{a y^{2} + b x y + c x^{2}}{x^{2} y^{2}} = \frac{a}{x^{2}} + \frac{b}{x y} + \frac{c}{y^{2}} \geq 0.$$
We show that this inequality is always satisfied for $b \leq 6$: indeed, if $x y > 0$, then the inequality is clearly verified. On the other hand, if $x y < 0$, then, using the equality $a c = 9$, we obtain
$$\frac{a y^{2} + b x y + c x^{2}}{x^{2} y^{2}} = \frac{(\sqrt{a} y + \sqrt{c} x)^{2} + x y (b - 2 \sqrt{a c})}{x^{2} y^{2}} = \frac{(\sqrt{a} y + \sqrt{c} x)^{2} + x y (b - 6)}{x^{2} y^{2}},$$
which is clearly non-negative for $b \leq 6$.

Moreover, 6 is the maximum value of $b$ for which this inequality holds for every $x$, $y$ with $x y \neq 0$: indeed, if $b > 6$, then for $y = \sqrt{c}$, $x = -\sqrt{a}$ we have that
$$\frac{a y^{2} + b x y + c x^{2}}{x^{2} y^{2}} = \frac{-\sqrt{a c}(b - 2 \sqrt{a c})}{a c} = \frac{-3(b - 6)}{9} \leq 0,$$
against the hypotheses of the problem.

Second solution: Multiplying the inequality in the statement by the positive number $y^{2}$ we obtain
$$a\left(\frac{y}{x}\right)^{2} + b\left(\frac{y}{x}\right) + c \geq 0$$
for every pair of real numbers $x$, $y$ both nonzero. In particular, since the ratio $t := y / x$ can take every real value different from 0, we have $a t^{2} + b t + c \geq 0$ for every $t$ in $\mathbb{R}$ (including $t = 0$: indeed for $t = 0$ we obtain $c$, which is positive by hypothesis). It is well known that a second-degree polynomial is positive for every value of the variable if and only if the following two conditions are satisfied: the coefficient of the degree-two term is positive (and this is verified in our case, since $a > 0$ by hypothesis) and the discriminant $b^{2} - 4 a c$ is less than or equal to 0. In our situation, this second condition translates into $b^{2} - 4 a c \leq 0$, that is $b^{2} \leq 4 a c = 36$, that is finally $b \leq 6$. The maximum possible value for $b$ is therefore 6.