8. Given three positive real numbers $a, b, c$ with $a c=9$. It is known that for all real numbers $x, y$ with $x y \neq 0$ the inequality $\frac{a}{x^{2}}+\frac{b}{x y}+\frac{c}{y^{2}} \geq 0$ holds. What is the maximum possible value for $b$?
(A) 1
(B) 3
(C) 6
(D) 9
(E) There is no such $b$.
8. The answer is $\mathbf{( C )}$. By hypothesis, for $x, y$ such that $x y \neq 0$, we have

$$
\frac{a y^{2}+b x y+c x^{2}}{x^{2} y^{2}}=\frac{a}{x^{2}}+\frac{b}{x y}+\frac{c}{y^{2}} \geq 0 \text {. }
$$

Let's prove that this inequality is always satisfied for $b \leq 6$: indeed, if $x y > 0$, then the inequality is clearly satisfied. On the other hand, if $x y < 0$, then for $y=\sqrt{c}, x=-\sqrt{a}$ we have that

$$
\frac{a y^{2}+b x y+c x^{2}}{x^{2} y^{2}}=\frac{-\sqrt{a c}(b-2 \sqrt{a c})}{a c}=\frac{-3(b-6)}{9} \leq 0,
$$

contradicting the problem's hypothesis.

Second solution: Multiplying the inequality in the text 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 non-zero. In particular, since the ratio $t:=y / x$ can take any real value except 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 get $c$, 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 met: the coefficient of the term of degree two is positive (which is true 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 to $b^{2}-4 a c \leq 0$, i.e., $b^{2} \leq 4 a c=36$, or finally $b \leq 6$. The maximum possible value for $b$ is therefore 6.