11. (This question is worth 20 points) Find all pairs of positive real numbers $(a, b)$ such that the function $f(x)=a x^{2}+b$ satisfies: for any real numbers $x, y$,
$$
f(x y)+f(x+y) \geq f(x) f(y) .
$$
Given conditions can be transformed into: for any real numbers $x, y$, we have
$$
\left(a x^{2} y^{2}+b\right)+\left(a(x+y)^{2}+b\right) \geq\left(a x^{2}+b\right)\left(a y^{2}+b\right).
$$

First, find the necessary conditions that $a, b$ must satisfy.

Substituting $y=0$ into (1), we get $b+\left(a x^{2}+b\right) \geq\left(a x^{2}+b\right) \cdot b$, which means for any real number $x$,
$$
(1-b) a x^{2}+b(2-b) \geq 0.
$$

Since $a>0$, $a x^{2}$ can take any large positive value, so it must be that $1-b \geq 0$, i.e., $0 < b \leq 1$. 

If $b=1$, then $g(x) = (1-a) x^{2} + (2-1) \geq 0$ for all real $x$. This implies $a \leq 1$. If $a=1$, then $g(x) = -2 b x^{2} + (2 b - b^2)$, where $b > 0$, so $g(x)$ can take negative values, which is a contradiction. Therefore, $a < 1$.

Thus,
$$
\begin{aligned}
g(x) & = \left(a - a^2\right) \left(x^2 - \frac{a b}{a - a^2}\right)^2 - \frac{(a b)^2}{a - a^2} + (2 b - b^2) \\
& = \left(a - a^2\right) \left(x^2 - \frac{b}{1 - a}\right)^2 + \frac{b}{1 - a} (2 - 2a - b) \geq 0
\end{aligned}
$$

for all real $x$. Therefore, it must be that $a - a^2 > 0$, i.e., $0 < a < 1$. Furthermore, since $g\left(\sqrt{\frac{b}{1-a}}\right) = \frac{b}{1-a} (2 - 2a - b) \geq 0$, we get $2a + b \leq 2$.

In summary, the necessary conditions that $a, b$ must satisfy are:
$$
0 < b \leq 1, \quad 0 < a < 1, \quad 2a + b \leq 2.
$$

To verify, consider $h(x, y) = \left(a - a^2\right) x^2 y^2 + a(1 - b)(x + y)^2 + 2a x y + (2b - b^2)$. Using $x^2 + y^2 \geq -2xy$, we get
$$
\begin{aligned}
h(x, y) & \geq \left(a - a^2\right) x^2 y^2 + a(1 - b)(-2xy) + 2a xy + (2b - b^2) \\
& = \left(a - a^2\right) x^2 y^2 + 2ab xy + (2b - b^2) \\
& = \left(a - a^2\right) \left(xy + \frac{b}{1 - a}\right)^2 + \frac{b}{1 - a} (2 - 2a - b) \geq 0.
\end{aligned}
$$

In conclusion, the set of all positive real number pairs $(a, b)$ is $\{(a, b) \mid 0 < b \leq 1, 0 < a < 1, 2a + b \leq 2\}$.
. 20 points