11. (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 $x, y$,
$$
f(x y)+f(x+y) \geqslant f(x) f(y) .
$$
11. From the problem, we have
$$
\begin{array}{l}
\left(a x^{2} y^{2}+b\right)+\left[a(x+y)^{2}+b\right] \\
\geqslant\left(a x^{2}+b\right)\left(a y^{2}+b\right) .
\end{array}
$$

First, find the necessary conditions that \(a\) and \(b\) must satisfy.
By setting \(y=0\) in equation (1), we get
$$
\begin{array}{l}
b+\left(a x^{2}+b\right) \geqslant\left(a x^{2}+b\right) b \\
\Rightarrow(1-b) a x^{2}+b(2-b) \geqslant 0 .
\end{array}
$$

Since \(a > 0\), \(a x^{2}\) can take any large positive value, so it must be that \(1-b \geqslant 0\), i.e., \(0 < b \leq 1\).
If \(b < 1\), then for \(x \neq 0\),
$$
(1-b) a x^{2} + b(2-b) > 0.
$$
If \(b = 1\), then
$$
(1-1) a x^{2} + 1(2-1) = 1 \geq 0.
$$
Thus, \(0 < b \leq 1\).

Next, consider the function
$$
g(x) = (a-a^{2}) x^{2} + a(1-b) x + (2b - b^{2}).
$$
Since \(a > 0\), we know \(a = 1\), in which case,
$$
\dot{g}(x) = -2 b x^{2} + (2 b - b^{2}).
$$

Then \(g(x)\) can take negative values, which is a contradiction.
Therefore, \(g(x)\)
$$
\begin{array}{l}
=\left(a-a^{2}\right)\left(x^{2}-\frac{a b}{a-a^{2}}\right)-\frac{(a b)^{2}}{a-a^{2}}+\left(2 b-b^{2}\right) \\
=\left(a-a^{2}\right)\left(x^{2}-\frac{b}{1-a}\right)^{2}+\frac{b}{1-a}(2-2 a-b) \\
\geqslant 0
\end{array}
$$

for all real \(x\).
Thus, \(a-a^{2} > 0\), i.e., \(0 < a < 1\), and
$$
g\left(\sqrt{\frac{b}{1-a}}\right)=\frac{b}{1-a}(2-2 a-b) \geqslant 0,
$$

which implies \(2a + b \leq 2\).
Therefore, the necessary conditions for \(a\) and \(b\) are
$$
\begin{array}{l}
0 < b \leq 1, \\
0 < a < 1, \\
2a + b \leq 2.
\end{array}
$$

Combining \(x^{2} + y^{2} \geq -2xy\), we get
$$
\begin{array}{l}
h(x, y) \\
\geqslant\left(a-a^{2}\right) x^{2} y^{2} + a(1-b)(-2 x y) + \\
2 a x y + \left(2 b - b^{2}\right) \\
=\left(a-a^{2}\right) x^{2} y^{2} + 2 a b x y + \left(2 b - b^{2}\right) \\
=\left(a-a^{2}\right)\left(x y + \frac{b}{1-a}\right)^{2} + \frac{b}{1-a}(2-2 a-b) \\
\geqslant 0 .
\end{array}
$$

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