11. Find all pairs of positive real numbers $(a, b)$ such that the function $f(x)=a x^{2}+b$ satisfies: for any real numbers $x, y$, we have $f(x y)+f(x+y) \geqslant f(x) f(y)$.
$$
\begin{array}{l}
f(x y)+f(x+y) \geqslant f(x) f(y) \Rightarrow a(x y)^{2}+b+a(x+y)^{2}+b \geqslant\left(a x^{2}+b\right)\left(a y^{2}+b\right) \\
\Rightarrow\left(a-a^{2}\right) x^{2} y^{2}+(a-a b)\left(x^{2}+y^{2}\right)+2 a x y+2 b-b^{2} \geqslant 0, \\
\text { thus } a-a^{2} \geqslant 0 \Rightarrow 0<a<1, \text { and for } b>0, \text { so } \Delta=4 a^{2} b^{2}-4\left(a-a^{2}\right)\left(2 b-b^{2}\right) \leqslant 0 \\
\Rightarrow 4 a b[a b-(1-a)(2-b)] \leqslant 0 \Rightarrow 2 a+b-2 \leqslant 0 .
\end{array}
$$

Thus, the conditions that $a, b$ should satisfy are $0<a<1, 0<b \leqslant 1, 2 a+b-2 \leqslant 0$.
On the other hand, under these conditions, $\left(a-a^{2}\right) x^{2} y^{2}+(a-a b)\left(x^{2}+y^{2}\right)+2 a x y+2 b-b^{2}$
$$
\begin{aligned}
\geqslant & \left(a-a^{2}\right) x^{2} y^{2}+(a-a b)(-2 x y)+2 a x y+2 b-b^{2} \\
& =\left(a-a^{2}\right) x^{2} y^{2}+2 a b x y+2 b-b^{2}=\left(a-a^{2}\right)\left(x y+\frac{b}{1-a}\right)^{2}-\frac{a b^{2}}{1-a}+2 b-b^{2} \\
& =\left(a-a^{2}\right)\left(x y+\frac{b}{1-a}\right)^{2}+\frac{b[(1-a)(2-b)-a b]}{1-a} \geqslant 0 .
\end{aligned}
$$

In summary, the pairs of positive real numbers $(a, b)$ that satisfy the conditions are $\{(a, b) \mid 0<a<1, 0<b \leqslant 1, 2 a+b-2 \leqslant 0\}$.