11. Taking the logarithm of both sides of the original inequality, we get $\lg \left(x+\sqrt{x^{2}+1}\right)+\lg \left(y+\sqrt{y^{2}+1}\right) \geqslant 0$.

Let $f(x)=\lg \left(x+\sqrt{x^{2}+1}\right)$. Then,
$$
f(-x)+f(x)=\lg \left(-x+\sqrt{x^{2}+1}\right)+\lg \left(x+\sqrt{x^{2}+1}\right)=\lg 1=0,
$$

so $f(x)$ is an odd function.
Assume $x_{1}>x_{2}$, then $x_{1}+\sqrt{x_{1}^{2}+1}-x_{2}-\sqrt{x_{2}^{2}+1}=\frac{x_{1}^{2}-x_{2}^{2}}{\sqrt{x_{1}^{2}+1}+\sqrt{x_{2}^{2}+1}}+\left(x_{1}-x_{2}\right)$
$$
=\left(x_{1}-x_{2}\right)\left(\frac{x_{1}+x_{2}}{\sqrt{x_{1}^{2}+1}+\sqrt{x_{2}^{2}+1}}+1\right) \text {. }
$$

Since
$$
\sqrt{x_{1}^{2}+1}+\sqrt{x_{2}^{2}+1}>\left|x_{1}\right|+\left|x_{2}\right| \geqslant-x_{1}-x_{2} \text {, }
$$

it follows that $\frac{x_{1}+x_{2}}{\sqrt{x_{1}^{2}+1}+\sqrt{x_{2}^{2}+1}}+1>0$. Therefore, $f\left(x_{1}\right)>f\left(x_{2}\right)$, which means $f(x)$ is an increasing function.
Since $f(x)+f(y) \geqslant 0$, we have $f(x) \geqslant-f(y)=f(-y)$. Hence, $x \geqslant-y$, which implies $x+y \geqslant 0$. Therefore, the minimum value of $x+y$ is 0.
11. Given real numbers $x, y$ satisfy $\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right) \geqslant 1$, find the minimum value of $x+y$.