For any real number $x$, we have
$$
|x+a|-|x+1| \leqslant 2 a \text{. }
$$

Then the minimum value of the real number $a$ is $\qquad$
1. $\frac{1}{3}$.

From $|x+a|-|x+1|$
$$
\leqslant|(x+a)-(x-1)|=|a-1|,
$$

we know that the maximum value of $f(x)=|x+a|-|x-1|$ is $|a-1|$. According to the problem, $|a-1| \leqslant 2 a \Rightarrow a \geqslant \frac{1}{3}$.

Therefore, the minimum value of the real number $a$ is $\frac{1}{3}$.