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. $\frac{1}{3}$.
$$
\begin{array}{l}
\text { Given }|x+a|-|x+1| \\
\leqslant|(x+a)-(x-1)|=|a-1|,
\end{array}
$$

we know that the maximum value of $f(x)=|x+a|-|x-1|$ is $|a-1|$.
From the problem, we have $|a-1| \leqslant 2 a \Rightarrow a \geqslant \frac{1}{3}$.
Therefore, the minimum value of the real number $a$ is $\frac{1}{3}$.