3. All real numbers $x$ that satisfy $2|x-| x+|x-1|||>|x+| x-|x+1|||$ are in ( ).
(A) $\left(\frac{1}{3}, 3\right)$
(B) $(3,+\infty)$
(C) $\left(-\infty, \frac{1}{3}\right)$
(D) $\left(-\infty, \frac{1}{3}\right) \cup(3,+\infty)$
3. (D).
$$
\begin{array}{l}
\text { Let } f(x)=-x+|x-1|, \\
g(x)=x-|x+1|, \\
f(x)=1+(x-1)+|x-1| \geqslant 1, \\
g(x)=(x+1)-|x+1|-1 \leqslant-1 .
\end{array}
$$

Given $2|x-1|>|x+1|$, squaring both sides, we get
$$
\begin{array}{l}
3 x^{2}-10 x+3>0, x \in\left(-\infty, \frac{1}{3}\right) \cup(3, \\
+\infty) .
\end{array}
$$