1. The solution to the inequality $\frac{x+1}{2 x-|x-1|} \geqslant 1$ is
$=$. $1 . x>\frac{1}{3}$.
When $x \geqslant 1$, the requirement is satisfied.
When $x<1$, the original inequality is equivalent to $\frac{x+1}{3 x-1} \geqslant 1$.
(1) When $\frac{1}{3}<x<1$, $x+1 \geqslant 3 x-1$, which means $x \leqslant 1$.

So, $\frac{1}{3}<x<1$ satisfies the requirement.
(2) When $x<\frac{1}{2}$, $x+1 \leqslant 3 x-1$, which means $x \geqslant 1$, a contradiction.