112 If $a>0$, the solution set of the inequality $|x-4|+|x-3|<a$ on $R$ is not empty, then the range of values for $a$ is $\qquad$
$112(1,+\infty)$. Since $a>|x-4|+|x-3|=|4-x|+|x-3| \geqslant |4-x+x-3|=1$
thus $a>1$.