Example 2 If the inequality $|x|+|x-3|<a$ always has a solution for $x$, then what is the range of real number $a$?
Analyzing, we use $|a|+|b| \geqslant|a+b|$ to cleverly solve the inequality, avoiding the conventional method.
Solving $|x|+|x-3|=|x|+|3-x| \geqslant|x+(3-x)|=3$. The equality holds if and only if $x(3-x) \geqslant 0$, which means $0 \leqslant x \leqslant 3$. Therefore, $|x|+|x-3|$ has a minimum value of 3. Hence, $a>3$.