The equation $||x-1|+a|=4$ can have real solutions for x if ‘a’ belongs to the interval

$||x-1|+a|=4\Rightarrow |x-1|+a=\pm 4\Rightarrow |x-1|=\pm 4-a$  
For real solution we have $\pm 4-a\ge 0$  
$\Rightarrow 4-a\ge 0$ and $-4-a\ge 0$ 
$\Rightarrow a\le 4$ and $a\le -4$ 
$\Rightarrow a\le -4\Rightarrow a\in (-\infty ,-4]$