当-1≤x≤3时,即x+1≥0,x-3≤0,则|x+1|+|x-3|=x+1+3-x=4;当x4;当x>3时,|x+1|+|x-3|=x+1+x-3=2x-2>4;∴对一切实数x,恒有|x+1|+|x-3|≥4;即原不等式有解,必须a≥4.故选B.
