(-∞，2+2√2]解析：(1) a≤0时，显然f(x)≥g(x)(x>1)成立(2) a>0时，欲使x²+1≥a|x-1|(x>1)，必须有a≤(x²+1)/(x-1)(x>1)a≤(x²+1)/(x-1)(x>1)a≤(x²-1+2)/(x-1)a≤x+1+2/(x-1)a≤(x-1)+2/(x-1)+2右边>2+2√2故，a≤2+2√2综上，a≤2+2√2
