任取1≤x1
=(x²1+3)/(x1+1)-(x²2+3)/(x2+1)
=[(x2+1)(x²1+3)-(x1+1)(x²2+3)]/[(x1+1)(x2+1)]
=[x²1x2-x1x²2+(x²1-x²2)+3(x2-x1)]/[(x1+1)(x2+1)]
=[x1x2(x1-x2)+(x1+x2)(x1-x2)-3(x1-x2)]/[(x1+1)(x2+1)]
=(x1-x2)(x1x2+x1+x2-3)/[(x1+1)(x2+1)]
=(x1-x2)[(x1+1)(x2+1)-4]/[(x1+1)(x2+1)]
∵1≤x1
∴(x1+1)(x2+1)>4
∴(x1+1)(x2+1)-4>0
又x1-x2<0
∴(x1-x2)[(x1+1)(x2+1)-4]/[(x1+1)(x2+1)]<0
∴f(x1)-f(x2)<0
∴f(x)在【1,+∞）上為∞增函數(shù)
（2）
y=f(x)在[1,2]上有最小值-1
即x∈[1,2],(x²+a)/(x+1)≥-1恒成立
即x²+a≥-x-1,a≥-x²-x-1恒成立
且等號(hào)能夠取得
設(shè)g(x)=-x²-x-1,所在二次函數(shù)對(duì)稱軸為x=-1/2,
∵x∈[1,2]
∴g(x)為減函數(shù)
∴x=-1時(shí),g(x)取得最大值-1
∴a≥-1
∵等號(hào)能夠取得
∴a=-1