(Ⅰ)当a=1时,f(x)=e x+
2
x -4,∴f′(x)=e x-
2
x 2 ,∴f′(1)=e-2,
∵f(1)=e-2,
∴f(x)在点(1,f(1))处的切线方程为:(e-2)x-y=0.
(Ⅱ)∵f(x)=a•e x+
a+1
x -2(a+1)(a>0) .
∴f′(x)=
a x 2 e x -(a+1)
x 2 ,
令g(x)=ax 2e x-(a+1),则g′(x)=ax(2+x)e x>0,
∴g(x)在(0,+∞)上单调递增,
∵g(0)=-(a+1)<0,当x→+∞时,g(x)>0,
∴存在x 0∈(0,+∞),使g(x 0)=0,且f(x)在(0,x 0)上单调递减,f(x)在(x 0,+∞)上单调递增,
∵g(x 0)= a x 0 2 e x 0 -(a+1)=0,∴ a x 0 2 e x 0 =a+1,即 a e x 0 =
a+1
x 0 2 ,
∵对于任意的x∈(0,+∞),恒有f(x)≥0成立,
∴f(x) min=f(x 0)= a e x 0 +
a+1
x 0 -2(a+1)≥0,∴
a+1
x 0 2 +
a+1
x 0 -2(a+1)≥0,
∴
1
x 0 2 +
1
x 0 -2≥0 ,∴ 2 x 0 2 - x 0 -1≤ 0,解得-
1
2 ≤x 0≤1,
∵ a x 0 2 e x 0 =a+1,∴ x 0 2 e x 0 =
a+1
a >1,
令h(x 0)= x 0 2 e x 0 ,而h(0)=0,当x 0→+∞时,h(x 0)→+∞,
∴存在m∈(0,+∞),使h(m)=1,
∵h(x 0)= x 0 2 e x 0 在(0,+∞)上,∴x 0>m,
∴m<x 0≤1,
∵h(x 0)= x 0 2 e x 0 在(m,1]上∴h(m)<h(x 0)≤h(1),
∴1<
a+1
a ≤e,∴a≥
1
e-1 .