(1)
f(x)≥1
e^x-(2a+e)x≥1
e^x≥(2a+e)x
因为x≥1,两边除以x
(e^x-ex)/2x≥a
令g(x)=(e^x-ex)/2x (x≥1)
g'(x)=[(e^x-e)*2x-2(e^x-ex)]/(4x^2)=2(x-1)e^x/(4x^2)
因为x≥1,g'(x)≥0恒成立
g(x)min=g(1)=0
0≥a
(2)
f(x)+b0
f'(x)min=f[ln(2a+e)]=2a+e-(2a+e)*ln(2a+e)
2a+e-(2a+e)*ln(2a+e)+