设f(x)=e^x-1/2x-ln(1/2x+1)-1,x≥0
则f'(x)=e^x-1/(x+2)-1/2.
f"(x)=e^x+1/(x+2)^2>0,所以f'(x)单调递增,所以f'(x)>f(0)=1-1/2-1/2=0.
所以f(x)单调递增,f(x)≥f(0)=0,所以当x≥0时e^x-1/2x≥ln(1/2x+1)+1.
所以x>2时e^x-1/2x≥ln(1/2x+1)+1成立 .