证明：f'(x)=[1/(x+1)]-1+2axa=1/2时,f'(x)=[1/(x+1)]-1+x=(1-x-1+x²+2x+1)/(x+1)=(x²+x+1)/(x+1)当x>=0时,x²+x+1>=1>0,x+1>=1>0故此时f(x)>0对于x>0恒成立,f(x)在x>=0时单调递增.由于x>=0,所以f(x)>...