∵f（x）=x2+2x+alnx，∴f(2t−1)≥2f(t)−3⇒2t2−4t+2≥2alnt−aln(2t−1)＝alnt22t−1当t≥1时，t2≥2t-1，∴lnt22t−1≥0．即t＞1时，a≤2(t−1)2lnt22t−1恒成立．又易证ln（1+x）≤x在x＞-1上恒成立，∴lnt22t...