微专题 7　导数与不等式的证明

微专题7导数与不等式的证明高考定位导数与不等式的交汇命题是高考的热点和难点，在利用导数证明不等式问题中，常用的方法有构造函数、适当换元、合理放缩、利用最值、不等式及其性质等.【难点突破】[高考真题](2023·新高考Ⅰ卷)已知函数f(x)＝a(ex＋a)－x.(1)讨论f(x)的单调性；(2)证明：当a>0时，f(x)>2lna＋eq\f(3,2).________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________样题1已知函数f(x)＝ex＋exlnx(其中e是自然对数的底数).求证：f(x)≥ex2.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________样题2已知函数f(x)＝lnx，g(x)＝ex，证明：f(x)＋eq\f(2,ex)>g(－x).________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________样题3已知函数f(x)＝eq\f(lnx,x)－k，(1)若f(x)≤0恒成立，求实数k的取值范围；(2)证明：lneq\f(1,2)＋lneq\f(1,3)＋…＋lneq\f(1,n)<eq\f(1,e)(eq\f(1,2)＋eq\f(1,3)＋…＋eq\f(1,n))(n>1).________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________规律方法利用导数证明不等式问题的基本方法(1)直接构造函数法：证明不等式f(x)＞g(x)(或f(x)＜g(x))转化为证明f(x)－g(x)＞0(或f(x)－g(x)＜0)，进而构造辅助函数h(x)＝f(x)－g(x)；(2)适当放缩构造法：一是根据已知条件适当放缩；二是利用常见放缩结论；(3)构造“形似”函数，稍作变形再构造，对原不等式同结构变形，根据相似结构构造辅助函数.训练(2024·长春调研)设函数f(x)＝ex－1，其中e为自然对数的底数.求证：(1)当x>0时，f(x)>x；(2)ex－2>lnx.__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________