2024年数学高考大一轮复习第三章 §3.6　利用导数证明不等式

§3.6　利用导数证明不等式

考试要求　导数中的不等式证明是高考的常考题型，常与函数的性质、函数的零点与极值、数列等相结合，虽然题目难度较大，但是解题方法多种多样，如构造函数法、放缩法等，针对不同的题目，灵活采用不同的解题方法，可以达到事半功倍的效果．

题型一　将不等式转化为函数的最值问题

例1 (2023·潍坊模拟)已知函数f(x)＝ex－ax－a，a∈R.

(1)讨论f(x)的单调性；

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(2)当a＝1时，令g(x)＝.证明：当x>0时，g(x)>1.

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思维升华 待证不等式的两边含有同一个变量时，一般地，可以直接构造“左减右”的函数，有时对复杂的式子要进行变形，利用导数研究其单调性和最值，借助所构造函数的单调性和最值即可得证．

跟踪训练1 设a为实数，函数f(x)＝ex－2x＋2a，x∈R.

(1)求f(x)的单调区间与极值；

(2)求证：当a>ln 2－1且x>0时，ex>x2－2ax＋1.

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题型二　将不等式转化为两个函数的最值进行比较

例2 (2023·苏州模拟)已知函数f(x)＝eln x－ax(a∈R)．

(1)讨论f(x)的单调性；

(2)当a＝e时，证明f(x)－＋2e≤0.

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思维升华 若直接求导比较复杂或无从下手时，可将待证式进行变形，构造两个函数，从而找到可以传递的中间量，达到证明的目标．本例中同时含ln x与ex，不能直接构造函数，把指数与对数分离两边，分别计算它们的最值，借助最值进行证明．

跟踪训练2 (2023·合肥模拟)已知函数f(x)＝ex＋x2－x－1.

(1)求f(x)的最小值；

(2)证明：ex＋xln x＋x2－2x>0.

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题型三　适当放缩证明不等式

例3 已知函数f(x)＝aex－1－ln x－1.

(1)若a＝1，求f(x)在(1，f(1))处的切线方程；

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(2)证明：当a≥1时，f(x)≥0.

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思维升华 导数方法证明不等式中，最常见的是ex和ln x与其他代数式结合的问题，对于这类问题，可以考虑先对ex和ln x进行放缩，使问题简化，简化后再构建函数进行证明．常见的放缩公式如下：

(1)ex≥1＋x，当且仅当x＝0时取等号；

(2)ln x≤x－1，当且仅当x＝1时取等号．

跟踪训练3 (2022·南充模拟)已知函数f(x)＝ax－sin x.

(1)若函数f(x)为增函数，求实数a的取值范围；

(2)求证：当x>0时，ex>2sin x.

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