2024年数学高考大一轮复习第十二章 §12.4　不等式的证明（附答单独案解析）

§12.4　不等式的证明

考试要求　1.通过一些简单问题了解证明不等式的基本方法：比较法、综合法、分析法、反证法与放缩法.2.掌握柯西不等式的用法．

知识梳理

1．比较法

(1)作差比较法

已知a>b⇔a－b>0，a<b⇔a－b<0，因此要证明a>b，只要证明________________即可，这种方法称为作差比较法．

(2)作商比较法

由a>b>0⇔>1且a>0，b>0，因此当a>0，b>0时，要证明a>b，只要证明__________即可，这种方法称为作商比较法．

2．综合法

从已知条件出发，利用定义、公理、定理、性质等，经过一系列的推理、论证而得出命题成立，这种证明方法叫做综合法，又叫顺推证法或由因导果法．

3．分析法

从要证的结论出发，逐步寻求使它成立的充分条件，直至所需条件为已知条件或一个明显成立的事实(定义、公理或已证明的定理、性质等)，从而得出要证的命题成立，这种证明方法叫做分析法，即“执果索因”的方法．

4．反证法

先假设要证的命题不成立，以此为出发点，结合已知条件，应用公理、定义、定理、性质等，进行正确的推理，得到和命题的条件(或已证明的定理、性质、明显成立的事实等)矛盾的结论，以说明假设不正确，从而证明原命题成立．

5．放缩法

证明不等式时，通过把不等式中的某些部分的值放大或缩小，简化不等式，从而达到证明的目的．

6．柯西不等式

(1)二维形式的柯西不等式：设a，b，c，d都是实数，则(a2＋b2)(c2＋d2)≥______________，当且仅当________________时，等号成立．

(2)一般形式的柯西不等式：设a1，a2，a3，…，an，b1，b2，b3，…，bn是实数，则(a＋a＋…＋a)(b＋b＋…＋b)≥(a1b1＋a2b2＋…＋anbn)2，当且仅当bi＝0(i＝1,2，…，n)或存在一个实数k，使得ai＝kbi(i＝1,2，…，n)时，等号成立．

(3)柯西不等式的向量形式：设α，β是两个向量，则|α·β|≤|α||β|，当且仅当β是零向量，或存在实数k，使α＝kβ时，等号成立．

思考辨析

判断下列结论是否正确(请在括号中打“√”或“×”)

(1)当a≥0，b≥0时，≥.(　　)

(2)用反证法证明命题“a，b，c全为0”的假设为“a，b，c全不为0”．(　　)

(3)若实数x，y满足不等式xy>1，x＋y>－2，则x>0，y>0.(　　)

(4)若m＝a＋2b，n＝a＋b2＋1，则n≥m.(　　)

教材改编题

1．若a>b>1，x＝a＋，y＝b＋，则x与y的大小关系是(　　)

A．x>y   B．x<y 

C．x≥y   D．x≤y

2．已知a，b∈R＋，a＋b＝2，则＋的最小值为(　　)

A．1  B．2  C．4  D．8

3．函数f(x)＝3＋的最大值为________．

题型一　综合法与分析法证明不等式

例1 已知f(x)＝|x＋1|＋|x－1|，不等式f(x)<4的解集为M.

(1)求M；

(2)当a，b∈M时，证明：2|a＋b|<|4＋ab|.

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思维升华 用综合法证明不等式是“由因导果”，用分析法证明不等式是“执果索因”，它们是两种思路截然相反的证明方法．综合法往往是分析法的逆过程，表述简单、条理清楚，所以在实际应用时，往往用分析法找思路，用综合法写步骤，由此可见，分析法与综合法相互转化，互相渗透，互为前提，充分利用这一辩证关系，可以增加解题思路，开阔视野．

跟踪训练1 已知a，b，c为正数，且满足a＋b＋c＝1.证明：(1)1－a≤；

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(2)2a2＋b2＋c2≥.

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题型二　放缩法证明不等式

例2 (1)设a>0，|x－1|<，|y－2|<，

求证：|2x＋y－4|<a.

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(2)设n是正整数，求证：≤＋＋…＋<1.

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思维升华 常用的放缩方法有

(1)①舍去或加上一些项，如2＋>2；

②将分子或分母放大(缩小)，如<，>，<，>(k∈N*，k>1)等．

(2)利用函数的单调性．

(3)真分数性质“若0<a<b，m>0，则<”．

跟踪训练2 求证：＋＋＋…＋<2.

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题型三　柯西不等式

例3 (2022·全国甲卷)已知a，b，c均为正数，且a2＋b2＋4c2＝3，证明：

(1)a＋b＋2c≤3；

(2)若b＝2c，则＋≥3.

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思维升华 (1)利用柯西不等式证明不等式，先使用拆项重组、添项等方法构造符合柯西不等式的形式及条件，再使用柯西不等式解决有关问题．

(2)利用柯西不等式求最值，实质上就是利用柯西不等式进行放缩，放缩不当则等号可能不成立，因此，一定不能忘记检验等号成立的条件．

跟踪训练3 (2022·咸阳模拟)已知函数f(x)＝|x＋2|＋2|x－1|(x∈R)的最小值为m.

(1)求m的值；

(2)设a，b，c均为正数，2a＋2b＋c＝m，求a2＋b2＋c2的最小值．

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