《高中數(shù)學 第二章 柯西不等式與排序不等式及其應用 2.1.2 柯西不等式的一般形式及其參數(shù)配方法的證明課件 新人教B版選修45》由會員分享，可在線閱讀，更多相關《高中數(shù)學 第二章 柯西不等式與排序不等式及其應用 2.1.2 柯西不等式的一般形式及其參數(shù)配方法的證明課件 新人教B版選修45（20頁珍藏版）》請在裝配圖網(wǎng)上搜索。

1、2 2.1 1.2 2柯西不等式的一般形式及其參數(shù)配方法的證明目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISH

2、ULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航1.認識一般形式的柯西不等式.2.理解一般形式的柯西不等式的幾何意義.3.會用一般形式的柯西不等式求解一些簡單問題.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLI

3、ANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航定理(柯西不等式的一般形式) (2)柯西不等式的一般形式的證明方法是參數(shù)配方法.名師點撥記憶柯西不等式的一般形式,一是抓住其結構特點:左邊是平方和再開方的積,右邊是積的和的絕對值;二是與二維形式的柯西不等式類比記憶.目標導航DIANLITOU

4、XI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONG

5、NANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHIS

6、HULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航【做一做1】 已知a,b,c(0,+),且a+b+c=1,則a2+b2+c2的最小值為()解析:由柯西不等式,得(a2+b2+c2)(12+12+12)(a1+b1+c1)2.答案:C 目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHIS

7、HULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航A.1B.-1C.2D.-2 (a1b1+a2b2+anbn)24.-2a1b1+anbn2.所求的最大值為2.答案:C目標導航DIANLITOUXI典例透析SUITAN

8、GLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚

9、焦ZHISHISHULI知識梳理目標導航1.一般形式的柯西不等式如何應用?剖析:我們主要利用柯西不等式來證明一些不等式或求值等問題,但往往不能直接應用,需要對數(shù)學式子的形式進行變形,拼湊出與一般形式的柯西不等式相似的結構,才能應用,因而適當變形是我們應用一般形式的柯西不等式的關鍵,也是難點.我們要注意在應用柯西不等式時,對于數(shù)學式子中數(shù)或字母的順序要對比柯西不等式中的數(shù)或字母的順序,以便能使其形式一致,然后應用解題.2.如何利用“1”?剖析:數(shù)字“1”的利用非常重要,為了利用柯西不等式,除了拼湊應該有的結構形式外,對數(shù)字、系數(shù)的處理往往起到某些用字母所代表的數(shù)或式子所不能起的作用.這要求在理論

10、上認識柯西不等式與實際應用時二者達到一種默契,即不因為“形式”與“面貌”的影響而不會用柯西不等式.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNAN

11、JUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三利用柯西不等式證明不等式【例1】 已知a1,a2,an都是正實數(shù),且a1+a2+an=1.分析:已知條件中a1+a2+an=1,可以看作“1”的代換,而要證明a1+a2+an=1應擴大2倍后再利用,本題還可以利用其他的方法證明.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUX

12、I典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三證明:證法一:根據(jù)柯西不等式,得 目標導航DIANLITO

13、UXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHON

14、GNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三當且僅當a1=a2=an時等號成立.故原不等式成立.證法二:a(0,+),目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUX

15、I典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三當且僅當a1=a2=an時等號成立.故原不等式成立.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導

16、航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三反思通過以上不同的證明方法可以看出,構造出所需要的某種結構是證題的難點,因此,對柯西不等式或其他重要不等式,要熟記公式的特點,能靈活變形,才能靈活應用.目標導航DIANLIT

17、OUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHO

18、NGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三利用柯西不等式求函數(shù)的最值 分析:將已知等式變形,直接應用柯西不等式求解.解:根據(jù)柯西不等式,得120=3(2x+1)+(3y+4)+(5z+6)目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJU

19、JIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三反思要求ax+by+z的最大值,利用柯西不等式(ax+by+z)2(a2+b2+12)(x2+y2+z2)的形式,再結合已知條件進行配湊,是常見的變形技巧.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難

20、聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題

21、型三易錯辨析易錯點:應用柯西不等式時,因忽略等號成立的條件而致錯.目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHI

22、SHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航題型一題型二題型三錯因分析:上題在求解時,由于等號不成立,因此求解過程錯誤,結果也不正確.正解:應用函數(shù)單調性的定義(或導數(shù))可證得f(x)在2,3上為增函數(shù),目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航

23、DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航1 2 3 4A.1B.-1C.3D.9 答案:D 目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標

24、導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航1 2 3 4A.1B.3C.6D.9 答案:D 目

25、標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIA

26、NXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航1 2 3 4答案:16 目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標導航1 2 3 44已知x+4y+9z=1,則x2+y2+z2的最小值為.解析:(x2+y2+z2)(12+42+92)(x+4y+9z)2=1,