高中數(shù)學(xué) 第二章 柯西不等式與排序不等式及其應(yīng)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

《高中數(shù)學(xué) 第二章 柯西不等式與排序不等式及其應(yīng)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(guān)《高中數(shù)學(xué) 第二章 柯西不等式與排序不等式及其應(yīng)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航1.認(rèn)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(biāo)導(dǎo)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(diǎn)撥記憶柯西不等式的一般形式,一是抓住其結(jié)構(gòu)特點(diǎn):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(biāo)導(dǎo)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(biāo)導(dǎo)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航1.一般形式的柯西不等式如何應(yīng)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(yīng)用,需要對數(shù)學(xué)式子的形式進(jìn)行變形,拼湊出與一般形式的柯西不等式相似的結(jié)構(gòu),才能應(yīng)用,因而適當(dāng)變形是我們應(yīng)用一般形式的柯西不等式的關(guān)鍵,也是難點(diǎn).我們要注意在應(yīng)用柯西不等式時,對于數(shù)學(xué)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(yīng)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(yīng)該有的結(jié)構(gòu)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(rèn)識柯西不等式與實(shí)際應(yīng)用時二者達(dá)到一種默契,即不因?yàn)椤靶问健迸c“面貌”的影響而不會用柯西不等式.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(shí)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(yīng)擴(kuò)大2倍后再利用,本題還可以利用其他的方法證明.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(biāo)導(dǎo)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航題型一題型二題型三當(dāng)且僅當(dāng)a1=a2=an時等號成立.故原不等式成立.證法二:a(0,+),目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航題型一題型二題型三當(dāng)且僅當(dāng)a1=a2=an時等號成立.故原不等式成立.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)

16、航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(gòu)造出所需要的某種結(jié)構(gòu)是證題的難點(diǎn),因此,對柯西不等式或其他重要不等式,要熟記公式的特點(diǎn),能靈活變形,才能靈活應(yīng)用.目標(biāo)導(dǎo)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(yīng)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(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(biāo)導(dǎo)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(jié)合已知條件進(jìn)行配湊,是常見的變形技巧.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難

20、聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航題型一題型二題

21、型三易錯辨析易錯點(diǎn):應(yīng)用柯西不等式時,因忽略等號成立的條件而致錯.目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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(jié)果也不正確.正解:應(yīng)用函數(shù)單調(diào)性的定義(或?qū)?shù))可證得f(x)在2,3上為增函數(shù),目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航

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

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

25、標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)航1 2 3 4答案:16 目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)航DIANLITOUXI典例透析SUITANGLIANXI隨堂演練ZHONGNANJUJIAO重難聚焦ZHISHISHULI知識梳理目標(biāo)導(dǎo)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(biāo)導(dǎo)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,