金霞 杜文發(fā)
一題多解,可以開闊學(xué)生思路、發(fā)散學(xué)生思維,讓學(xué)生學(xué)會(huì)多角度分析和解決問題;能夠促進(jìn)學(xué)生思維的靈活性. 多題一解,能夠加深學(xué)生的思維深度,分析問題時(shí)學(xué)會(huì)由表及里,抓住問題的本質(zhì),找出問題間內(nèi)在的聯(lián)系,能夠檢驗(yàn)學(xué)生思維的成熟性. 下面我們看一個(gè)問題的演繹:
若正數(shù)a、b滿足a+b=1,求1a+2b的最小值.
師生能熟練的運(yùn)用代數(shù)換元法、均值換元法、三角換元法、求導(dǎo)數(shù)法、柯西不等式、均值不等式(1的妙用)等解決.
下面就用均值不等式(1的妙用)統(tǒng)求“Aan+Bbn” 型結(jié)構(gòu)的最小值.
知識(shí)點(diǎn):
1.均值定理:n個(gè)正數(shù)的算術(shù)平均數(shù)不小于它的幾何平均數(shù).
2.二項(xiàng)式定理:(a+b)n=C0nan+C1nan-1b+C2nan-2b2+…+Crnan-rbr+…+Cnnbn(n≥1,n∈N*,r=0,1,2,…n).
基本問題 若正數(shù)a、b滿足a+b=1,求1a+2b的最小值.
解 1a+2b=(1a+2b)(a+b)=3+ba+2ab≥3+22,當(dāng)且僅當(dāng)b=2a時(shí)等號(hào)成立. 所以1a+2b的最小值為3+22=(1+2)2,當(dāng)且僅當(dāng)b=2-2,a=2-1時(shí)取到.
結(jié)論 若正數(shù)a、b滿足a+b=1,則Aa+Bb(A>0,B>0)的最小值為(A+B)2,當(dāng)且僅當(dāng)Ba2=Ab2時(shí)取到.
1 問題展開、歸納總結(jié)
例1 若正數(shù)a、b滿足a+b=1,求1a2+2b2的最小值.
解 1a2+2b2=(1a2+2b2)(a+b)2=(1a2+2b2)(a2+2ab+b2)=3+2ba+b2a2+2a2b2+4ab=3+(b2a2+4ab)+(2ba+2a2b2).
b2a2+4ab= b2a2+2ab+2ab≥334. 當(dāng)且僅當(dāng)b3=2a3時(shí)等號(hào)成立.
2ba+2a2b2=ba+ba+2a2b2≥332. 當(dāng)且僅當(dāng)b3=2a3時(shí)等號(hào)成立.
兩式等號(hào)同時(shí)成立,所以1a2+2b2的最小值為3+334+332=(1+32)3 ,當(dāng)且僅當(dāng)b3=2a3時(shí)取到.
錯(cuò)誤做法:1a2+2b2=(1a2+a+a)+(2b2+b+b)-2(a+b)=(1a2+a+a)+(2b2+b+b)-2.
1a2+a+a≥3,當(dāng)且僅當(dāng)a=1時(shí)等號(hào)成立.
2b2+b+b≥332,當(dāng)且僅當(dāng)b=32時(shí)等號(hào)成立.
所以1a2+2b2的最小值為6+332 ,當(dāng)且僅當(dāng)a=1,b=32時(shí)取到.
此解法的錯(cuò)誤在于兩個(gè)不等式成立的條件不滿足題設(shè)a+b=1,也就是說兩個(gè)不等式的等號(hào)不能同時(shí)成立.
結(jié)論 若正數(shù)a、b滿足a+b=1,則Aa2+Bb2(A>0,B>0)的最小值為A+33A2B+33AB2+B=(3A+3B)3,當(dāng)且僅當(dāng)Ba3=Ab3時(shí)取到.
例2 若正數(shù)a、b滿足a+b=1,求1a3+2b3的最小值.
解 1a3+2b3=(1a3+2b3)(a+b)3
=3+3ba+3b2a2+b3a3+2a3b3+6a2b2+6ab
=3+(b3a3+6ab)+(3b2a2+6a2b2)+(3ba+2a3b3).
b3a3+6ab=b3a3+2ab+2ab+2ab≥448,當(dāng)且僅當(dāng)b4=2a4時(shí)等號(hào)成立.
3b2a2+6a2b2≥218=62,當(dāng)且僅當(dāng)b4=2a4時(shí)等號(hào)成立.
3ba+2a3b3=ba+ba+ba+2a3b3≥442,當(dāng)且僅當(dāng)b4=2a4時(shí)等號(hào)成立.
上述三式等號(hào)同時(shí)成立,所以1a3+2b3的最小值為3+448+442+62=(1+42)4 ,當(dāng)且僅當(dāng)b4=2a4時(shí)取到.
結(jié)論 若正數(shù)a、b滿足a+b=1,則Aa3+Bb3(A>0,B>0)的最小值為A+44A3B+6AB+44AB3+B=(4A+4B)4,當(dāng)且僅當(dāng)Ba4=Ab4時(shí)取到.
是否有什么規(guī)律?請(qǐng)看例3的解法
例3 若正數(shù)a、b滿足a+b=1,求1a4+2b4的最小值.
解 1a4+2b4=(1a4+2b4)(a+b)4=3+4ba+6b2a2+4b3a3+b4a4+2a4b4+8a3b3+12a2b2+8ab
=3+(b4a4+8ab)+(4b3a3+12a2b2)+(6b2a2+8a3b3)+(4ba+2a4b4).
b4a4+8ab=b4a4+2ab+2ab+2ab+2ab≥5516,當(dāng)且僅當(dāng)b5=2a5時(shí)等號(hào)成立.
4b3a3+12a2b2=2b3a3+2b3a3+4a2b2+4a2b2+4a2b2≥
55256 當(dāng)且僅當(dāng)b5=2a5時(shí)等號(hào)成立.
6b2a2+8a3b3=2b2a2+2b2a2+2b2a2+4a3b3+4a3b3≥
55128,當(dāng)且僅當(dāng)b5=2a5時(shí)等號(hào)成立.
4ba+2a4b4=ba+ba+ba+ba+2a4b4≥552,當(dāng)且僅當(dāng)b5=2a5時(shí)等號(hào)成立.
上述四式等號(hào)同時(shí)成立,所以1a4+2b4的最小值為3+5516+55256+55128+552 =(1+52)5,當(dāng)且僅當(dāng)b5=2a5時(shí)取到.
結(jié)論 若正數(shù)a、b滿足a+b=1,則Aa4+Bb4(A>0,B>0)的最小值為A+55A4B+105A3B2+105A2B3+55AB4+B=(5A+5B)5,當(dāng)且僅當(dāng)Ba5=Ab5時(shí)取到.
說明 要注意例2與例3在構(gòu)造基本不等式時(shí)的不同之處,構(gòu)造規(guī)律十分明確. 例3不能如下分組構(gòu)造:
1a4+2b4=(1a4+2b4)(a+b)4=3+4ba+6b2a2+4b3a3+b4a4+2a4b4+8a3b3+12a2b2+8ab
=3+(b4a4+8ab)+(4b3a3+8a3b3)+(6b2a2+12a2b2)+(4ba+2a4b4).
如果如此構(gòu)造,四個(gè)基本不等式“等號(hào)”成立的條件就不完全相同了.
2 再看一例
例4 若正數(shù)a、b滿足a+b=1,求1a5+2b5的最小值.
解 1a5+2b5=(1a5+2b5)(a+b)5=(1a5+2b5)(a5+5a4b+10a3b2+10a2b3+5ab4+b5)
=3+5ba+10b2a2+10b3a3+5b4a4+b5a5+2a5b5+10a4b4+20a3b3+20a2b2+10ab
=3+(b5a5+10ab)+(5b4a4+20a2b2)+(10b3a3+20a3b3)+(10b2a2+10a4b4)+(5ba+2a5b5).
b5a5+10ab =b5a5+2ab+2ab+2ab+2ab+2ab≥
6632,當(dāng)且僅當(dāng)b6=2a6時(shí)等號(hào)成立.
5b4a4+20a2b2 =5(b4a4+2a2b2+2a2b2)≥1534,當(dāng)且僅當(dāng)b6=2a6時(shí)等號(hào)成立.
10b3a3+20a3b3≥202,當(dāng)且僅當(dāng)b6=2a6時(shí)等號(hào)成立.
10b2a2+10a4b4=5b2a2+5b2a2+10a4b4≥1532,當(dāng)且僅當(dāng)b6=2a6時(shí)等號(hào)成立.
5ba+2a5b5=ba+ba+ba+ba+ba+2a5b5≥662,當(dāng)且僅當(dāng)b6=2a6時(shí)等號(hào)成立.
上述五式等號(hào)同時(shí)成立,所以1a5+2b5的最小值為3+6632+1534+202+1532+662=(1+62)6 ,當(dāng)且僅當(dāng)b6=2a6時(shí)取到.
現(xiàn)在可以看出分組的規(guī)律了.
結(jié)論 若正數(shù)a、b滿足a+b=1,則Aa5+Bb5(A>0,B>0)的最小值為A+66A5B+156A4B2+206A3B3+155A2B4+66AB5+B=(6A+6B)6,當(dāng)且僅當(dāng)Ba6=Ab6時(shí)取到.
到此,得到優(yōu)美的結(jié)論:
若正數(shù)a、b滿足a+b=1,則Aan+Bbn(A>0,B>0,n≥1,n∈N*)的最小值為(n+1A+n+1B)n+1,當(dāng)且僅當(dāng)Ban+1=Abn+1時(shí)取到.
證明Aan+Bbn=(Aan+Bbn)(a+b)n
=(Aan+Bbn)(C0nan+C1nan-1b+C2nan-2b2+…Cr-1nan-r+1br-1+Crnan-rbr+…+Cnnbn)
=A(C0n+C1nba+C2nb2a2+…+Cr-1nbr-1ar-1+Crnbrar+…+Cnnbnan)+B(C0nanbn+C1nan-1bn-1+…+Cr-1nan-r+1bn-r+1+Crnan-rbn-r+…+Cn-1nab+Cnn)
=AC0n+(C1nAba+C0nBanbn)+(C2nAb2a2+C1nBan-1bn-1)+…+(CrnAbrar+Cr-1nBan-r+1bn-r+1)+…+(CnnAbnan+Cn-1nBab)+BCnn.
又CrnAbrar+Cr-1nBan-r+1bn-r+1=Cr-1nr((n-r+1)Abrar+rBan-r+1bn-r+1)=Cr-1nr(Abrar+Abrar+Abrar+…+Abrarn-r+1項(xiàng)+Ban-r+1bn-r+1+Ban-r+1bn-r+1+…+Ban-r+1bn-r+1r項(xiàng))≥Cr-1nr(n+1)n+1An-r+1Br=Crn+1n+1An-r+1n+1Br.
當(dāng)且僅當(dāng)Abn+1=Ban+1時(shí)等號(hào)成立(其中n≥1,n∈N*,r=,1,2,…n).
所以Aan+Bbn≥A+∑nr=1Crn+1n+1An-r+1n+1Br+B=(n+1A+n+1B)n+1,當(dāng)且僅當(dāng)Abn+1=Ban+1時(shí)等號(hào)成立.
所以Aan+Bbn的最小值為(n+1A+n+1B)n+1,當(dāng)且僅當(dāng)Ban+1=Abn+1時(shí)取到.
此類型題結(jié)構(gòu)更一般形式為:
若正數(shù)a、b滿足pa+qb=d,則Aan+Bbn(p>0,q>0,d>0,A>0,B>0,n≥1,n∈N*)的最小值為1dn(pn+1A+qn+1B)n+1,當(dāng)且僅當(dāng)Bpan+1=Aqbn+1時(shí)取到.
問題深入研究:
若正數(shù)a、b滿足Aa+Bb=d,則pan+qbn(A>0,B>0,p>0,q>0,d>0,n≥1,n∈N*)的最小值為何?
通過多題一解,化靜為動(dòng),提高學(xué)生分析問題、解決問題的能力,滲透類比思想、轉(zhuǎn)化思想. 多題一解,更好地踐行了劃歸與轉(zhuǎn)化、類比思想,是學(xué)好數(shù)學(xué)的一個(gè)重要手段. 與此同時(shí),對(duì)一題多解和多題一解的運(yùn)用,要注意相互結(jié)合,靈活運(yùn)用,不可只求一技,失之偏頗. 對(duì)于學(xué)生而言,他們應(yīng)該更傾向于“多題一解”.