20xx高考數(shù)學(xué)文人教a版一輪復(fù)習(xí)學(xué)案：74-直接證明與間接證明-【含解析】-資料下載頁(yè)
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【正文】 證明(1)要證6+102+14,只要證(6+10)2(2+14)2,即260228,顯然成立的,所以,原不等式成立.(2)當(dāng)a0,b0時(shí),要證lga+b2≥lga+lgb2,只要證lga+b2≥lgab,因?yàn)楹瘮?shù)y=lgx在(0,+∞)上單調(diào)遞增,即證a+b2≥ab0,此不等式顯然成立,當(dāng)且僅當(dāng)a=+b2≥lga+lgb2.例3(1)解設(shè){an}的前n項(xiàng)和為Sn,則當(dāng)q=1時(shí),Sn=a1+a1+…+a1=na1。當(dāng)q≠1時(shí),Sn=a1+a1q+a1q2+…+a1qn1,①qSn=a1q+a1q2+…+a1qn,②①②得,(1q)Sn=a1a1qn,∴Sn=a1(1qn)1q,∴Sn=na1,q=1,a1(1qn)1q,q≠1.(2)證明假設(shè){an+1}是等比數(shù)列,則對(duì)任意的k∈N*,(ak+1+1)2=(ak+1)(ak+2+1),ak+12+2ak+1+1=akak+2+ak+ak+2+1,a12q2k+2a1qk=a1qk1a1qk+1+a1qk1+a1qk+1,∵a1≠0,∴2qk=qk1+qk+1.∵q≠0,∴q22q+1=0,∴q=1,這與已知矛盾.∴假設(shè)不成立,故{an+1}不是等比數(shù)列.對(duì)點(diǎn)訓(xùn)練3證明(1)(反證法)假設(shè)結(jié)論不成立,即有1+2yx≥3,且1+2xy≥3,由已知x0,y0,所以有1+2y≥3x,且1+2x≥3y,故2+2x+2y≥3x+3y,化簡(jiǎn)得2≥x+y,與已知x+y2矛盾,+2yx與1+2xy中至少有一個(gè)小于3成立.(2)(分析法)要證a2+b2≥22(a+b),只需證(a2+b2)2≥[22(a+b)]2,即證a2+b2≥12(a2+b2+2ab),即證a2+b2≥+b2≥2ab對(duì)一切實(shí)數(shù)恒成立,所以a2+b2≥22(a+b)成立.7
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