第1課時(shí)基本不等式課后·訓(xùn)練提升基礎(chǔ)鞏固1.不等式a2+1≥2a中等號(hào)成立的條件是()A.a=±1B.a=B1 C.a=1D.a=D0答案B解析a2+12a=(a1)2≥0,當(dāng)a=1時(shí),等號(hào)成立.2.已知m=a+1a+1(a>0),0<n<3,則m,n之間的大小關(guān)系是(A.m>nB.mB<nC.m=nCD.mD≤n答案A解析因?yàn)閍>0,所以m=a+1a+1≥2a·1a+1=3,當(dāng)且僅當(dāng)a=1時(shí)等號(hào)成立3.(多選題)下列不等式一定成立的是()A.x2+14>x(x>0) B.2x+2x≥4(C.x2+1≥2|x|(x∈R) D.1x2+1>1(x答案BC解析∵當(dāng)x=12時(shí),x2+14∴A不一定成立;∵當(dāng)x>0時(shí),有2x+2x≥4x·1x=4,當(dāng)且僅當(dāng)x=1時(shí)等號(hào)成立,∵x2+12|x|=(|x|1)2≥0,即x2+1≥2|x|恒成立,∴C一定成立;∵x2+1≥1,∴0<1x2+1≤1,故4.(多選題)下列條件中,能使ba+ab≥2A.ab>0B.abB<0C.a>0,b>0D.aD<0,b<0答案ACD解析當(dāng)ba,ab均為正數(shù)時(shí),ba+ab≥2,故只需a5.已知a>b>c,則(a-b)(答案(解析∵a>b>c,∴ab>0,bc>0.∴a-c2=(a-b)+(6.不等式x2+2x2+1≥2恒成立,當(dāng)且僅當(dāng)答案0解析x2+2x2+1=(x2+1)+1x2+1=x2+1+1x2+17.若a>0,b>0,a+b=2,則下列不等式對(duì)一切滿足條件的a,b恒成立的是(填序號(hào)).

①ab≤1;②a+b≤2;③a2+b2≥2;④a3+b3≥3;⑤答案①③⑤解析令a=b=1,排除②④;由2=a+b≥2ab?ab≤1,①中不等式恒成立;a2+b2=(a+b)22ab=42ab≥42×a+b22=2,③中不等式恒成立;1a8.設(shè)a,b,c都是正數(shù),求證:bca+ac證明因?yàn)閍,b,c都是正數(shù),所以bca,所以bca+acb≥2c,acb+abc≥三式相加得2bca+acb+abc≥2(a+b+c),即bc9.已知a,b,c∈R,求證:a2+b2證明∵a+b22-a2+b∴a2+b2≥a+同理b2+c2≥22a2+c2≥22三式相加得a2+b2+b2+c2+a2+c2≥22(a+b)+能力提升1.已知a,b是正數(shù),則a+b2,abA.aB.aC.aD.a答案D解析∵a,b是正數(shù),∴a+b2≥ab(當(dāng)且僅當(dāng)a=b時(shí),取等號(hào)∵a+b22-a2+b22=2.如果正數(shù)a,b,c,d滿足a+b=cd=4,那么()A.ab≤c+d,且當(dāng)?shù)忍?hào)成立時(shí),a,b,c,d的取值唯一B.ab≥c+d,且當(dāng)?shù)忍?hào)成立時(shí),a,b,c,d的取值唯一C.ab≤c+d,且當(dāng)?shù)忍?hào)成立時(shí),a,b,c,d的取值不唯一D.ab≥c+d,且當(dāng)?shù)忍?hào)成立時(shí),a,b,c,d的取值不唯一答案A解析因?yàn)閍,b,c,d均為正數(shù),且a+b=cd=4,所以由基本不等式得a+b≥2ab,故ab≤4.又因?yàn)閏d≤(c+d)24,所以c+d≥4,所以ab≤c+d,當(dāng)且僅當(dāng)3.(多選題)已知a>0,b>0,則下列不等式恒成立的是()A.a+b+1ab≥22B.(a+b)1a+1C.a2+b2ab≥答案ABC解析對(duì)A,a+b+1ab≥2ab+1ab≥22,當(dāng)a=b,2ab=1時(shí),取等號(hào),故A恒成立;對(duì)B,(a+b)(1a+1b)=2+ba+ab≥4,當(dāng)a=b時(shí),取等號(hào),故B恒成立;對(duì)C,a2+b2ab≥2abab=2ab,4.已知a,b是不相等的正數(shù),x=a+b2,y=a+b,則x,答案x<y解析∵a,b是不相等的正數(shù),∴x2y2=a+b22(a+b)2=a+b+2ab2ab=2ab-(a+b)2<5.已知a>0,b>0,且a+b=1,求證:(a+1)2+(b+1)2≥92證明∵a>0,b>0,∴a+1>0,b+1>0,∴(a+1)2+(b+1)2≥2(a+1)(b+1),∴2(a+1)2+2(b+1)2≥(a+1)2+(b+1)2+2(a+1)(b+1)=[(a+1)+(b+1)]2,∴2(a+1)2+2(b+1)∴(a+1)2+(b+1)2≥92當(dāng)且僅當(dāng)a=b=12時(shí),等號(hào)成立6.已知a>0,b>0,a+b=1,求證:(1)1a+(2)1+1a1+證明(1)1a+1b+1∵a+b=1,a>0,b>0,∴1a+1b=a+ba+∴1a+