2021-2021版高中數(shù)學(xué)第三章柯西不等式與排序不等式3.2一般形式的柯西不等式試題新人教A版選修4-5

.PAGE下載后可自行編輯修改，頁(yè)腳下載后可刪除。二一般形式的柯西不等式課后篇穩(wěn)固探究A組1.a,b,c均大于0,A=a2+b2+c23,B=A.A>B B.A≥BC.A<B D.A≤B解析因?yàn)?12+12+12)·(a2+b2+c2)≥(a+b+c)2,所以a2+b2又a,b,c均大于0,所以a+b+c>0,所以a2答案B2.假設(shè)x2+y2+z2=1,那么x+y+2z的最大值等于()C.2解析由柯西不等式,可得[12+12+(2)2](x2+y2+z2)≥(x+y+2z)2,即(x+y+2z)2≤4,因此x+y+2z≤2當(dāng)且僅當(dāng)x=y=z2,即x=12,y=12,z=22時(shí),等號(hào)成立答案A3.a12+a22+…+an2=1,x12+x22+…+xn2=1,那么a解析∵(a1x1+a2x2+…+anxn)2≤(a12+a22+…+an2)×(x12+x22+…+xn2)=1×1=1,∴答案A4.設(shè)a,b,c均為正數(shù)且a+b+c=9,那么4a+9解析由柯西不等式,可得4a+9b+16c=19(a+b+c)·4a+答案C5.x,y是實(shí)數(shù),那么x2+y2+(1-x-y)2的最小值是 ()A.16 B.解析由柯西不等式,得(12+12+12)[x2+y2+(1-x-y)2]≥[x+y+(1-x-y)]2=1,即x2+y2+(1-x-y)2≥13當(dāng)且僅當(dāng)x=y=1-x-y,即x=y=13時(shí),x2+y2+(1-x-y)2取得最小值1答案B6.a,b,c>0,且a+b+c=1,那么4a+1+解析由柯西不等式,得(4a+1=(1×4a+1+1×4b+1+1≤(12+12+12)(4a+1+4b+1+4c=3[4(a+b+c)+3]=21.當(dāng)且僅當(dāng)a=b=c=13時(shí),取等號(hào)故4a+1+答案217.設(shè)a,b,c是正實(shí)數(shù),且a+b+c=9,那么2a+2解析因?yàn)?a+b+c)2=[(a)2+(b)2+(c)2]22c2所以2a+2b+2c≥2當(dāng)且僅當(dāng)a答案28.設(shè)a,b,c,x,y,z都是正數(shù),且a2+b2+c2=25,x2+y2+z2=36,ax+by+cz=30,那么a+b+解析由柯西不等式知25×36=(a2+b2+c2)·(x2+y2+z2)≥(ax+by+cz)2=302=25×36,當(dāng)且僅當(dāng)ax=b由k2(x2+y2+z2)2=25×36,解得k=56所以a+b+答案59.a+b+c=1,且a,b,c是正數(shù),求證2a+證明左邊=[2(a+b+c)]1a+b+1b+c+1c+a=[(a+b)+(b+c)+(c+a)]1a10.x,y,z∈R,且x-2y-3z=4,求x2+y2+z2的最小值.解由柯西不等式,得[x+(-2)y+(-3)z]2≤[12+(-2)2+(-3)2](x2+y2+z2),即(x-2y-3z)2≤14(x2+y2+z2),所以16≤14(x2+y2+z2).因此x2+y2+z2≥87,當(dāng)且僅當(dāng)x=y-2=z-3,即當(dāng)x=27,y=-47,z=-67時(shí)B組1.x2+y2+z2=1,那么x+2y+2z的最大值為()A.1 B.2 解析由柯西不等式,得(x+2y+2z)2≤(12+22+22)(x2+y2+z2)=9,所以-3≤x+2y+2z≤3.當(dāng)且僅當(dāng)|x|=y2=所以x+2y+2z的最大值為3.答案C2.導(dǎo)學(xué)號(hào)26394054a,b,c為正實(shí)數(shù),且a+2b+3c=9,那么3a+2A.39 B.13解析3a+2b+c=答案A3.設(shè)a,b,c為正數(shù),那么(a+b+c)4a+9解析(a+b+c)4=[(a)2+(b)2+(c)2]·2≥a·2a+b·3b+c當(dāng)且僅當(dāng)a2=答案1214.設(shè)x,y,z∈R,2x+2y+z+8=0,那么(x-1)2+(y+2)2+(z-3)2的最小值為.

解析2x+2y+z+8=0?2(x-1)+2(y+2)+(z-3)=-9.考慮以下兩組向量:u=(2,2,1),v=(x-1,y+2,z-3),由柯西不等式,得(u·v)2≤|u|2·|v|2,即[2(x-1)+2(y+2)+(z-3)]2≤(22+22+12)·[(x-1)2+(y+2)2+(z-3)2].所以(x-1)2+(y+2)2+(z-3)2≥(-9)29=9,當(dāng)且僅當(dāng)x=-1,y=-答案95.導(dǎo)學(xué)號(hào)26394055x1,x2,x3,x4為實(shí)數(shù),且x1+x2+x3+x4=6,x12+x22+x3證明由柯西不等式,得(x2+x3+x4)2≤(1+1+1)(x2由題設(shè)條件,得x2+x3+x4=6-x1,x22+x3代入上式,得(6-x1)2≤3(12-x1∴36-12x1+x12≤36-3∴4x12-12x1≤0,∴0≤x同理可證0≤xi≤3(i=2,3,4).綜上所述,0≤xi≤3(i=1,2,3,4).6.導(dǎo)學(xué)號(hào)26394056設(shè)實(shí)數(shù)a,b,c,d,e滿足a+b+c+d+e=8,且a2+b2+c2+d2+e2=16,試確定e的最大值.解由,得a+b+c+