5.4.2排序不等式同步测控我夯基，我达标1.已知a、b、c∈R+,则a3+b3+c3与a2b+b2c+c2a的大小为()A.a3+b3+c3>a2b+b2c+c2aB.a3+b3+c3≥a2b+b2c+c2aC.a3+b3+c3
0,则a2≥b2≥c2,∴a2b+b2c+c2a≤a3+b3+c3.答案:B2.已知a、b、c∈R+,则a2(a2-bc)+b2(b2-ac)+c2(c2-ab)的正负情况为()A.大于零B.大于等于零C.小于零D.小于等于零解析:不妨设a≥b≥c>0,则a2≥b2≥c2,ab≥ac≥bc.∴a4+b4+c4=a2·a2+b2·b2+c2·c2≥a2b2+b2c2+c2a2=(ab)·(ab)+(bc)·(bc)+(ac)·(ac)≥(ab)·(ac)+(ac)·(bc)+(bc)·(ab)=a2bc+abc2+ab2c.∴a4-a2bc+b4-ab2c+c4-abc2≥0,即a2(a2-bc)+b2(b2-ac)+c2(c2-ab)≥0.答案:B3.设x1,x2,…,xn是不同的正整数,则m=222121xx+…+2nxn的最小值是()A.1B.2C.1+21+31+…+n1D.1+221+231+…+21n解析:∵x1,x2,…,xn是不同的正整数,设b1,b2,…,bn是x1,x2,…,xn的一个排列且b1≤b2≤b3≤…≤bn,则b1≥1,b2≥2,…,bn≥n.又∵211>221>…>21n,∴m=2222121nxxxn≥2222121nbbbn≥1+21+31+…+n1.答案:C4.已知cab212121logloglog,则()A.2b>2a>2cB.2a>2b>2cC.2c>2b>2aD.2c>2a>2b解析:∵b21loga>c.∴2b>2a>2c.答案:A15.设b、c为互不相等的正整数,则2232cb的最小值为()A.3613B.3617C.1D.41解析:∵b、c为互不相等的正整数,若bc,则b≥2,c≥1,又∵221>231,∴2232cb≥2232bc≥223221=3617.∴最小值为3617.答案:B6.若a、b、c为实数,则a2+b2+c2与ab+bc+ca的大小关系为_______________.解析:不妨设a≥b≥c,则ab+bc+ca≤a·a+b·b+c·c=a2+b2+c2.答案:a2+b2+c2≥ab+bc+ca7.比较a4+b4与a3b+ab3的大小关系为_______________.解析:不妨设a≥b,则a3≥b3,∴a4+b4≥a3b+b3a.答案:a4+b4≥a3b+ab3我综合，我发展8.若a、b为正数,则21a+21b与33abba的大小关系为_______________.解析:不妨设a≥b>0,则a3≥b3>0,∴31a≤31b.∴31a·a+31b·b≤33baab,即21a+21b≤33baab.答案:21a+21b≤33baab9.若a、b为正数,则24ba-a2与b2-24ab的大小关系为_______________.解析:不妨设a≥b>0,则a2≥b2>0,21b≥21a>0,a4≥b4>0.∴2424abba≥2424aabb=a2+b2.∴24ba-a2≥b2-24ab.2答案:24ba-a2≥b2-24ab10.设a、b、c都是正数,求证:cba+acb+bac≥23.证明:不妨设a≥b≥c>0,则a+b≥a+c≥b+c>0,∴ba1≤ca1≤cb1.∴acbcbabac≥baaacccbb,acbcbabac≥babacacbc.两式相加,得2(acbcbabac)≥3,即acbcbabac≥23成立.11.设a、b、c∈R+,求证:a1+b1+c1≤333888cbacba.分析:本题可多次利用排序不等式证明,不等式的右边=335335335bacacbcba.证明:不妨设a≥b≥c>0,则c1≥b1≥a1.∴331cb≥331ac≥331ba,且a5≥b5≥c5.∴335335335bacacbcba≥323232335335335abcabcbabacacbc.∵a≥b≥c>0,∴a2≥b2≥c2,333111cba.∴323232abcabc≥.111323232cbaccbbaa∴333888cbacba≥a1+b1+c1成立.我创新，我超越12.设a+b>0,n为偶数,求证:nnab1+nnba1≥a1+b1.3分析:本题将an-1与bn-1交换一下就得到右边,可用排序不等式解答,情况不定可分类讨论.证明:∵a+b>0,∴a>-b,共有四种情况.(1)当a≥b>0时,an≥bn>0,an-1≥bn-1,∴na1≤nb1.∴nnnnbbaa11≤nnnnbaab11,即nnnnbaab11≥a1+b1成立.(2)当b≥a>0时,bn≥an>0,bn-1≥an-1,∴nb1≤na1.∴nnnnaabb11≤nnnnbaab11,即nnnnbaab11≥a1+b1.(3)当a>-b>0时,∵n为偶数,∴an>(-b)n=bn>0,且a>b.∴an-1>bn-1,且na1a>-b时,则b>-a>0,∵n为偶数,∴bn>(-a)n=an>0且bn-1>an-1.∴na1>nb1.∴nnnnbbaa11≤nnnnbaab11,即nnnnbaab11≥a1+b1.综上,nnnnbaab11≥a1+b1.13.设x1≥x2≥…≥xn,y1≥y2≥…≥yn,z1,z2,…,zn是y1,y2,…,yn的任意一个排列,求证:4niiiyx12)(≤niiizx12)(.分析:本题可利用排序不等式解答,要证niiiyx12)(≤niiizx12)(成立,只需证niiiyx122)(-niiiyx1)2(≤niiizx122)(-niiizx1)2(,由排序不等式证明出.证明:∵x1≥x2≥…≥xn,y1≥y2≥…≥yn,z1,z2,…,zn是y1,y2,…,yn的任意一个排列,∴x1y1+x2y2+…+xnyn≥x1z1+x2z2+…+xnzn,即niiiniiizxyx11.∴niiiniiizxyx11,22且x12+y12+x22+y22+…+xn2+yn2=x12+z12+x22+z22+…+xn2+zn2.∴nininiiiiiiiniiizxzxyxyx1112212,2)(2)(即niiiniiizxyx1212)()(成立.5
