習(xí)題2：排序不等式

排序不等式解答題1．若a1≤a2≤…≤an，而b1≥b2≥…≥bn或a1≥a2≥…≥an而b1≤b2≤…≤bn，證明：eq\f(a1b1＋a2b2＋…＋anbn,n)≤eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a1＋a2＋…＋an,n)))·eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(b1＋b2＋…＋bn,n))).當(dāng)且僅當(dāng)a1＝a2＝…＝an或b1＝b2＝…＝bn時(shí)等號(hào)成立．證明不妨設(shè)a1≤a2≤…≤an，b1≥b2≥…≥bn.則由排序原理得：a1b1＋a2b2＋…＋anbn＝a1b1＋a2b2＋…＋anbna1b1＋a2b2＋…＋anbn≤a1b2＋a2b3＋…＋anb1a1b1＋a2b2＋…＋anbn≤a1b3＋a2b4＋…＋an－1b1＋anb2……a1b1＋a2b2＋…＋anbn≤a1bn＋a2b1＋…＋anbn－1.將上述n個(gè)式子相加，得：n(a1b1＋a2b2＋…＋anbn)≤(a1＋a2＋…＋an)(b1＋b2＋…＋bn)上式兩邊除以n2，得：eq\f(a1b1＋a2b2＋…＋anbn,n)≤eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a1＋a2＋…＋an,n)))eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(b1＋b2＋…＋bn,n))).等號(hào)當(dāng)且僅當(dāng)a1＝a2＝…＝an或b1＝b2＝…＝bn時(shí)成立．2．設(shè)a1，a2，…，an為實(shí)數(shù)，證明：eq\f(a1＋a2＋…＋an,n)≤eq\r(\f(a\o\al(2,1)＋a\o\al(2,2)＋…＋a\o\al(2,n),n)).證明不妨設(shè)a1≤a2≤a3≤…≤an由排序原理得aeq\o\al(2,1)＋aeq\o\al(2,2)＋aeq\o\al(2,3)＋…＋aeq\o\al(2,n)＝a1a1＋a2a2＋a3a3＋…＋anan.aeq\o\al(2,1)＋aeq\o\al(2,2)＋aeq\o\al(2,3)＋…＋aeq\o\al(2,n)≥a1a2＋a2a3＋a3a4＋…＋ana1aeq\o\al(2,1)＋aeq\o\al(2,2)＋aeq\o\al(2,3)＋…＋aeq\o\al(2,n)≥a1a3＋a2a4＋a3a5＋…＋ana2aeq\o\al(2,1)＋aeq\o\al(2,2)＋aeq\o\al(2,3)＋…＋aeq\o\al(2,n)≥a1an＋a2a1＋a3a2＋…＋anan－1以上n個(gè)式子兩邊相加n(aeq\o\al(2,1)＋aeq\o\al(2,2)＋aeq\o\al(2,3)＋…＋aeq\o\al(2,n))＝(a1＋a2＋a3＋…＋an)2兩邊同除以n2得eq\f(a\o\al(2,1)＋a\o\al(2,2)＋a\o\al(2,3)＋…＋a\o\al(2,n),n)≥eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a1＋a2＋a3＋…＋an,n)))2所以eq\r(\f(a\o\al(2,1)＋a\o\al(2,2)＋a\o\al(2,3)＋…＋a\o\al(2,n),n))≥eq\f(a1＋a2＋a3＋…＋an,n)結(jié)論得證．3．設(shè)a1，a2，…，an為正數(shù)，求證：eq\f(a\o\al(2,1),a2)＋eq\f(a\o\al(2,2),a3)＋…＋eq\f(a\o\al(2,n－1),an)＋eq\f(a\o\al(2,n),a1)≥a1＋a2＋…＋an.證明不妨設(shè)a1>a2>…>an>0，則有aeq\o\al(2,1)>aeq\o\al(2,2)>…>aeq\o\al(2,n)也有eq\f(1,a1)<eq\f(1,a2)<…<eq\f(1,an)，由排序原理：亂序和≥反序和，得：eq\f(a\o\al(2,1),a2)＋eq\f(a\o\al(2,2),a3)＋…＋eq\f(a\o\al(2,n),a1)≥eq\f(a\o\al(2,1),a1)＋eq\f(a\o\al(2,2),a2)＋…＋eq\f(a\o\al(2,n),an)＝a1＋a2＋…＋an.4．設(shè)A、B、C表示△ABC的三個(gè)內(nèi)角的弧度數(shù)，a，b，c表示其對(duì)邊，求證：eq\f(aA＋bB＋cC,a＋b＋c)≥eq\f(π,3).證明法一不妨設(shè)A>B>C，則有a>b>c由排序原理：順序和≥亂序和∴aA＋bB＋cC≥aB＋bC＋cAaA＋bB＋cC≥aC＋bA＋cBaA＋bB＋cC＝aA＋bB＋cC上述三式相加得3(aA＋bB＋cC)≥(A＋B＋C)(a＋b＋c)＝π(a＋b＋c)∴eq\f(aA＋bB＋cC,a＋b＋c)≥eq\f(π,3).法二不妨設(shè)A>B>C，則有a>b>c，由排序不等式eq\f(aA＋bB＋cC,3)≥eq\f(A＋B＋C,3)·eq\f(a＋b＋c,3)，即aA＋bB＋cC≥eq\f(π,3)(a＋b＋c)，∴eq\f(aA＋bB＋cC,a＋b＋c)≥eq\f(π,3).5．設(shè)a，b，c為正數(shù)，利用排序不等式證明a3＋b3＋c3≥3abc.證明不妨設(shè)a≥b≥c>0，∴a2≥b2≥c2，由排序原理：順序和≥反序和，得：a3＋b3≥a2b＋b2a，b3＋c3≥b2c＋c2bc3＋a3≥a2c＋c2a三式相加得2(a3＋b3＋c3)≥a(b2＋c2)＋b(c2＋a2)＋c(a2＋b2)．又a2＋b2≥2ab，b2＋c2≥2bc，c2＋a2≥2ca.所以2(a3＋b3＋c3)≥6abc，∴a3＋b3＋c3≥3abc.當(dāng)且僅當(dāng)a＝b＝c時(shí)，等號(hào)成立．6．設(shè)a，b，c是正實(shí)數(shù)，求證：aabbcc≥(abc)eq\f(a＋b＋c,3).證明不妨設(shè)a≥b≥c>0，則lga≥lgb≥lgc.據(jù)排序不等式有：alga＋blgb＋clgc≥blga＋clgb＋algcalga＋blgb＋clgc≥clga＋algb＋blgcalga＋blgb＋clgc＝alga＋blgb＋clgc上述三式相加得：3(alga＋blgb＋clgc)≥(a＋b＋c)(lga＋lgb＋lgc)即lg(aabbcc)≥eq\f(a＋b＋c,3)lg(abc)故aabbcc≥(abc)eq\f(a＋b＋c,3).7．設(shè)xi，yi(i＝1,2，…，n)是實(shí)數(shù)，且x1≥x2≥…≥xn，y1≥y2≥…≥yn，而z1，z2，…，zn是y1，y2，…，yn的一個(gè)排列．求證：eq\o(∑,\s\up12(n),\s\do4(i＝1))(xi－yi)2≥eq\o(∑,\s\up12(n),\s\do4(i＝1))(xi－zi)2.證明要證eq\o(∑,\s\up12(n),\s\do4(i＝1))(xi－yi)2≥eq\o(∑,\s\up12(n),\s\do4(i＝1))(xi－zi)2只需證eq\o(∑,\s\up12(n),\s\do4(i＝1))yeq\o\al(2,i)－2eq\o(∑,\s\up12(n),\s\do4(i＝1))xiyi≥eq\o(∑,\s\up12(n),\s\do4(i＝1))zeq\o\al(2,i)－2eq\o(∑,\s\up12(n),\s\do4(i＝1))xizi.因?yàn)閑q\o(∑,\s\up12(n),\s\do4(i＝1))yeq\o\al(2,i)＝eq\o(∑,\s\up12(n),\s\do4(i＝1))zeq\o\al(2,i)，∴只需證eq\o(∑,\s\up12(n),\s\do4(i＝1))xizi≤eq\o(∑,\s\up12(n),\s\do4(i＝1))xiyi.而上式左邊為亂序和，右邊為順序和．由排序不等式得此不等式成立．故不等式eq\o(∑,\s\up12(n),\s\do4(i＝1))(xi－yi)2≥eq\o(∑,\s\up12(n),\s\do4(i＝1))(xi－zi)2成立．8．已知a，b，c為正數(shù)，且兩兩不等，求證：2(a3＋b3＋c3)>a2(b＋c)＋b2(a＋c)＋c2(a＋b)．證明不妨設(shè)a>b>c>0.則a2>b2>c2，a＋b>a＋c>b＋c，∴a2(a＋b)＋b2(a＋c)＋c2(b＋c)