SOS方法证明不等式

1、数学竞赛讲座SOS方法证明不等式(sum of squares)S A2BSab cSfjc a $ Sca2b0性质一：若 Sa，Sb，Sc0，则SABSab2 cS c2 2aSc a b0性质二：右 a, b, c, Sa, Sb, Sc2且满足(1)SaSb，&Sc，&Sa0 , (2) 若abc或a bc，则 Sb0 ,那么S2A B Sa b c2Sb c aScab 20性质三：若a,b,c, Sa, Sb, Sc扌且满足若a bc或abc,则 Sa，Sc0 且 Sa 2Sb,Sc2Sb 0 ,那么S AB Sa bc22Sb c aSc a b 20性质四：若 a,b,cRSa

2、 , Sb, Sci.且满足若a bc或abc,则 Sa，Sb20 且 b Sc c2Sb0 ,那么222S AB Sa bcSb c aSc a b0性质五：若a,b,c, Sa,Sb, Scg且满足(1)SaSb，SbSc，QSa0，(2)SaSbSbScScSa0 那么2 2 2SAB Sa beSb c aSc a b 0性质六：若a,b,c, Sa,Sb, Scg且满足(2) 存在0,使得若Sa2Sca bc b(3)Sc20或1ScSd02(1)若 Sa或 a b c21Sb0 ；SaSb0Sa1Sb2 2 2S=A-B= Sa(b c)Sb(c a)Sc(b a) 0(1)a2b

3、22 cabbcaca3b33 ca2b b2 : c aca3b33 c3abc(a b(4)a2bb2c2 c aab2bc2(5)a4b44 c3. a bb33 c c a(3a22abb2)(ab)2(3b2.常见的恒等式2cb2c a2 2 2(a b) (a c) (c b)2(2a b)(a b)2 (2b c)(c b)2 (2c a)(a c)23c)(a b)2 (a c)2 (c b)22(a b)3 (a c)3 (c b)33, 33, 333(6) a b b c c a b a c b a cc2)(c b)2 (3c2 2ac a2)(a c)24a b c_

4、3333(a b) (a c) (c b)(7)a4 b4 c4 a2b21-(ab)2(a b)2 (c b)2(c b)2 (a c)2(a c)2三例题1.已知正数a,b,c,求证:19ab( )=；2.已知正数a,b,c满足mina,b, cmaxa,b,c求证(ab (J)=9 +丄(a b)24 16(a-b)2(a b)2正实数宀八二满足亦二al证明:525252兀-x丁 - 丁二-二5, 2丄 2十 5丄 2丄 2十 5丄 2X 十丁 十二丁 十二十；V二十X r J4已知正数a,b,c,试求最优常数k，使1 1 1 (a b c)(a b c)bc acb2 c25.已知正数a, b, c,试求最优常数k,使得bcb2c2_3ka256.已知正数a,b,c为三角形三边，求证abcacb、c-+)2()3b c ac b a7已知正数x,y,z,求证:x2 y2 z2 xy xz yz 2( (xy2 z2)8.已知正数a,b,c, 且 abc=1，求证:3(丄丄 1