原文地址:大于33个代数不等式作者:sqing551. with prove that: 2.for prove that 3.let ,prove that: 4.let .prove that: 5.let .prove that: 6.let and prove that: 7.let . prove that: 8.let .prove that: 9..let .prove that: 10..let .with prove that: 11.let. prove that: 12. prove that: 13. prove that: 14.for prove that: 15. prove that: 16. prove that: 17. prove that: 18. prove that: 20.let prove that: 21.let prove that: 22.for ,show that: 23.for prove that: 24..let , prove that: 25..let prove that: 26.let prove that: 27.let ,show that: 28.let prove that: (1)if  (2)if  29.let prove that: 30. find the 31.find the best val e of for 32.let prove that: 33. prove that: pxchg1200"ineq alities from hang jin j n s book" 对宋庆老师大于33个代数不等式中第15个不等式的改进 张老师 解大于33个代数不等式第16个的变式 再证大于33个代数不等式中的一不等式 切线法证大于33个代数不等式中的一不等式 证大于33个代数不等式中的第7个三元不等式 关中狂人 问题38:大于33个代数不等式第7个三元不等式的加强及证 参考文献: sqing55: 大于33个代数不等式 熊昌进:证大于33个代数不等式第7个三元不等式 证大于33个代数不等式中的第24个三元不等式 参考文献:宋庆 大于33个代数不等式张云华:证一个三元无理不等式 张云华:证一个4元分式不等式 大于33个代数不等式 2013年07月25日 wayo94 在链接[1]里,宋老师一气登载了33个不等式,个个都长相凶恶,几乎没有勇气去做,然艺高人胆大的张云华老师依然无所畏惧,证明了一些不等式(例如文[2][3]等).在张老师的冲锋掩护下,我也壮着胆子,试着做了一道(用了调和平均=平方平均,实际用权方和不等式也立刻可得): 相关链接: [1]大于 33个代数不等式 [2]张云华:证一个4元分式不等式 [3]张云华:证一个三元无理不等式 用两种方法再证一个四元分式不等式wayo94 下面这道题已有不少证明方法(链接[2]-[4]),下面再给出两种证明方法: 需要说明的是,方法一用了柯西不等式的变式,方法二也一样. 相关链接: [1]大于33个代数不等式 [2]用调和平均与平方平均的关系证明一个四元分式不等式 [3] 张云华:证一个4元分式不等式 [4]一个四元分式不等式的另证 再证大于33个代数不等式中的一个三元条件不等式 证明大于33个代数不等式中的一个三元不等式 参考文献: 宋庆 大于33个代数不等式作者:sqing55 大于33个代数不等式(待续) reny 出自宋庆老师大于33个代数不等式一文. 1,已知 ,,证明. 推广:, ,则有. 证明:见漂亮的不等式(更新)第6题. 条件略微有差别,不过没影响. 2,已知,求证. 证明1: (*) 不妨设,则,, 由排序不等式有 (顺序乱序),(乱序逆序) 因此, 结合(*)式,我们有,得证. 证明2:因为 所以. 3,已知,求证. 证明:注意到,,要使用伯努利不等式,我们得有 ,于是,同理有 两式相加即证. 4,已知,证明. 证明:两边展开得,显然成立. 15,已知,证明. 链接:在帖子maxim m val e中有:相同条件下,求 的最值. 证明:由有. 由三元均值不等式有 化简得. 注:检验可知,等号取不到,下确界是多少呢? 回应:reny老师关于大于33个不等式中的第15个问题 张老师 here i post some ineq alites from my book,i think they are interesting,hope yo enjoy it.problem 1: (uk 2008) for ,find the min of (hint: there exist a ca chy-schwarz sol tion) sol tion so we have so the min of problem 2: (iran 2008) with prove that: (hint: both am-gm and cs can solve it)problem 3:(brazilian 2007),prove  4:(t rkey 2007)for ,prove that: problem 5:(india 2007) prove that: (there even exist a geometry sol tion )problem 6:(poland 2007) problem 7:(uk.2007) for are real n mbers. problem 8:(china 2007) with pxchg1200"some ineq alies from my book" problem 1 for . proof:(by leedt26) we only need to prove that which is tr e by am-gm.problem 2 by tom_damrong let be positive real n mbers s ch that prove that potla wrote: from the ca chy-schwarz ineq ality, we obtain note that from the ca chy-schwarz ineq ality again, we have so that we get that d e to eq ality holds in the original ineq ality iff problem 3 given and show that we always have sol tion 1 (abdek):let if according to the ca chy schwarz ineq ality th s since so we have done if so which is eq ivalent to the original ineq ality the proof is completed sol tion 2:(by sayan) this is dedicated to yo , abdek; beca se yo are the am-gm master. as in my previo s sol tion, we can rephrase the given ineq ality as note that after m ltiplying both sides by and sing the identity it is s fficient to check that now, the seq ences and are oppositely sorted, so sing the rearrangement ineq ality, we obtain hence it wo ld s ffice to check that which, again, is a direct application of the am-gm ineq ality. sol tion 3:(sayan) adding to each fractions it is s fficient to show that note that we have the well-known ineq alities: so that it is s fficient to check that which is act ally and fact ally a direct application of the am-gm ineq ality. i sed the ineq ality which is etremelty sef l in these types of ineq alities. hence we are done, eq ality occ rs iff problem 4 abch42 wrote: let be positive real n mbers. prove that: i think the best proof belong to abdek: abdek wrote: eq ality occ rs when problem 5 lawas wrote: with prove that proof:(mathuniverse) ineq ality is eq ivalent to: pxchg1200"some colllections abo t interesting ineq alities" example: 1. mari sstanean wrote: let be positive real n mbers. prove the ineq ality proof: se this identity,we can prove this ineq ality easy. j st like 2. leedt26 wrote: let satisfy . prove that: 3.