条件为abc=1的不等式 - 副本

题1、（2005第16届波罗的海数学奥林匹克（Baltic Way）正数满足，求证：题2、（1995年第36届IMO第2题）正实数满足，求证：题3、正数满足，求证：题4、正数满足，求证：题5、正数满足，求证：题6、正数满足，求证：题7、（2000年香港数学奥林匹克）正数满足，求证：题8、（1998第39届IMO备选题）正数满足，求证：题11、（2000第届IMO）正数满足，求证：题13、（1997罗马尼亚数学奥林匹克）题14、（1997保加利亚数学奥林匹克）题16、（2003第14届波罗的海（Baltic Way）数学奥林匹克）题18、（Crux第2023题）题19、（2005年中国国家集训队测试题）设a，b，c，d+ 且abcd1求证：$frac1(1a)2 frac1(1b)2 frac1(1c)2frac1(1d)2$1（2007年中国东南地区数学竞赛题）设a，b，c，+ 且abc1 m为大于1的整数 求证：am/(a+b) +bm/(b+c) +cm/(c+a)3/2设正数$a、b、c、d$满足$abcd1$, 则$frac 9(2 + a)2 + frac 9(2 + b)2 + frac 9(2 + c)2 + frac 9(2 + d)2 geq frac 64(7 + a)2 + frac 64(7+ b)2 + frac 64(7+ c)2 + frac 64(7 + d)2$,等号成立当且仅当$a = b = c = d = 1$. 2008第49届IMO第1天第2题（7月16日，西班牙马德里）2、English Version：(i)If x，y and z are real numbers，different from 1，such that xyz1。prove that$fracx2(x1)2fracy2(y1)2fracz2(z1)21$。 (ii) Prove that equality case is achieved for infinitely many triples of rational numbers x，y and z。中文版：（）均不为1的实数x、y、z满足xyz1，求证$fracx2(x1)2fracy2(y1)2fracz2(z1)21$。（）求证有无穷多组三元有理数数组(x，y，z)使得上式的等号成立。以下是引用李启印在2008-7-16 23:20:00的发言：均不为1的实数x、y、z满足xyz1，求证$fracx2(x1)2fracy2(y1)2fracz2(z1)21$。$frac x2(x - 1)2 + frac y2(y - 1)2 + frac z2(z - 1)2 - 1equivfrac a6(a3 - abc)2 + frac b6(b3 - abc)2 + frac c6(c3 - abc)2 - 1$ $= frac (bc + ca + ab)2(b2c2 + c2a2 + a2b2 - a2bc - b2ca - c2ab)2(a2 - bc)2(b2 - ca)2(c2 - ab)2geq 0.$ Let Then we have : Therefore : So problem a claim . The equality hold if and only if This is equivalent : From we have We only chose then the equation has rational solution Because so it also a rational . Problem claim .设正数$a、b、c$满足$abc=1$, 则$(b+frac1c-frac13)(c+frac1a-frac13)(a+frac1b-frac13)leqfraca+b+c9(frac1a+frac1b+frac1c)+frac9827$,当$a=1,b=2,c=frac12$时等号成立.设正数$a、b、c、d$满足$abcd=1$, 则 $(a+frac1b-frac12)(b+frac1c-frac12)(c+frac1d-frac12)(d+frac1a-frac12)leqfraca+b+c+d4(frac1a+frac1b+frac1c+frac1d)+frac1716$ 等号成立当且仅当$a=b=c=d=1$.已知a，b，cR+， abc1，求证：a2b2c232(1/a1/b1/c)。证明：分析法倒推，要证abc32(1/a1/b1/c)，去分母得(abc3)abc2ab2bc2ca，（用条件abc1）即abc32ab2bc2ca，即abc(2ab2bc2ca)34ab4bc4ca，即(abc)34(abbcca)，即(abc)34(abbcca)，即(abc)3(abc)4(abbcca)(abc)，而(abc)3(abc)(abc)3(三次根号abc)，即证：(abc)3(三倍的三次根号abc)4(abbcca)(abc)，即(abc)9abc4(abbcca)(abc)，而此式即为Schur不等式（的特例）。Community Blogs kuings blog sum1/a+k/suma=3+k/3 kuings blogsum1/a+k/suma=3+k/3by kuing, Sep 07, 2010, 10:30 am Find the maximum of such that the following inequality holds:SolutionThe inequality can written asIf are all equal, then , inequality obvious true for all ;If are not all equal, then , inequality equivalent toWhen we fixed , by theorem, we know that get minimum only when two of are equal, so we just need to find the infimum of Because and is continuous function in , andso if , then is a roo