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沧海杯的一个不等式

陈洪葛 posted @ 11 年前 in 不等式 , 1210 阅读

a,b,c>0,有
a3a2+ab+b2+b3b2+bc+c2+c3c2+ca+a2a+b+c3
证明:作替换x2=a,y2=b,z2=c,不等式变成
cycx6x4+x2y2+y4+2cycx3y3(x4+x2y2+y4)(y4+y2z2+z4)x2+y2+z2+2xy+2yz+2xz3
注意到
cycx6x4+x2y2+y4=cycy6x4+x2y2+y4
因此,不等式变成
12cycx6+y6x4+x2y2+y4+2cycx3y3(x4+x2y2+y4)(y4+y2z2+z4)x2+y2+z2+2xy+2yz+2xz3
6cycx3y3(x4+x2y2+y4)(y4+y2z2+z4)12cyc(x2+y2+4xy3(x6+y6)x4+x2y2+y4)

12cyc(x2+y2+4xy3(x6+y6)x4+x2y2+y4)=cyc6x3y3(xy)4(x+y)2x4+x2y2+y4

6cycx3y3(x4+x2y2+y4)(y4+y2z2+z4)cyc6x3y3(xy)4(x+y)2x4+x2y2+y4
注意到两组顺序
x3y3x4+x2y2+y4,y3z3y4+y2z2+z4,z3x3z4+z2x2+x4
1x4+x2y2+y4,1y4+y2z2+z4,1z4+z2x2+x4
由排序不等式有
cycx3y3(x4+x2y2+y4)(y4+y2z2+z4)cycx3y3x4+x2y2+y4
Hence we are done!
 


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