沧海杯的一个不等式
陈洪葛
posted @ 11 年前
in 不等式
, 1210 阅读
设a,b,c>0,有
√a3a2+ab+b2+√b3b2+bc+c2+√c3c2+ca+a2≥√a+√b+√c√3
证明:作替换x2=a,y2=b,z2=c,不等式变成
∑cycx6x4+x2y2+y4+2∑cycx3y3√(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
12∑cycx6+y6x4+x2y2+y4+2∑cycx3y3√(x4+x2y2+y4)(y4+y2z2+z4)≥x2+y2+z2+2xy+2yz+2xz3
⇔6∑cycx3y3√(x4+x2y2+y4)(y4+y2z2+z4)≥12∑cyc(x2+y2+4xy−3(x6+y6)x4+x2y2+y4)
而
12∑cyc(x2+y2+4xy−3(x6+y6)x4+x2y2+y4)=∑cyc6x3y3−(x−y)4(x+y)2x4+x2y2+y4
故
⇔6∑cycx3y3√(x4+x2y2+y4)(y4+y2z2+z4)≥∑cyc6x3y3−(x−y)4(x+y)2x4+x2y2+y4
注意到两组顺序
x3y3√x4+x2y2+y4,y3z3√y4+y2z2+z4,z3x3√z4+z2x2+x4
1√x4+x2y2+y4,1√y4+y2z2+z4,1√z4+z2x2+x4
由排序不等式有
⇔∑cycx3y3√(x4+x2y2+y4)(y4+y2z2+z4)≥∑cycx3y3x4+x2y2+y4
Hence we are done!