沧海杯的一个不等式

陈洪葛 posted @ Oct 11, 2013 09:43:18 AM in 不等式 , 933 阅读

设$a,b,c>0$,有
\[ \sqrt{\frac{a^3}{a^2+ab+b^2}}+\sqrt{\frac{b^3}{b^2+bc+c^2}}+\sqrt{\frac{c^3}{c^2+ca+a^2}}\geq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{3}} \]
证明:作替换$x^2=a,y^2=b,z^2=c$,不等式变成
\[ \sum_{cyc}\frac{x^6}{x^4+x^2y^2+y^4}+2\sum_{cyc}\frac{x^3y^3}{\sqrt{(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)}}\geq \frac{x^2+y^2+z^2+2xy+2yz+2xz}{3} \]
注意到
\[ \sum_{cyc}\frac{x^6}{x^4+x^2y^2+y^4}=\sum_{cyc}\frac{y^6}{x^4+x^2y^2+y^4}\]
因此,不等式变成
\[ \frac{1}{2}\sum_{cyc}\frac{x^6+y^6}{x^4+x^2y^2+y^4}+2\sum_{cyc}\frac{x^3y^3}{\sqrt{(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)}}\geq \frac{x^2+y^2+z^2+2xy+2yz+2xz}{3} \]
\[ \Leftrightarrow 6\sum_{cyc}\frac{x^3y^3}{\sqrt{(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)}}\geq \frac{1}{2}\sum_{cyc}\left(x^2+y^2+4xy-\frac{3(x^6+y^6)}{x^4+x^2y^2+y^4}\right)\]

\[ \frac{1}{2}\sum_{cyc}\left(x^2+y^2+4xy-\frac{3(x^6+y^6)}{x^4+x^2y^2+y^4}\right)=\sum_{cyc}\frac{6x^3y^3-(x-y)^4(x+y)^2}{x^4+x^2y^2+y^4}\]

\[ \Leftrightarrow 6\sum_{cyc}\frac{x^3y^3}{\sqrt{(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)}}\geq\sum_{cyc}\frac{6x^3y^3-(x-y)^4(x+y)^2}{x^4+x^2y^2+y^4}\]
注意到两组顺序
\[ \frac{x^3y^3}{\sqrt{x^4+x^2y^2+y^4}},\frac{y^3z^3}{\sqrt{y^4+y^2z^2+z^4}},\frac{z^3x^3}{\sqrt{z^4+z^2x^2+x^4}}\]
\[ \frac{1}{\sqrt{x^4+x^2y^2+y^4}},\frac{1}{\sqrt{y^4+y^2z^2+z^4}},\frac{1}{\sqrt{z^4+z^2x^2+x^4}}\]
由排序不等式有
\[  \Leftrightarrow \sum_{cyc}\frac{x^3y^3}{\sqrt{(x^4+x^2y^2+y^4)(y^4+y^2z^2+z^4)}}\geq \sum_{cyc}\frac{x^3y^3}{x^4+x^2y^2+y^4}\]
Hence we are done!
 


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