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一題不等式

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\(x+y+z=0\)
若\(x=y=z=0\),不等式成立
若\(x,y,z\)不全為\(0\),則三者中至少有一正一負
不失一般性,令\(y>0,z<0\)
\(\begin{align}
  & \left( x+y+z \right)\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx \right)=0 \\
& {{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz \\
&  \\
& 6{{\left( {{x}^{3}}+{{y}^{3}}+{{z}^{3}} \right)}^{2}} \\
& =6{{\left( 3xyz \right)}^{2}} \\
& =27\times 2{{x}^{2}}{{y}^{2}}{{z}^{2}} \\
& \le 27\times {{\left( \frac{2{{x}^{2}}+\left| yz \right|+\left| yz \right|}{3} \right)}^{3}} \\
& ={{\left( 2{{x}^{2}}+2\left| yz \right| \right)}^{3}} \\
& ={{\left[ {{x}^{2}}+{{\left( -x \right)}^{2}}+2\left| yz \right| \right]}^{3}} \\
& ={{\left[ {{x}^{2}}+{{\left( y+z \right)}^{2}}-2yz \right]}^{3}} \\
& ={{\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}^{3}} \\
\end{align}\)

[ 本帖最後由 thepiano 於 2018-7-4 21:52 編輯 ]

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