Ta có:
\(x+y+x=0\)
<=>\(x+y=-z\)
\(\Leftrightarrow\left(x+y\right)^2=\left(-z\right)^2\\ \Leftrightarrow x^2+2xy+y^2=z^2\\ \Leftrightarrow x^2+y^2-z^2=-2xy\)
\(\Leftrightarrow\left(x^2+y^2-z^2\right)^2=\left(-2xy\right)^2\\ \Leftrightarrow x^4+y^4+z^4+2x^2y^2-2y^2z^2-2z^2x^2=4x^2y^2\\ \Leftrightarrow x^4+y^4+z^4=4x^2y^2-2x^2y^2+2y^2z^2+2z^2y^2\\ \Leftrightarrow x^4+y^4+z^4=2x^2y^2+2y^2z^2+2z^2x^2\\ \Leftrightarrow2\left(x^4+y^4+z^4\right)=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2\\ \Leftrightarrow2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\)
Ta có:
\(x+y+z=0\)
\(\Leftrightarrow x+y=-z\)
Bình phương 2 vế:
\(\Leftrightarrow x^2+2xy+y^2=z^2\)
\(\Leftrightarrow x^2+y^2-z^2=-2xy\)
Bình phương 2 vế thêm lần nữa:
\(\Leftrightarrow x^4+y^4+z^4+2x^2y^2-2x^2z^2-2y^2z^2=4x^2y^2\)
\(\Leftrightarrow x^4+y^4+z^4=2x^2y^2+2y^2z^2+2x^2z^2\)
Cộng 2 vế cho \(x^4+y^4+z^4\) , ta có:
\(\Rightarrow2\left(x^4+y^4+z^4\right)=x^4+y^4+z^4+2\left(x^2y^2+y^2z^2+z^2x^2\right)\)
\(\Leftrightarrow2\left(x^4+y^4+z^4\right)=\left(x^2+y^2+z^2\right)^2\) (đpcm)