Do \(x+y+z=0\) nên \(\left(x+y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=4\left(xy+yz+xz\right)^2\)
\(\Leftrightarrow x^4+y^4+z^4+2x^2y^2+2y^2z^2+2x^2z^2=4\left(x^2y^2+y^2z^2+x^2z^2+2xy^2z+2x^2yz+2xyz^2\right)\)
\(\Leftrightarrow x^4+y^4+z^4=2x^2y^2+2y^2z^2+2x^2z^2+8xyz\left(x+y+z\right)\)
Vì \(x+y+z=0\)
\(\Leftrightarrow x^4+y^4+z^4=2\left(x^2y^2+y^2z^2+x^2z^2\right)\)
\(\Leftrightarrow2\left(x^4+y^4+z^4\right)=4\left(x^2y^2+y^2z^2+x^2z^2\right)\left(1\right)\)
Mặt khác, ta có: \(\left(x^2+y^2+z^2\right)^2=4\left(xy+yz+xz\right)^2=4\left[x^2y^2+y^2z^2+x^2z^2+2xyz\left(x+y+z\right)\right]\)
\(\Leftrightarrow\left(x^2+y^2+z^2\right)^2=4\left(x^2y^2+y^2z^2+x^2z^2\right)\)(Do \(x+y+z=0\)) \(\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrowđpcm\)
chả hiểu Phước Nguyễn làm kiểu gì #phuocnguyen
