Theo đề bài ta có : \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2+4ab^2c+4abc^2+4a^2bc\right)\left(1\right)\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\left(2\right)\)
Thế(2) vào (1) Ta được \(2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Leftrightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ca\right)^2\)( ĐPCM)
