Ta có : \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)\)
Bình phương hai vế , ta được :
\(\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ac\right)\right]^2\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2\right)\) \(\left(1\right)\)
\(\Rightarrow a^4+b^4+c^4=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\) ( vì \(a+b+c=0\) ) \(\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\):
\(2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ca\right)^2\)
