Từ a+b+c=0 có b+c =-a
Suy ra (b+c)^2 = (-a)^2 hay b^2 + c^2 +2bc = a^2
hay b^2 + c^2 -a^2 = -2bc
Suy ra (b^2 + c^2 - a^2)^2 = (-2bc)^2
<=> b^4 + c^4 + a^4 +2b^2.c^2 - 2a^2.b^2 - 2a^2.c^2 = 4b^2.c^2
<=> a^4 + b^4 + c^4 = 2a^2.b^2 + 2b^2.c^2 + 2c^2.a^2
<=> 2(a^4 + b^4 + c^4) =a^4 + b^4 + c^4 + 2a^2.b^2 + 2b^2.c^2 + 2c^2.a^2
<=> 2(a^4 + b^4 + c^4 ) =(a^2 + b^2 + c^2) ( Đpcm)
Ta có: \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow a^2+b^2+c^2=-2.\left(ab+bc+ca\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=\left[-2.\left(ab+bc+ca\right)\right]^2\)
\(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4a^2b^2+4b^2c^2+4c^2a^2+4abc\left(a+b+c\right)\)
\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)
\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+a^4+b^4+c^4\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=2.\left(a^4+b^4+c^4\right)\)
đpcm
Tham khảo nhé~
