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Ta có:
(a + b + c)2 = 0 => a2 + b2 + c2 + 2(ab + bc + ca) = 0
=> a2 + b2 + c2 = -2(ab + bc + ca)
=> (a2 + b2 + c2)2 = 4(ab + bc + ca)2
=> a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 4[a2b2 + b2c2 + c2a2 + 2(ab2c + bc2a + ca2b)
=> a4 + b4 + c4 = 2(a2b2 + b2c2 + c2a2) + 8abc(a + b + c)
=> a4 + b4 + c4 = 2(a2b2 + b2c2 + c2a2) (vì a + b + c = 0) (1)
Có: \(\left\{{}\begin{matrix}2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2a^2bc+2abc^2\right)\\2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(ab+bc+ca\right)^2\left(2\right)\\a^4+b^4+c^4=\dfrac{\left(a^2+b^2+c^2\right)}{2}\left(3\right)\end{matrix}\right.\)
Từ (1); (2) và (3) ta có đpcm
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\(a+b+c=0\)
\(\Rightarrow a^2+b^2+c^2\)= \(2ab+2bc+2ca=0\)
\(\Rightarrow a^2+b^2+c^2\)= \(-2\left(ab+bc+ca\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2\)= \(\left(-2ab-2bc-2ca\right)^2\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2a^2\)= \(4a^2b^2+4b^2c^2+4c^2a^2+4abc\left(a+b+c\right)\)= \(4a^2b^2+4b^2c^2+4c^2a^2\)( Do a + b + c = 0 )
\(\Rightarrow a^4+b^4+c^4\)= \(2\left(a^2b^2+b^2c^2+c^2a^2\right)\).
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Từ \(a+b+c=0\) \(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)+8abc\left(a+b+c\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)(vì a+b+c=0)
\(\Rightarrow a^4+b^4+c^4+a^4+b^4+c^4=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)
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~