Biến đổi từ giả thuyết:
a + b + c = 0
<=> (a + b + c)² = 0
<=> a² + b² + c² + 2(ab + bc + ca) = 0
<=> a² + b² + c² = -2(ab + bc + ca) (1)
Ta cần chứng minh
\(a^4+b^4+c^4=\dfrac{\left(a^2+b^2+c^2\right)^2}{2}\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)
\(\Leftrightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(a^2b^2+b^2c^2+a^2c^2\right)\) ( Cộng cả hai vế cho 2(a2b2 + b2c2 + c2a2
\(\Leftrightarrow\left[-2\left(ab+bc+ac\right)\right]^2=4\left(a^2b^2+b^2c^2+a^2c^2\right)\) ( Do 1 )
\(\Leftrightarrow4\left(a^2b^2+b^2c^2+a^2c^2\right)+8\left(ab^2c+bc^2a+a^2bc\right)=4\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow8\left(ab^2c+bc^2a+a^2bc\right)=0\)
\(\Leftrightarrow8abc\left(a+b+c\right)=0\)
\(\Leftrightarrow0=0\) (đúng), vì a + b + c = 0
=> Đpcm
