Question

Cho a + b + c = 0. CMR \(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(ab+bc+ca\right)^2=\dfrac{\left(a^2+b^2+c^2\right)^2}{2}\)

Answer 1

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