Question

[Thanh Tu Nguyen]

cho \(a+b+c=0,a^2+b^2+c^2=1\)

chứng minh rằng \(a^4+b^4+c^4=\dfrac{1}{2}\)

Answer 1

[ミ★Zero ❄ ( Hoàng Nhật )]

Từ đề bài ta có :

\(a+b+c=0< =>\left(a+b+c\right)^2=0< =>a^2+b^2+c^2+2ab+2ac+2bc=0\)

Mà \(a^2+b^2+c^2=1\)  < = > 1 + 2 ( ab + ac + bc ) = 0

< = > 2 ( ab + ac + bc ) = -1 

< = > ab + ac + bc = -1/2

\(< =>\left(ab+ac+bc\right)^2=\left(-\dfrac{1}{2}\right)^2< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\)

\(< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(< =>\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2=\dfrac{1}{4}\)

Lại có từ \(a^2+b^2+c^2=1\)

\(< =>\left(a^2+b^2+c^2\right)^2=1< =>a^4+b^4+c^4+2\left[\left(ab\right)^2+\left(ac\right)^2+\left(bc\right)^2\right]=1\)

\(< =>a^4+b^4+c^4+2.\dfrac{1}{4}=1< =>a^4+b^4+c^4+\dfrac{1}{2}=1< =>a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}\left(đpcm\right)\)