Question

Cho a+b+c=0 và \(a^2+b^2+c^2=1\)

Chứng minh \(a^4+b^4+c^4=\frac{1}{2}\)

Nhanh nhaaa

Answer 1

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Rightarrow1+2\left(ab+bc+ca\right)=0\Rightarrow ab+bc+ca=-\frac{1}{2}\Rightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc.0=\frac{1}{4}\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)

Xét: \(a^2+b^2+c^2=1\Rightarrow\left(a^2+b^2+c^2\right)^2=1\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)

\(\Rightarrow a^4+b^4+c^4+2.\frac{1}{4}=1\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)(đpcm)