Question

Cho a+b+c=0. Chứng minh \(a^4+b^4+c^4\)bằng biểu thức sau đây:

\(2\left(ab+bc+ca\right)^2\)

Answer 1

Ta có \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2a^2bc+2acb^2+2abc^2\)

\(=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2\)

Ta lại có 

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

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(ab+bc+ca\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(ab+bc+ca\right)^2=4\left(ab+bc+ca\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)

Answer 2

Ta có (ab+bc+ca)2=a2b2+b2c2+c2a2+2a2bc+2acb2+2abc2

=a2b2+b2c2+c2a2+2abc(a+b+c)=a2b2+b2c2+c2a2

Ta lại có 

(a+b+c)2=a2+b2+c2+2(ab+bc+ca)=0

⇔(a2+b2+c2)2=4(ab+bc+ca)2

⇔a4+b4+c4+2(a2b2+b2c2+c2a2)=4(ab+bc+ca)2

⇔a4+b4+c4+2(ab+bc+ca)2=4(ab+bc+ca)2

⇔a4+b4+c4=2(ab+bc+ca)2