Ta có :

\(\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

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

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

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

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\left(1\right)\)

\(\Leftrightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\left(2\right)\) (vì \(a+b+c=0\))

\(\left(1\right)+\left(2\right)\Rightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

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

\(\Rightarrow dpcm\)

\(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+c^2a^2\\ \Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+abc^2+a^2bc\right)=a^2b^2+b^2c^2+c^2a^2\\ \Leftrightarrow2\left(ab^2c+abc^2+a^2bc\right)=0\\ \Leftrightarrow abc\left(a+b+c\right)=0\left(đpcm;a+b+c=0\right)\)

Ta có : a + b + c = 0

( a + b + c )\(^2\) = 0

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

Nên : \(a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

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

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

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

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+8ab^2c+8abc^2+8a^2bc\)

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+8abc\left(b+c+a\right)\)

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)

Lại có : \(2\left(ab+bc+ca\right)^2\)

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

\(=2a^2b^2+2b^2c^2+2c^2a^2+4ab^2c+4abc^2+4a^2bc\)

\(=2a^2b^2+2b^2c^2+2c^2a^2+4abc\left(b+c+a\right)\)

\(=2a^2b^2+2b^2c^2+2c^2a^2\)

Vì : \(2a^2b^2+2b^2c^2+2c^2a^2=2a^2b^2+2b^2c^2=2c^2a^2\)

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

Bất đẳng thức trên đúng với mọi số thực a, b, c. Ai có thể chứng minh?

Sửa lại đề: \(a+b+c=0\)

a) Ta có:

\(A=a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)\)

\(=[(a+b+c)^2-2(ab+bc+ac)]^2-2(a^2b^2+b^2c^2+c^2a^2)\)

\(=4(ab+bc+ac)^2-2(a^2b^2+b^2c^2+c^2a^2)\)

\(=4(ab+bc+ac)^2-2(a^2b^2+b^2c^2+c^2a^2)-4abc(a+b+c)\)

(do \(a+b+c=0\))

\(A=4(ab+bc+ac)^2-2[a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)]\)

\(=4(ab+bc+ac)^2-2(ab+bc+ac)^=2(ab+bc+ac)^2\)

Ta có đpcm

b) Ta có:

\(\frac{(a^2+b^2+c^2)^2}{2}=\frac{[(a+b+c)^2-2(ab+bc+ac)]^2}{2}=\frac{[-2(ab+bc+ac)]^2}{2}=2(ab+bc+ac)^2\)

Kết hợp với kết quả phần a ta có đpcm.

\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)

\(\Leftrightarrow a^4+b^4+c^4=2\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(ab+bc+ac\right)\right]\)\(\Leftrightarrow a^4+b^4+c^4=2\left(ab+bc+ac\right)^2\)