How prove $\frac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$ Let $a,b,c\in R$,and such $ab+bc+ac=0,a+b+c\neq 0$
show that
$$\dfrac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$$

WLOG, assume $a+b+c=1 \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2)=2$, since $1=(a+b+c)^2 = a^2+b^2+c^2$, and also $2((a-b)(b-c))^2 = 2(ab-ac-b^2+bc)^2=2(-2ac-b^2)^2=2(b^4+4acb^2+4a^2c^2)$. Thus:$\displaystyle \sum_{cyclic}2((a-b)(b-c))^2=2(a^4+b^4+c^4)+8abc(a+b+c)+8(a^2b^2+b^2c^2+c^2a^2)=2(1-2(a^2b^2+b^2c^2+c^2a^2))+8abc+8(a^2b^2+b^2c^2+c^2a^2)=2+4(a^2b^2+b^2c^2+c^2a^2)+8abc=8abc+2+4(0-2abc(a+b+c))=2\Rightarrow \displaystyle \sum_{cyclic} (a-b)^4 =\left(\displaystyle \sum_{cyclic} (a-b)^2\right)^2-\displaystyle \sum_{cyclic} 2((a-b)(b-c))^2= 2^2-2 = 4-2=2$. Done.
