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
$$\dfrac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$$

Hint:  Express the numerator in terms of elementary symmetric polynomials, as the denominator and constraint already are.  You get
$$(a-b)^4+(b-c)^4+(c-a)^4 \\= 2(a+b+c)^4 - 12(ab + bc + ca) (a+b+c)^2 + 18(ab+bc+ca)^2$$
