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Let $a,b,c\in R$,and such $ab+bc+ac=0,a+b+c\neq 0$

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

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    Expand the numerator using binomial formula and show that it equals $2(a+b+c)^4$. – Miz Jun 15 '15 at 09:37

2 Answers2

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

Macavity
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  • How find this indentity? –  Jun 15 '15 at 09:55
  • Any symmetric polynomial can be expressed in terms of elementary symmetric forms, and there is a straightforward algorithm. Google for Newton's identities or check http://math.stackexchange.com/questions/14051/symmetric-polynomials-and-the-newton-identities – Macavity Jun 15 '15 at 09:57
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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.

DeepSea
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