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If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove

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

What is a good way to do this?

This question came from answering this slightly harder question. Those answers were somewhat hard to understand for me. To get something easier to digest I made a very similar but easier (lower exponent) question.

Martin Sleziak
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3 Answers3

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Let $e_1 = a+b+c, e_2 = ab+bc+ac, e_3 = abc$.

Since $A = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$ is symmetric and homogeneous of degree $4$, it is a linear combination of $e_1^4, e_1^2e_2, e_1e_3, e_2^2$, and when $e_1=0$ the only nonzero one is possibly $e_2^2$ :

There is a coefficient $k$ such that if $e_1=0$ then $A = ke_2^2$.

To check that $k$ is $0$, we only need to check that the identity is true for one case where $e_1=0$ and $e_2 \neq 0$.

Picking for example $a=0,b=1,c=-1$, you get $4=4$ so things work out.

mercio
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$$(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)=$$ $$=\sum_{cyc}(2a^2b^2-a^4)=(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0.$$ Because $$\sum_{cyc}(2a^2b^2-a^4)=4a^2b^2-(a^4+b^4+c^4-2a^2c^2-2b^2c^2+2a^2b^2)=$$ $$=(2ab)^2-(a^2+b^2-c^2)^2=(2ab-a^2-b^2+c^2)(2ab+a^2+b^2-c^2)=$$ $$=(c^2-(a-b)^2)((a+b)^2-c^2)=(c-a+b)(c+a-b)(a+b-c)(a+b+c).$$

Michael Rozenberg
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  • @Peter Sheldrick I knew the Heron formula for an area $S$ of the triangle. If $p=\frac{a+b+c}{2}$ then $S=\sqrt{p(p-a)(p-b)(p-c)}=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}=\frac{1}{4}\sqrt{\sum\limits_{cyc}(2a^2b^2-a^4)}$, but there is another way. I'll post it. – Michael Rozenberg Oct 13 '17 at 01:43
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Put $c=-a-b$. Then $$ (a^2+b^2+c^2)^2=4(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4)=2(a^4+b^4+c^4). $$ A very similar calculation was given here.

Dietrich Burde
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