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If $a+b+c+d=0=a^7+b^7+c^7+d^7$, Then prove $a(a+b)(a+c)(a+d)=0$ and all are real numbers.

I am confused and don't even know which to tag for help.

DeepSea
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thebeatle
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1 Answers1

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Yeah, I saw the duplicate question. But here is another useful application of the AM-GM inequality(didn't think of it, did you) Tricky factorization: $$0=-(a^7+b^7+c^7+d^7)$$$$=7(b+c)(c+d)(d+a)E$$ where $E=(b^2+c^2+d^2+bc+cd+da)^2+bcd(b+c+d)$. However, $$4E=((b+c)^2+(c+d)^2+(b+d)^2)^2)-4abcd$$$$=((b+a)^2+(c+a)^2+(a+d)^2)^2)-4abcd$$$$=(3a^2+2a(b+c+d)+b^2+c^2+d^2)^2-4abcd$$ The last expression reduces to $$(a^2+b^2+c^2+d^2)^2-4abcd$$ However we see that $$(a^2+b^2+c^2+d^2)^2-4abcd \geq 12|abcd| \geq 0$$ from the AM-GM inequality. We conclude that $4E \geq 0$ with equality if and only if $|a|=|b|=|c|=|d|=0$. Otherwise one of $b+c,b+d,c+d$ must be $0$. Hence, the result follows.