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Let $a,b,c,d,e,f$ be real numbers,such $a+b+c+d+e+f=a^3+b^3+c^3+d^3+e^3+f^3=0$ show that $$\dfrac{a^7+b^7+c^7+d^7+e^7+f^7}{(a^5+b^5+c^5+d^5+e^5+f^5)(a^2+b^2+c^2+d^2+e^2+f^2)}=\dfrac{7}{10}$$ or

if we let $S_{i}=a^i+b^i+c^i+d^i+e^i+f^i$,and such $$S_{1}=S_{3}=0$$ then we $$\dfrac{S_{7}}{S_{5}S_{2}}=\dfrac{7}{10}$$I read about it in a book, and it felt amazing, and I tried for a long time, and I didn't prove it.

Martin Sleziak
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math110
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    Is there anything special about the fact that you have six numbers? Does the same (possibly with a different value than $\frac{7}{10}$) hold for just $a,b,c$, for instance? Or do you have to use $\frac{S_4}{S_2S_2}$ instead? Or does it just not work? These are the kinds of things you ought to explore if you want to try finding a solution on your own. You can't just stare at it and hope that the answer pops into your head. – Arthur May 04 '18 at 15:16
  • https://math.stackexchange.com/questions/851985/big-dfracx7y7z77-big2-big-dfracx5y5z55-big2-cdot-big – lab bhattacharjee May 04 '18 at 15:19
  • @labbhattacharjee,but my problem are six numbers,so maybe it's ugly than three numbers – math110 May 04 '18 at 15:21
  • This might be not the shortest way to solve this, but $$x^n+y^n=(x+y)(x^{n-1}+y^{n-1})-xy(x^{n-2}+y^{n-2})$$$$x^n+y^n+z^n=(x+y+z)(x^{n-1}+y^{n-1}+z^{n-1})-(xy+yz+zx)(x^{n-2}+y^{n-2}+z^{n-2})+xyz(x^{n-3}+y^{n-3}+z^{n-3})$$ and so on. Then you can use elementary symmetric polynomials to expand your power sum and evaluate each symmetric polynomial of 6 variables using the given details. – Bumblebee May 04 '18 at 15:45
  • This is better stated in the form of Karn Watcharasupat's answer, because it's possible that the sum of the fifth powers is $0$ also, as when $a=b=c=1,\space d=e=f=-1$ – saulspatz May 04 '18 at 18:00
  • @Arthur Lol I always fall for that trap... I have of course seen this advice before, but for some reason I always forget to use it... – Ovi Jan 20 '19 at 15:31

1 Answers1

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Note: The original, more step-by-step approach is below the line.

Notations used in this answer: \begin{align} S_r&=\sum_{1\le i \le 6}{x_i^r}\\ e_r&=\displaystyle{\sum_{{{1\le \underbrace{i\le j \le k \le \ldots}_{\displaystyle{{r \text{ indices}}}}} \le 6}}}\underbrace{x_ix_jx_k\cdots x_{\dots}}_{r \text{ factors}}\\ e_1&=x_1+\dots+x_6\\ e_2&=x_1x_2+x_1x_3+\dots+x_2x_3+\dots+x_5x_6\\ e_6&=x_1x_2\cdots x_6\\ \left(e_r\right)_{r>6}&=0 \end{align} Identities: refer to Newton's identities

\end{align} The 'one-shot' proof: \begin{align} & \frac{{{S}_{7}}}{{{S}_{2}}{{S}_{5}}}=\frac{\overbrace{{{e}_{1}}}^{={{S}_{1}}=0}{{S}_{6}}-{{e}_{2}}{{S}_{5}}+{{e}_{3}}{{S}_{4}}-{{e}_{4}}\overbrace{{{S}_{3}}}^{=0}+{{e}_{5}}{{S}_{2}}-{{e}_{6}}\overbrace{{{S}_{1}}}^{=0}+7\overbrace{{{e}_{7}}}^{=0}}{\left( \underbrace{{{e}_{1}}}_{=0}\underbrace{{{S}_{1}}}_{=0}-2{{e}_{2}} \right)\left( \underbrace{{{e}_{1}}}_{=0}{{S}_{4}}-{{e}_{2}}\underbrace{{{S}_{3}}}_{=0}+{{e}_{3}}{{S}_{2}}-{{e}_{4}}\underbrace{{{S}_{1}}}_{=0}+5{{e}_{5}} \right)} \\ & =\frac{-{{e}_{2}}{{S}_{5}}+{{e}_{3}}{{S}_{4}}+{{e}_{5}}{{S}_{2}}}{\left( -2{{e}_{2}} \right)\left( {{e}_{3}}{{S}_{2}}+5{{e}_{5}} \right)} \\ & =\frac{-{{e}_{2}}{{S}_{5}}+0\cdot {{S}_{4}}+{{e}_{5}}{{S}_{2}}}{\left( -2{{e}_{2}} \right)\left( 0\cdot {{S}_{2}}+5{{e}_{5}} \right)}\qquad\qquad\qquad\because 3{{e}_{2}}={{e}_{2}}\underbrace{{{S}_{1}}}_{=0}-\underbrace{{{e}_{1}}}_{=0}{{S}_{2}}+\underbrace{{{S}_{3}}}_{=0}=0 \\ & =\frac{-{{e}_{2}}{{S}_{5}}+{{e}_{5}}{{S}_{2}}}{\left( -2{{e}_{2}} \right)\left( 5{{e}_{5}} \right)} \\ & =\frac{-\frac{1}{2}\left( {{e}_{1}}{{S}_{1}}-{{S}_{2}} \right){{S}_{5}}+{{e}_{5}}{{S}_{2}}}{\left( -2\left( {{e}_{1}}{{S}_{1}}-{{S}_{2}} \right) \right)\left( 5{{e}_{5}} \right)}\qquad\qquad\qquad\because 2{{e}_{2}}=\underbrace{{{e}_{1}}{{S}_{1}}}_{=0}-{{S}_{2}}\Rightarrow {{S}_{2}}=-\frac{1}{2}{{e}_{2}} \\ & =\frac{\frac{1}{2}{{S}_{2}}{{S}_{5}}+{{e}_{5}}{{S}_{2}}}{5{{e}_{5}}{{S}_{2}}} \\ & =\frac{{{S}_{5}}+2{{e}_{5}}}{10{{e}_{5}}}\qquad\qquad\because 5{{e}_{5}}={{e}_{4}}\underbrace{{{S}_{1}}}_{=0}-\underbrace{{{e}_{3}}}_{=0}{{S}_{2}}+{{e}_{2}}\underbrace{{{S}_{3}}}_{=0}-\underbrace{{{e}_{1}}}_{=0}{{S}_{4}}+{{S}_{5}}\Rightarrow {{S}_{5}}=5{{e}_{5}} \\ & =\frac{5{{e}_{5}}+2{{e}_{5}}}{10{{e}_{5}}} \\ & =\frac{7}{10} \end{align}

Consider \begin{align} \sum_{1\le i\le6}{x^2_i} &=\left(\sum_{1\le i\le6}{x_i}\right)^2 -2(\underbrace{x_1x_2+x_2x_3+\ldots+x_5x_6}_\text{all permuations of pairs})\\ &=0-2(\underbrace{x_1x_2+x_2x_3+\ldots+x_5x_6}_\text{all permuations of pairs})\\ \frac12S_2&=-\sum_{1\le i\le j \le6}{x_ix_j}=-e_2 \end{align}

Using Newton's identities (which you can sort of observe yourself also) \begin{align} S_5 &=\underbrace{\left(\sum_{1\le i\le6}{x_i}\right)}_{=0}S_4 -\left(\overbrace{\sum_{1\le i\le j\le6}{x_ix_j}}^{e_2}\right)\underbrace{S_3}_{=0} +\left(\overbrace{\sum_{1\le i\le j\le k \le6}{x_ix_jx_k}}^{e_3}\right)S_2 -\left(\overbrace{\sum_{1\le i\le j\le k \le k \le 6}{x_ix_jx_kx_l}}^{e_4}\right)\underbrace{S_1}_{=0}+e_5\\ &=0-0+e_3S_2-0\\ &=e_3S_2+5e_5 \end{align} and \begin{align} S_7 &=\underbrace{e_1}_{=S_1=0}S_6 -e_2S_5 +e_3S_4 -e_4S_3 +e_5S_2-e_6S_1+7\underbrace{e_7}_{=0}\\ &=-e_2S_5 +e_3S_4 +e_5S_2\\ \end{align}

Then we settle some unfinished business with our dear $e_3$, \begin{align} S_3&=e_1S_2-e_2S_1+3e_3\\ 0&=0-0+3e_3\\ e_3&=0 \end{align} so we have \begin{align} S_5&=e_3S_2+5e_5\\ S_5&=0+5e_5\\ e_5&=\frac{1}{5}S_5 \end{align} and \begin{align} S_7 &=-e_2S_5+e_3S_4+e_5S_2\\ &=\frac12S_2S_5+0+\frac15S_2S_5\\ &=\frac7{10}S_2S_5 \end{align}

Karn Watcharasupat
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