Prove $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$ if $\frac{1}{a+w}+\frac{1}{b+w}+\frac{1}{c+w}+\frac{1}{d+w}=\frac{2}{w}$

Question

If $\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}$, where $a,b,c,d\in\mathcal{R}$ and $\omega$ is a non-real cube root of unity, then prove that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+d}=2$

As it was asked as a multiple choice question(this expression was one option to pick) I really do not see an easy way to prove this. $$ \frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}\\ \frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}+\frac{1}{d+\omega^2}=\frac{2}{\omega^2}\\ \frac{1}{a\omega^2+\omega}+\frac{1}{b\omega^2+\omega}+\frac{1}{c\omega^2+\omega}+\frac{1}{d\omega^2+\omega}=2\\ \frac{1}{a\omega+1}+\frac{1}{b\omega+1}+\frac{1}{c\omega+1}+\frac{1}{d\omega+1}=2\\ $$

Is it no coincidence that replacing $\omega$ with other two cube root of unity satisfy the equation ?

Answer 1

I don't think its that different from @Luis solution, but it seems more convincing to me.

$$ \frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}+\frac{1}{d+\omega}=\frac{2}{\omega}\\ \implies \frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}+\frac{1}{d+\omega^2}=\frac{2}{\omega^2}\\ $$ $\omega,\omega^2$ are the solutions of the equation,

$$ \frac{1}{a+x}+\frac{1}{b+x}+\frac{1}{c+x}+\frac{1}{d+x}=\frac{2}{x}\\ \tfrac{(b+x)(c+x)(d+x)+(a+x)(c+x)(d+x)+(a+x)(b+x)(d+x)+(a+x)(b+x)(c+x)}{(a+x)(b+x)(c+x)(d+x)}=\tfrac{2}{x}\\ \frac{4x^4+2x^3\sum a+3x^2\sum ab+x\sum abc}{x^4+x^3\sum a+x^2\sum ab+x\sum abc+abcd}=2\\ \boxed{2x^4+x^3\sum a+0.x^2-x.\sum abc-2abcd=0} $$ $$ \alpha\beta+\beta\gamma+\gamma\eta+\eta\alpha+\alpha\gamma+\beta\eta=0\quad\&\quad\gamma=\omega,\;\eta=\omega^2\\ \alpha\beta+\omega\beta+\omega^3+\omega^2\alpha+\omega\alpha+\omega^2\beta=0\\ \alpha\beta-\alpha-\beta+1=0\implies \alpha(\beta-1)-(\beta-1)=(\alpha-1)(\beta-1)=0\\ \alpha=1\quad\&\quad\beta=1 $$

Answer 2

Note that $(1+a_1)(1+a_2)...(1+a_n)= 1 + SP + DP + TP + ... nP$, where:

• SP stands for "single products": $a_1 + a_2 + ... + a_n$,

• DP for "double products" $a_1a_2 + a_1a_3 + ... $,

• TP for "triple products" $a_1a_2a_3 + ...$ and so on until nP which is $a_1a_2... a_n$

And, for the same reasons, $(\omega +a_1)(\omega+a_2)...(\omega+a_n)= \omega^n + \omega^{n-1}SP + \omega^{n-2}DP + \omega^{n-3}TP + ... nP$.

Then, you just want to sum everything:

$$0 = \dfrac{1}{a+\omega} + \dfrac{1}{b+\omega} +\dfrac{1}{c+\omega} +\dfrac{1}{d+\omega} - \dfrac{2}{w} = \dfrac{m\omega - n}{(a+\omega)(b+\omega)(c+\omega)(d+\omega)\omega},$$ where $m= 2 - abc - abd - acd - bcd$ and $n = 2abcd - a-b-c-d$ are real numbers. Please note that the numerator was reduced by using $\omega^3=1$ and $\omega^4=\omega$ when needed, and the double products cancel due to the "-2" term.

If $m\neq 0$, then $\omega = \dfrac{n}{m} \in \mathbb{R}$ which contradicts the hypotesis. Thus, $m=0$, and it follows that $n=0$ too.

In particular, $$0 = m-n = 2 - abc - abd - acd - bcd - (2abcd - a-b-c-d)$$ Then, $$ 0 = \dfrac{2 - abc - abd - acd - bcd - (2abcd - a-b-c-d)}{(1+a)(1+b)(1+c)(1+d)} $$ $$= \dfrac{1}{1+a} + \dfrac{1}{1+b} + \dfrac{1}{1+c} + \dfrac{1}{1+d} - 2, $$ so we finish.

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PS. It would be nice to see proofs using other methods less rustic than this :)