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

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 ?

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.

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

• It's smooth enough. I think this is the way to solve this problem.+1 – Michael Rozenberg Feb 21 '19 at 1:45
• @MichaelRozenberg I was wondering why replacing $\omega$ with other two cube roots of unity also satisfies the equation ?. Got to be some property to identify the solution ? – Sooraj S Feb 21 '19 at 8:32
• @ss1729 I still see a proof of your statements by the Luis's way only. – Michael Rozenberg Feb 21 '19 at 8:37

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,

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