# A proof about complex number

If $a, b, c\in \mathbb{C}$, and if $\left \| a \right \|=\left \| b \right \|=\left \| c \right \|=1$, prove $(a+b)(b+c)(c+a)/(abc)\in \mathbb{R}$.

I have thought this Q for a long time, but I can only get something long and troublesome but not the answer. Can anyone help me please? THANK YOU! ~

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show it is invariant under complex conjugation. –  i. m. soloveichik Jul 26 '12 at 13:35
To i. m. soloveichik , can you show me how? –  ᴊ ᴀ s ᴏ ɴ Jul 26 '12 at 13:38
If $a=cosα+isinα=exp(iα)$ and so on, then $a+b=2cos\frac{α-β}{2} exp(\frac{i(α+β)}{2})$ . Now $abc= exp(i(α+β+γ))$ –  lab bhattacharjee Jul 26 '12 at 13:41
Draw a picture, and figure out the angle (i.e. argument) of $a+b,~b+c$ and $c+a$, nevermind their length. Figure out the argument of $abc$, and you're done. –  Olivier Bégassat Jul 26 '12 at 13:42
@jasoncube--since the absolute values are 1 then $\bar{a}=1/a$, etc. so the conjugate of your expression is $(1/a+1/b)(1/b+1/c)(1/c+1/a)abc$ which simplifies to the original expressenion –  i. m. soloveichik Jul 29 '12 at 13:56

Let us denote the quantity by $q$. Since $a$, $b$ and $c$ are all of unit modulus, $${\overline q} = \overline{(a+b)(a+c)(b+c)/(abc)} = (1/a + 1/b)(1/a + 1/c)(1/b + 1/c)(abc)$$ now add the fractions and get $${\overline q} = (abc){a + b\over ab}{a + c\over ac}{b + c\over bc} = (a+b)(b+c)(a+c)/(abc) = q.$$

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Since $|a|=|b|=1$, we have that $$\left|\frac{a}{b}\right|=1 \quad \Rightarrow \quad \overline{\dfrac{a}{b}} =\left.\left|\dfrac{a}{b}\right|^2 \middle/\dfrac{a}{b}\right. =\frac{b}{a}$$ Therefore, \begin{align} \frac{(a+b)^2}{ab} &=\frac{a}{b}+2+\frac{b}{a}\\ &=2+2\,\mathrm{Re}\left(\frac{a}{b}\right)\\ &\ge0 \end{align} This means that $$\left(\frac{(a+b)(b+c)(c+a)}{abc}\right)^2=\frac{(a+b)^2}{ab}\frac{(b+c)^2}{bc}\frac{(c+a)^2}{ca}\ge0$$ which leads immediately to $$\frac{(a+b)(b+c)(c+a)}{abc}\in\mathbb{R}$$

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Here's a nice geometric argument (with some holes and assumptions left to fill up):

\begin{align} \arg \frac {(a+b)(b+c)(a+c)}{abc} &= \arg (a+b)+\arg(b+c)+\arg(a+c) - \arg (a) -\arg( b) - \arg( c) \\ &= \frac{\arg (a)+arg(b)} 2 + \frac{\arg (b)+arg(c)} 2 + \frac{\arg (a)+arg(c)} 2 \\ & \qquad\qquad - \arg (a) -\arg( b) - \arg( c) \\&= 0 \end{align}

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Hint 1. Since $\Vert a\Vert=\Vert b\Vert=\Vert c\Vert=1$, then $$a=e^{i\alpha}\qquad b=e^{i\beta}\qquad c=e^{i\gamma}\tag{1}$$ Hint 2 A complex number $z\in\mathbb{C}$ is real iff $$z+\overline{z}=2z\tag{2}$$

Substitute $(1)$ into $(2)$ and check that it is correct.

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a = exp(iα) = exp(iβ) = exp(iγ) ??? –  ᴊ ᴀ s ᴏ ɴ Jul 26 '12 at 13:45
Sorry, that was a typo –  Norbert Jul 26 '12 at 13:54