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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! ~

  • 0
    @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 expressenion2012-07-29

4 Answers 4

11

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 ${\overline q} = (abc){a + b\over ab}{a + c\over ac}{b + c\over bc} = (a+b)(b+c)(a+c)/(abc) = q.$

3

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} $

2

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}

1

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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    Sorry, that was a typo2012-07-26