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Let $n$ positive integer, $n\geqslant3$ and $a_1, a_2, ... , a_n, b_1, b_2, ..., b_n$ complex numbers such that $a_1\not=b_1$ and $|a_1|, |a_2|, .., |a_n|, |b_1|, |b_2|, ..., |b_n| \in (0, 1]$ . Show that $|a_1a_2...a_n-b_1b_2...b_n|=|a_1-b_1| + |a_2-b_2| +...+|a_n-b_n|$ if anf only if $a_k=b_k$ and $|a_k|=1$ ,where $k\geqslant2$.

Again, problem from G.M. 11/2016. I can't solve it. I tried the case $n=3$ but that didnt work as well. I think that we can show that $|a_1a_2...a_n-b_1b_2...b_n|\leqslant|a_1-b_1|$ and the problem would be solved, but i don't know if it is correct or how to demonstrate it.

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An answer for the direction "if $a_k=b_k, |a_k|=1, k>1$ then the equation holds":

Since $a_k=b_k, k>1$, let $X=\prod_{k>1}a_k$. From $|a_k|=1$ we get $|X|=1$. Further, $|a_k-b_k|=0, k>1$. Then

$$|a_1X-b_1X|=|a_1-b_1||X|=|a_1-b_1|$$

which proves this direction.

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  • $\begingroup$ Yes, this is right. The Other direction is harder $\endgroup$ – razvanelda Mar 17 '17 at 11:52
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This can be proved through repeated application of the triangle inequality.

Let the difference $D_k = a_k\cdots a_n-b_k\cdots b_n$, so

Note that $$ \begin{align} D_1 & = a_1a_2\cdots a_n-b_1b_2\cdots b_n \\ & = (a_1 - b_1)b_2\cdots b_n + a_1(a_2\cdots a_n -b_2\cdots b_n) \\ & = (a_1 - b_1)b_2\cdots b_n + a_1 D_2 \end{align} $$ which allows for a recursion, i.e. an equivalent formula can be constructed again for $D_2$ etc.

For the absolute values, we use the triangle inequality: $$ \begin{align} |D_1| & = | (a_1 - b_1)b_2\cdots b_n + a_1 D_2| \\ & \leq | a_1 - b_1 | |b_2\cdots b_n| + |a_1| | D_2| \end{align} $$

with equality if and only if the phase of $(a_1 - b_1)b_2\cdots b_n$ equals the phase of $a_1 D_2$. Repeated use results in

$$ D_1 = \sum_{k= 1}^n c_k ( a_k - b_k) $$

with $c_k = a_1 \cdots a_{k-1} \cdot b_{k+1} \cdots b_n$. Hence

$$ |D_1| \leq \sum_{k= 1}^n |c_k | \; | a_k - b_k| $$

Since all $|a_1|, |a_2|, .., |a_n|, |b_1|, |b_2|, ..., |b_n| \in (0, 1]$, we have all $|c_k | \leq 1$, and therefore

$$ |D_1| \leq \sum_{k= 1}^n | a_k - b_k| $$

with equality if and only if for all terms with $a_k - b_k \neq 0$, the $|c_k| = 1$ and the phases of the corresponding $c_k (a_k - b_k) $ are equal.

Since the term $a_1 - b_1 \neq 0$, $c_1 (a_1 - b_1)= (a_1 - b_1) b_2\cdots b_n$ sets the phase and $|c_1| =1 $ requires that for all $k \geq 2$, all $|b_k| =1 $.

Now suppose also $a_2 - b_2 \neq 0$. Then $c_2 = a_1 b_3\cdots b_n$ and $|c_2| =1 $ is only possible if $|a_1| =1 $. So if $|a_1| < 1 $, we have a contradiction. Hence for $|a_1| < 1 $, the OP's desired equation is only possible when $a_2 - b_2 = 0$. Continuing with the following terms, the same argument gives that for all $k \geq 2$, $a_k - b_k = 0$. By exchange of $a_k$ and $b_k$, the same argument holds for $|b_1| < 1 $. This proves the OP's claim for $|a_1| < 1 $ or $|b_1| < 1 $.

$\quad \quad \Box$

P.S.: For $|a_1| = |b_1| = 1 $, I am still lacking an argument for the proof.

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  • $\begingroup$ From the first inequality with $|c_k|$ you obtain that $|c_k|=1$ and therefor all $|a_k|$ and $|b_k|$ are equal with 1. $\endgroup$ – razvanelda Apr 9 '17 at 15:38

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