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Suppose $a_1,\ldots,a_k$ and $b_1,\ldots,b_k$ are complex numbers bounded in absolute value by $1$. Is it true that $$ \left| \prod_{i=1}^k a_i - \prod_{i=1}^k b_i\right|\leq \sum_{i=1}^k |a_i-b_i|? $$

justhereforonequestion
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Consider the telescoping sum $$a_1\cdots a_k-b_1\cdots b_k=\sum_{i=1}^k a_1\cdots a_{i-1}(a_i-b_i)b_{i+1}\cdots b_k. $$

  • What you wrote doesn't make sense. In the RHS, you have a sum over $i$ from 1 and you write $a_{i-1}$ which is not even defined – Apprentice Oct 17 '22 at 06:47
  • @Apprentice Your bounties are placed on two low-quality questions, but it's really better to start a new question following the guide in How to ask a good question. – Ѕᴀᴀᴅ Oct 17 '22 at 07:13
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    @Apprentice: An empty product is usually understood as the value $1$, so that the term for $i=1$ on the RHS is just $(b_1-a_1)b_2 \cdots b_n$. – Btw, this has been asked and answered more than once on this site, I have added some links below the question. – Martin R Oct 17 '22 at 08:14