Open Problem Garden - Inequality of complex numbers - Comments

Yes, such exists, say (re: Inequality of complex numbers)

fedorpetrov — Wed, 05 Jan 2011 13:01:40 +0100

Yes, such exists, say works. Assume the contrary and consider the counterexample. Without loss of generality, , else multiple all 's to some so that this bacomes true, LHS is multiplied by , while RHS only by . So, we again get a counterexample. Denote , , . Since LHS does not exceed 2, we have (else RHS is too large). Hence for s.t. . Then we have , where , by some easy estimate. Finally, LHS equals and we just use estimate for small enough ( is small enough, since ).

A hint (re: Inequality of complex numbers)

Christian Blatter — Sun, 01 Aug 2010 14:15:48 +0200

I am the author of this hint and somehow mismanaged the posting. So here it is again:

Let with small and . Then and .

Now from it follows that This shows that your inequality has the "right order of magnitude" with .

A hint: \par Let with (re: Inequality of complex numbers)

Anonymous — Thu, 29 Jul 2010 12:22:45 +0200

A hint: \par Let with small and . Then and . \par Now from it follows that This shows that your inequality has ``the right order of magnitude'' with .

Requesting background information (re: Inequality of complex numbers)

Carolus — Wed, 21 Apr 2010 03:38:28 +0200

To feanor, the author of this conjecture:

What is the motivation for this conjecture ? The selected importance "medium" let me assume the verification or falsification of this conjecture would bring some benefit. If possible, describe that benefit ("practical" applications or consequences), please.

Inequality of complex numbers

feanor — Wed, 07 Apr 2010 13:06:06 +0200

Author(s):

Subject: Analysis

Conjecture There exists a real positive , such that for any and any where for and , the following holds: