180. Own proposed problem from CRUX (with complex numbers).
by Virgil Nicula, Dec 1, 2010, 10:27 AM
Quote:
Proposed problem (own). Let 
and 
, where 
. Prove that 
(CRUX, 3242/4/2007).
and 
, where 
. Prove that 
(CRUX, 3242/4/2007).Proof 1 (directly - own). Show easily that
.
for which
. Then 
.![$ \sum_{k = 0}^{n - 1}r^{2k}\ge n\cdot\sqrt [n]{\prod_{k = 0}^{n - 1}r^{2k}} = n\cdot\sqrt [n] {r^{n(n - 1)}} = n\cdot r^{n - 1}$](http://latex.artofproblemsolving.com/6/e/e/6eea9010e5196e159a6da1ed43feaf211f67a5b1.png)
. Thus, 
.Denote
for which 
. Thus, 
. In conclusion,
. For 
obtain : 
. Using the upperrelations
and 
obtain that 
.Remark. This problem points out a very interesting property of the structure of complex numbers :
Quote:
Denote the set 
of the complex numbers 
for which 
and for 
denote
the set
of the complex numbers 
for which 
, where 
. Then 
.
of the complex numbers 
for which and for 
denotethe set
of the complex numbers 
for which 
, where 
. Then 
.Proof 2. Denote
and 
. Suppose 
. Have 
. Thus, 
.Denote
. Then we have 
. We know that 
. By induction we can show that 
. Indeed, if 
, then
. Therefore 
. Thus, 
.Proof 3. Put
. Suppose contrary to all reason 
. Have 
. Hence 
with 
. From 
obtain that 
for any
. On the other hand 
. Thus 
forall 
, what is absurd. Extension. Let
, 
and 
. Then 
.Proof. At first I"ll show that
.
,
and aplying the inequality of means obtain :![$ r^{2n}-(ab)^n\stackrel{AM-GM}{\ge} (r^2-ab)\cdot n\cdot\sqrt[n]{r^{(n-1)n}\cdot (ab)^{\frac{(n-1)n}{2}}}\ \implies\color{white}{.}$](http://latex.artofproblemsolving.com/1/2/9/1292859ab070af6cfa2e16d4583feb9e30f86e6d.png)
.Denote
for which 
.Thus,
.In conclusion,
. Particularly, for obtain
. Using the upper relations si 
obtain :
Therefore,
.This post has been edited 36 times. Last edited by Virgil Nicula, Nov 26, 2015, 7:00 PM
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