391. CRUX Mathematicorum.
by Virgil Nicula, Dec 21, 2013, 9:58 AM
Some own special inequalities from "CRUX Mathematicorum with Mathematical Mayhem - CANADA".
Proposed problem 1. Let
and 
, where 
. Prove that 
(CRUX, 3242/4/2007).
Proof 1 (directly - own). Show easily that
.
Denote
for which 
. Thus, 
. In conclusion,
. For 
obtain : 
. Using the upper
relations
and 
obtain that 
.
Remark. This problem points out a very interesting property of the structure of complex numbers
Let the set 
of for which
. For 
denote the set 
of the complex numbers 
for which 
, where 
. Then 
.
Proof 2 (induction - leva1980). 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 (induction - mathlinks). Put
. Suppose contrary to all reason 
. Have 
. Hence 
with 
. From 
obtain that 
for any . On the other hand 
. Thus 
forall 
, what is absurd.
Extension (Mateescu Constantin). Let
, 
and 
. Then 
.
Proof. At first I"ll show that
Denote
for which 
. Thus, 
. In conclusion, 
.
Particularly, for obtain 
. Using the upper relations si 
obtain :
. Therefore, 
.
Proposed problem 2.
(CRUX, 3241/4/2007).
Proof. I"ll suppose w.l.o.g. that
. Thus, the problem reduces to prove that 
. If 
, then 
and 
.
If
, then 
and 
, i.e. 
. Thus, are remaining only two cases.
If 
or 
, then 
and 
. But 
, which is true. Thus, 
, i.e. 
. We have equality iff 
and 
.
If 
, then 
, i.e. 
, 
and 
, i.e. 
. But 
,
i.e.
. Is sufficient to prove 
, i.e. ![$f(a)=2a^6-9a^4+(3a-1)^2\le 0\ ,\ (\forall )\ a\in \left(0 , \sqrt 3\right]$](//latex.artofproblemsolving.com/b/0/1/b014b31c26f14f80ce2c55dbb0b89e7d26d18987.png)
.
![$2.1\blacktriangleright\ \ :\ \ a\in \left(0, \frac 13 \right]$](//latex.artofproblemsolving.com/c/2/c/c2cf5654b542505146d49e5dda10fbc33facb3a3.png)
the relation 
is truly, i.e. 
.
![$2.2\blacktriangleright\ \ :\ \ a\in \left(\frac 13 ,1\right]$](//latex.artofproblemsolving.com/a/d/2/ad22618ff1cccd8c3e9a88da2986ad557178d9ed.png)
.
![$2.3\blacktriangleright\ \ :\ \ a\in \left(1, \sqrt{\frac 32}\right]$](//latex.artofproblemsolving.com/1/8/2/182d937e9d7d8416c52cb0c1e7443c7bbcd9e900.png)
.
![$2.4\blacktriangleright\ \ :\ \ a\in \left(\sqrt {\frac 32} ,\sqrt 3 \right]$](//latex.artofproblemsolving.com/5/9/1/591b47b335ca96544c7f92c4186e5b3573b241fe.png)
.
Application. Prove that
. Indeed, can suppose w.l.o.g. . Thus,
![$\underline{2(a+b+c)}=\left[2\cdot b+2\cdot c+(-1)\cdot a\right]+3a\stackrel{(C.B.S)}{\le}$](//latex.artofproblemsolving.com/7/6/c/76c7d174f694c6695558554d386d79a649687d0c.png)
![$\sqrt {\left[2^2+2^2+(-1)^2\right]\cdot\left[b^2+c^2+a^2\right]}+3a=9+3a\stackrel{(*)}{\le}\underline{10+abc}$](//latex.artofproblemsolving.com/8/8/0/8800ac7421f8f99e5334983aa8cba26ca617f68a.png)
.
Proposed problem 3. Let
(CRUX, 3260/4/2007).
Proof. I"ll use the well-known Bernoulli's inequality :
.
The first case.
. Thus, 
![$\left[ 1+\left( \frac 1a -1\right)\right]^b\le $](//latex.artofproblemsolving.com/e/3/f/e3f6a0b06c43bae04bd7dc6bb46ec3a1ce792bc3.png)
. Analogously, 
.
Thus,
. Since 
, 
,
what is truly. Thus,
. Therefore, 
.
The second case.
. Thus, 
, i.e. 
.
Analogously,
. Thus, 
.
Proposed Problem 4. Let
be a complex number for which exists 
so that 
and 
. Prove that 
.
Proof. Observe that
. Suppose w.l.o.g. 
. Denote 
. Thus, 
and 
. Suppose 
that 
. Thus, 
because 
. From the relation 
and the well-known inequality 
obtain
, i.e. 
, what is absurd . In concluzion, . Here is an equivalent enunciation :
An easy extension. Let
be two complex numbers for which exists so that and 
. Prove that 
.
Proposed problem 1. Let
and 
, where 
. Prove that 
(CRUX, 3242/4/2007).Proof 1 (directly - own). Show easily that
.
for which
. Then 
.![$ \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}$](http://latex.artofproblemsolving.com/6/e/e/6eea9010e5196e159a6da1ed43feaf211f67a5b1.png)
. Thus, 
Denote
for which 
. Thus, 
. In conclusion,
. For 
obtain : 
. Using the upperrelations
and 
obtain that 
.Remark. This problem points out a very interesting property of the structure of complex numbers
Let the set 
of for which
. For 
denote the set 
of the complex numbers 
for which 
, where 
. Then 
.Proof 2 (induction - leva1980). Denote
and 
. Suppose 
. Have 
. Thus, 
. Denote 
.Then we have
. We know that 
. By induction we can show that 
. Indeed, if 
, then 
.Therefore
. Thus, 
.Proof 3 (induction - mathlinks). Put
. Suppose contrary to all reason 
. Have 
. Hence 
with 
. From 
obtain that 
for any
. On the other hand 
. Thus 
forall 
, what is absurd. Extension (Mateescu Constantin). Let
, 
and 
. Then 
.Proof. At first I"ll show that
.
, 
![$ 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}{.}$](http://latex.artofproblemsolving.com/1/2/9/1292859ab070af6cfa2e16d4583feb9e30f86e6d.png)
.Denote
for which 
. Thus, 
. In conclusion, 
.Particularly, for
obtain 
. Using the upper relations si 
obtain :
. Therefore, 
.Proposed problem 2.
(CRUX, 3241/4/2007).Proof. I"ll suppose w.l.o.g. that
. Thus, the problem reduces to prove that 
. If 
, then 
and 
.If
, then 
and 
, i.e. 
. Thus, are remaining only two cases.
If 
or 
, then 
and 
. But 
, which is true. Thus, 
, i.e. 
. We have equality iff 
and 
.
If 
, then 
, i.e. 
, 
and 
, i.e. 
. But 
,i.e.
. Is sufficient to prove 
, i.e. ![$f(a)=2a^6-9a^4+(3a-1)^2\le 0\ ,\ (\forall )\ a\in \left(0 , \sqrt 3\right]$](http://latex.artofproblemsolving.com/b/0/1/b014b31c26f14f80ce2c55dbb0b89e7d26d18987.png)
.![$2.1\blacktriangleright\ \ :\ \ a\in \left(0, \frac 13 \right]$](http://latex.artofproblemsolving.com/c/2/c/c2cf5654b542505146d49e5dda10fbc33facb3a3.png)
the relation 
is truly, i.e. 
.![$2.2\blacktriangleright\ \ :\ \ a\in \left(\frac 13 ,1\right]$](http://latex.artofproblemsolving.com/a/d/2/ad22618ff1cccd8c3e9a88da2986ad557178d9ed.png)
.![$2.3\blacktriangleright\ \ :\ \ a\in \left(1, \sqrt{\frac 32}\right]$](http://latex.artofproblemsolving.com/1/8/2/182d937e9d7d8416c52cb0c1e7443c7bbcd9e900.png)
.![$2.4\blacktriangleright\ \ :\ \ a\in \left(\sqrt {\frac 32} ,\sqrt 3 \right]$](http://latex.artofproblemsolving.com/5/9/1/591b47b335ca96544c7f92c4186e5b3573b241fe.png)
.Application. Prove that
. Indeed, can suppose w.l.o.g. . Thus,![$\underline{2(a+b+c)}=\left[2\cdot b+2\cdot c+(-1)\cdot a\right]+3a\stackrel{(C.B.S)}{\le}$](http://latex.artofproblemsolving.com/7/6/c/76c7d174f694c6695558554d386d79a649687d0c.png)
![$\sqrt {\left[2^2+2^2+(-1)^2\right]\cdot\left[b^2+c^2+a^2\right]}+3a=9+3a\stackrel{(*)}{\le}\underline{10+abc}$](http://latex.artofproblemsolving.com/8/8/0/8800ac7421f8f99e5334983aa8cba26ca617f68a.png)
.Proposed problem 3. Let
(CRUX, 3260/4/2007).Proof. I"ll use the well-known Bernoulli's inequality :
.The first case.
. Thus, 
![$\left[ 1+\left( \frac 1a -1\right)\right]^b\le $](http://latex.artofproblemsolving.com/e/3/f/e3f6a0b06c43bae04bd7dc6bb46ec3a1ce792bc3.png)
. Analogously, 
.Thus,
. Since 
, 
,what is truly. Thus,
. Therefore, 
.The second case.
. Thus, 
, i.e. 
.Analogously,
. Thus, 
.Proposed Problem 4. Let
be a complex number for which exists 
so that 
and 
. Prove that 
.Proof. Observe that
. Suppose w.l.o.g. 
. Denote 
. Thus, 
and 
. Suppose 
that 
. Thus, 
because 
. From the relation 
and the well-known inequality 
obtain
, i.e. 
, what is absurd . In concluzion, . Here is an equivalent enunciation :An easy extension. Let
be two complex numbers for which exists so that and 
. Prove that 
.This post has been edited 19 times. Last edited by Virgil Nicula, Jun 14, 2015, 2:33 AM
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