Community

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
Login Register Memberlist Search AoPS Blogs Contests Galleries Forum Index
The time now is Mon Dec 07, 2009 3:22 pm
All times are UTC - 8
 		View posts since last visit
View unanswered posts
Home » Mathematics » High School Pre-Olympiad (Ages 14+)
 
Inequalities Marathon
Moderators: Pre-Olympiad Moderators
Post new topic   Reply to topic View previous topicView next topic
382 Posts • Page 3 of 20 • Previous 1, 2, 3, 4, 5, ..., 18, 19, 20 Next
Author Message
geniusbliss
Riemann Hypothesis
Riemann Hypothesis


Offline
Joined: 09 Feb 2009
Posts: 270
Location: chennai,india
India
To rate posts you must be logged in
#41
hasan4444 wrote:
Problem 20. Let a, b, c be the lengths of the sides of a triangle. Prove that:
a^2b(a - b) + b^2c(b - c) + c^2a(c - a) \ge 0

solution to Problem 20

we know from triangle inequality that b \ge (a - c) and c\ge(b - a) and a\ge(c - b)
therefore,
a^2b(a - b) + b^2c(b - c) + c^2a(c - a)\ge a^2(a - c)(a - b) + b^2(b - a)(b - c) + c^2(c - b)(c - a)\ge0
and the last one is schur's inequality for r = 2 so proved with the equality holding when a = b = c or for and equilateral triangle Smile

P.S. this is IMO 1983
_________________
Quis custodiet ipsos custodes
Mathematical Dreams Mr. Green

PostPosted: Sat Sep 12, 2009 8:35 am  Back to top 
  ProfilePMBlog
enndb0x
Yang-Mills Theory
Yang-Mills Theory


Offline
Joined: 20 Jan 2009
Posts: 519
Location: Kosovo ,Norway
AlbaniaEuropean Union
To rate posts you must be logged in
#42
alex2008 wrote:


Problem 21.Let x,y,z \in \left[\frac {1}{3} , \frac {2} {3}\right] . Show that :
1 \ge \sqrt [3] {xyz} + \frac {2} {3(x + y + z)}



solution to problem 21


By Am-Gm x + y + z \geq 3\sqrt [3]{xyz} ,then

\frac {2}{3(x + y + z)} + \sqrt [3]{xyz} \leq \frac {2}{9} \sqrt [3]{xyz} } + \sqrt [3]{xyz}

Since x,y,z \in \left[\frac {1}{3} ,\frac {2}{3} \right] ,then \frac {1}{3} \leq \sqrt [3]{xyz} \leq \frac {2}{3}

Let a = \sqrt [3]{xyz} ,then a + \frac {2}{9a} \leq 1 \iff 9a^2 - 9a + 2 \leq 0 \iff 9\left(a - \frac {1}{3} \right)\left(a - \frac {2}{3} \right) \leq 0 ,we're done since \frac {1}{3} \leq a \leq \frac {2}{3}


@Fantasy Lover : Please explain your solution

Problem 22 .Let a,b,c be side lengths of a triangle,and \beta is the angle between a and c.Prove that
\frac {b^2 + c^2}{a^2} > \frac {2\sqrt 3 c\sin \beta - a }{b + c}
Last edited by enndb0x on Sat Sep 12, 2009 5:36 pm; edited 1 time in total 
PostPosted: Sat Sep 12, 2009 10:42 am  Back to top 
  ProfilePMMSN
FantasyLover
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 26 Mar 2008
Posts: 1901
Location: AAST
To rate posts you must be logged in
#43
alex2008 wrote:
Fantasylover , how does u\le 3 imply
u^7\cdot\frac {9 - u^2}{2} + \frac {u^9}{3^2}\le 3^7

Wow, somehow I thought u\le 3 was enough to prove that above inequality...

Meh, differentiating, u achieves its maximum when \frac{7u^6(9-u^2)}{2}=0.

Since a, b, c are positive, u cannot be 0, and the only possible value for u is 3.

Since u\le 3, the above inequality is true.
_________________
AAST 2013

PostPosted: Sat Sep 12, 2009 10:55 am  Back to top 
  ProfilePMBlog
zserf
Poincare Conjecture
Poincare Conjecture

Offline
Joined: 10 Feb 2007
Posts: 177
To rate posts you must be logged in
#44
Dimitris X wrote:


PROBLEM 10
Let a,b,c,d be REAL numbers such that a^2 + b^2 + c^2 + d^2 = 4.
Prove that a^3 + b^3 + c^3 + d^3 \le 8


solution to 10


2 (a^2 + b^2 + c^2 + d^2) = 8

a^3 + b^3 + c^3 + d^3 \le 2 (a^2 + b^2 + c^2 + d^2)

a^3=(a^2)a and so on with the other terms. If no cubed term is more than twice as much as it's respective squared term, than the sum of all the cubed terms can't be more then twice as much as the sum of all the squared terms. Because any squared term is positive, the largest possible value for a^2 is 4. and the largest possible value of a is a=2. Therefore, the sum of the cubed terms can't be larger then twice as much as the sum of the squared terms, and must be less than or equal to 8.

PostPosted: Sun Sep 13, 2009 1:07 am  Back to top 
  ProfilePM
geniusbliss
Riemann Hypothesis
Riemann Hypothesis


Offline
Joined: 09 Feb 2009
Posts: 270
Location: chennai,india
India
To rate posts you must be logged in
#45
zserf wrote:
Dimitris X wrote:


PROBLEM 10
Let a,b,c,d be REAL numbers such that a^2 + b^2 + c^2 + d^2 = 4.
Prove that a^3 + b^3 + c^3 + d^3 \le 8


solution to 10


2 (a^2 + b^2 + c^2 + d^2) = 8

a^3 + b^3 + c^3 + d^3 \le 2 (a^2 + b^2 + c^2 + d^2)

a^3 = (a^2)a and so on with the other terms. If no cubed term is more than twice as much as it's respective squared term, than the sum of all the cubed terms can't be more then twice as much as the sum of all the squared terms. Because any squared term is positive, the largest possible value for a^2 is 4. and the largest possible value of a is a = 2. Therefore, the sum of the cubed terms can't be larger then twice as much as the sum of the squared terms, and must be less than or equal to 8.


that is essentially what i posted as solution,isnt it?
_________________
Quis custodiet ipsos custodes
Mathematical Dreams Mr. Green

PostPosted: Sun Sep 13, 2009 1:46 am  Back to top 
  ProfilePMBlog
hasan4444
Riemann Hypothesis
Riemann Hypothesis


Offline
Joined: 28 Nov 2008
Posts: 486
Saudi Arabia
To rate posts you must be logged in
#46
Problem 23
Dinu Serbanescu
Solution to Problem 22

According to the Weitzenbock's inequality we have
a^{2} + b^{2} + c^{2}\geq 4\sqrt3 S and S = \frac {ac\sin\beta}{2}
Then
b^{2} + c^{2}\geq 2\sqrt 3 ac\sin\beta - a^{2} ,dividing by a^2 , we have
\frac {b^{2} + c^{2}}{a^{2}}\geq\frac {2\sqrt 3 c\sin\beta - a}{a}
Since a < b + c \implies \frac {b^{2} + c^{2}}{a^{2}}\geq\frac {2\sqrt 3 c\sin\beta - a}{a} > \frac {2\sqrt 3 c\sin\beta - a}...

Important Notes

I'm planning to make a big PDF with all the questions here and solutions so please if you know a source of a problem you gave PM it to me to make a reference in the PDF and if you make any own problems please provide me with its number.
Thanks for your attention

Problem 23. If a,b,c \in (0,1) Prove that:
\sqrt {abc} + \sqrt {(1 - a)(1 - b)(1 - c)} < 1
_________________
New "Inequalities Marathon" join it now Welcome Police

PostPosted: Sun Sep 13, 2009 8:07 am  Back to top 
  ProfilePMWWWMSN
FantasyLover
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 26 Mar 2008
Posts: 1901
Location: AAST
To rate posts you must be logged in
#47
hasan4444 wrote:
Problem 23. If a,b,c \in (0,1) Prove that:
\sqrt {abc} + \sqrt {(1 - a)(1 - b)(1 - c)} < 1


Solution to Problem 23
Since a, b, c\in (0,1), let us have a=\sin^2 A, b=\sin^2 B, c=\sin^2C where A, B, C\in (0,\frac{\pi}{2}).

Then, we are to prove that \sin A\sin B\sin C+\cos A\cos B\cos C<1.

Now noting that \sin C, \cos C<1, we have \sin A\sin B\sin C+\cos A\cos B\cos C< \sin A\sin B+\cos A\cos B=\cos(A-B)\le 1. \blacksquare


Problem 24.

For all positive real numbers a, b, c, prove the following:
\frac{1}{\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}}-\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge \frac{1}{3}
_________________
AAST 2013

PostPosted: Sun Sep 13, 2009 8:36 am  Back to top 
  ProfilePMBlog
alex2008
Yang-Mills Theory
Yang-Mills Theory


Offline
Joined: 26 Oct 2008
Posts: 749
Location: Tulcea , Romania
Romania
To rate posts you must be logged in
#48
hasan4444 wrote:
Problem 23. If a,b,c \in (0,1) Prove that:
\sqrt {abc} + \sqrt {(1 - a)(1 - b)(1 - c)} < 1


Another Solution to problem 23


Cauchy-Schwartz and AM-GM works fine :

\sqrt {abc} + \sqrt {\left( 1 - a\right) \left( 1 - b\right) \left( 1 - c\right) } = \sqrt {a}\sqrt {bc} + \sqrt {1 - a}\sqrt... \sqrt {a + \left( 1 - a\right) }\sqrt {bc + \left( 1 - b\right) \left( 1 - c\right) } = \sqrt {bc + \left( 1 - b\right) \left...
FantasyLover wrote:


Problem 24.

For all positive real numbers a, b, c, prove the following:
\frac {1}{\frac {1}{a + 1} + \frac {1}{b + 1} + \frac {1}{c + 1}} - \frac {1}{\frac {1}{a} + \frac {1}{b} + \frac {1}{c}}\ge ...


Solution to problem 24
Using p,q,r substitution (p=a+b+c\ ,\ q=ab+bc+ca\ ,\ r=abc) the inequality becomes :

\frac{3(p+q+r+1)}{2p+q+r}\ge \frac{9+3r}{q}\Leftrightarrow pq+2q^2\ge 6pr+9rwhich is true because is well-known that pq\ge 9r and q^2\ge 3pr


Problem 25.Let a,b,c>0 such that abc=1 . Prove that :

\frac{ab}{a^2+b^2+\sqrt{c}}+\frac{bc}{b^2+c^2+\sqrt{a}}+\frac{ca}{c^2+a^2+\sqrt{b}}\le 1
_________________
own problems are the best

PostPosted: Sun Sep 13, 2009 9:36 am  Back to top 
  ProfilePMBlog
FantasyLover
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 26 Mar 2008
Posts: 1901
Location: AAST
To rate posts you must be logged in
#49
alex2008 wrote:
Problem 25.Let a,b,c > 0 such that abc = 1 . Prove that :
\frac {ab}{a^2 + b^2 + \sqrt {c}} + \frac {bc}{b^2 + c^2 + \sqrt {a}} + \frac {ca}{c^2 + a^2 + \sqrt {b}}\le 1


Solution to Problem 25
We have a^2 + b^2 + \sqrt {c}\ge 2ab + \sqrt {c} = \frac {2}{c} + \sqrt {c}.

Hence, it suffices to prove that \displaystyle\sum_{\text{cyc}}\frac {\frac {1}{a}}{\frac {2}{a} + \sqrt {a}} = \displaystyle\sum_{\text{cyc}}\frac {1}{2 + a\....

Reducing to a common denominator, we prove that \displaystyle\sum_{\text{cyc}}\frac {1}{2 + a\sqrt {a}} = \frac {4(a\sqrt {a} + b\sqrt {b} + c\sqrt {c}) + ab\sqrt {ab} + bc\....

Rearranging, it remains to prove that ab\sqrt {ab} + bc\sqrt {bc} + ca\sqrt {ca}\ge 3.

Applying AM-GM, we have ab\sqrt {ab} + bc\sqrt {bc} + ca\sqrt {ca}\ge 3\sqrt [3]{a^2b^2c^2\sqrt {a^2b^2c^2}} = 3, and we are done. \blacksquare


Problem 26.

x, y, z are real numbers satisfying the condition 3x + 2y + z = 1.

Find the maximum value of \frac {1}{1 + |x|} + \frac {1}{1 + |y|} + \frac {1}{1 + |z|}.

EDIT: Source: Korea 2006 First Examination
_________________
AAST 2013

PostPosted: Sun Sep 13, 2009 10:41 am  Back to top 
  ProfilePMBlog
dgreenb801
Navier-Stokes Equations
Navier-Stokes Equations

Online
Joined: 08 Sep 2007
Posts: 1239
Location: Florida
United States
To rate posts you must be logged in
#50
Solution to Problem 26

We can assume x,y, and z are all positive, because if one was negative we could just make it positive, which would allow us to lessen the other two variables, making the whole sum larger.
Let 3x = a, 2y = b, z = c, then a + b + c = 1 and we have to maximize
\frac {3}{a + 3} + \frac {2}{b + 2} + \frac {1}{c + 1}
Note that
(\frac {3}{a + c + 3} + 1) - (\frac {3}{a + 3} + \frac {1}{c + 1}) = \frac {a^2c + ac^2 + 6ac + 6c}{(a + 3)(c + 1)(a + c + 3)...
So for fixed a + c, the sum is maximized when c = 0.
We can apply the same reasoning to show the sum is maximized when b = 0.
So the maximum occurs when a = 1, b = 0, c = 0, and the sum is \frac {11}{4}.


Problem 27
\frac {1}{a(1 + b)} + \frac {1}{b(1 + c)} + \frac {1}{c(1 + a)} \ge \frac {3}{1 + abc} for all positive reals.

PostPosted: Sun Sep 13, 2009 3:58 pm  Back to top 
  ProfilePM
alex2008
Yang-Mills Theory
Yang-Mills Theory


Offline
Joined: 26 Oct 2008
Posts: 749
Location: Tulcea , Romania
Romania
To rate posts you must be logged in
#51
dgreenb801 wrote:


Problem 27
\frac {1}{a(1 + b)} + \frac {1}{b(1 + c)} + \frac {1}{c(1 + a)} \ge \frac {3}{1 + abc} for all positive reals.


Solution to problem 27
AM-GM works :

\displaystyle \left( 1 + abc\right) LHS + 3= \sum_{cyc}\frac {1 + abc + a + ab}{a + ab}\displaystyle = \sum_{cyc}\frac {1 + a...


Problem 28.Let a,b,c>0 such that ab+bc+ca=3. Show that :

\frac{1}{1+a^2(b+c)}+\frac{1}{1+b^2(c+a)}+\frac{1}{1+c^2(a+b)} \leq \frac{1}{abc}
_________________
own problems are the best

PostPosted: Sun Sep 13, 2009 8:45 pm  Back to top 
  ProfilePMBlog
enndb0x
Yang-Mills Theory
Yang-Mills Theory


Offline
Joined: 20 Jan 2009
Posts: 519
Location: Kosovo ,Norway
AlbaniaEuropean Union
To rate posts you must be logged in
#52
alex2008 wrote:


Problem 28.Let a,b,c > 0 such that ab + bc + ca = 3. Show that :
\frac {1}{1 + a^2(b + c)} + \frac {1}{1 + b^2(c + a)} + \frac {1}{1 + c^2(a + b)} \leq \frac {1}{abc}


solution to problem 28


\sum {\frac{1}{1+a^2(b+c)}} =\sum {\frac{1}{1 +3a-abc}}

Since ab +bc +ca =3 \implies abc \leq 1

Then 1+ 3a -abc \geq 3a

Then

\sum {\frac{1}{1+3a-abc}} \leq \frac{1}{3} \left(\frac{1}{a} +\frac{1}{b} +\frac{1}{c} \right) = \frac{ab +bc +ca}{3abc} =\fr...



Problem 29 .Let a,b,c be positive real numbers .Prove that

\frac{a^3}{a^2 +ab +b^2 } + \frac{b^3}{b^2 +bc+c^2} +\frac{c^3}{c^2 +ca +a^2} \geq \frac{a+b+c}{3}

PostPosted: Mon Sep 14, 2009 1:50 pm  Back to top 
  ProfilePMMSN
dgreenb801
Navier-Stokes Equations
Navier-Stokes Equations

Online
Joined: 08 Sep 2007
Posts: 1239
Location: Florida
United States
To rate posts you must be logged in
#53
Solution to Problem 29

By Cauchy,
\sum \frac {a^3}{a^2 + ab + b^2} = \sum \frac {a^4}{a^3 + a^2b + ab^2} \ge \frac {(a^2 + b^2 + c^2)^2}{a^3 + b^3 + c^3 + a^2b...
This is \ge \frac {a + b + c}{3} if
(a^4 + b^4 + c^4) + 2(a^2b^2 + b^2c^2 + c^2a^2) \ge (a^3b + a^3c + b^3a + b^3c + c^3a + c^3b) + (a^2bc + ab^2c + abc^2)
This is equivalent to
(a^2 + b^2 + c^2)(a^2 + b^2 + c^2 - ab - bc - ca) \ge 0
Which is true as a^2 + b^2 + c^2 - ab - bc - ca \ge 0 \iff (a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0

Problem 30
Given ab + bc + ca = 1
Show that
\frac {\sqrt {3a^2 + b^2}}{ab} + \frac {\sqrt {3b^2 + c^2}}{bc} + \frac {\sqrt {3c^2 + a^2}}{ca} \ge 6\sqrt {3}

PostPosted: Mon Sep 14, 2009 5:09 pm  Back to top 
  ProfilePM
great math
Riemann Hypothesis
Riemann Hypothesis


Online
Joined: 07 Mar 2008
Posts: 345
Location: University of Auckland
Viet NamNew Zealand
To rate posts you must be logged in
#54
Dear friends, I am so glad that from now on, I will be on track to keep this marathon up. Smile

Solution to problem 19 (proposed and solution by Hoang Quoc Viet)

Let A=(abc)^2(a^3+b^3+c^3)

Therefore, we only need to maximize the following expression

A^3=(abc)^6(a^3+b^3+c^3)^3

Using Cauchy inequality as follows, we get

A^3 & = \frac{1}{3^6}\left(3a^2b\right)\left(3a^2c\right)\left(3b^2a\right)\left(3b^2c\right)\left(3c^2a\right)\left(3c^2...

It is fairly straightforward that

a+b+c \le \sqrt{3(a^2+b^2+c^2)}=3

Therefore, A^3 \le 3^3

which leads to A \le 3 as desired.

The equality case happens \iff a=b=c=1





Solution to problem 30


Let's make use of Cauchy-Schwarz as demonstrated as follows

\frac{\sqrt{(3a^2+b^2)(3+1)}}{2ab} \ge \frac{3a+b}{2ab}

Thus, we have the following estimations

\sum_{cyc}\frac{\sqrt{3a^2+b^2}}{ab} \ge 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)

Finally, we got to prove that

\sum_{cyc}\frac{1}{a} \ge 3\sqrt{3}

However, from the given condition, we derive

abc \le \frac{1}{3\sqrt{3}}

and \sum_{cyc}\frac{1}{a} \ge 3\sqrt[3]{\frac{1}{abc}} \ge 3\sqrt{3}






Problem 31 ( Komal Magazine). Let a,b,c be real numbers. Prove that the following inequality holds

(a^2+2)(b^2+2)(c^2+2) \ge 3(a+b+c)^2
_________________
Math is a holy subject Very Happy

\left(\frac{a_{1}+a_{2}+...+a_{n}}{n} \right) \geq \sqrt[n]{a_1a_{2}a_{3}....a_{n}}

PostPosted: Mon Sep 14, 2009 8:31 pm  Back to top 
  ProfilePMYMMSNBlog
alex2008
Yang-Mills Theory
Yang-Mills Theory


Offline
Joined: 26 Oct 2008
Posts: 749
Location: Tulcea , Romania
Romania
To rate posts you must be logged in
#55
Great solution , Viet to problem 19 !

alex2008 wrote:


Problem 25.Let a,b,c > 0 such that abc = 1 . Prove that :
\frac {ab}{a^2 + b^2 + \sqrt {c}} + \frac {bc}{b^2 + c^2 + \sqrt {a}} + \frac {ca}{c^2 + a^2 + \sqrt {b}}\le 1


Another solution to problem 25

LHS\le\sum_{cyc} \frac {ab}{2ab + \sqrt {c}} = \sum_{cyc}\frac {1}{2 + c\sqrt {c}} = \sum_{cyc}\frac {1}{2 + \frac {x}{y}}\le...


enndb0x wrote:


Problem 29 .Let a,b,c be positive real numbers .Prove that
\frac {a^3}{a^2 + ab + b^2 } + \frac {b^3}{b^2 + bc + c^2} + \frac {c^3}{c^2 + ca + a^2} \geq \frac {a + b + c}{3}


Another solution to problem 29
Since :
\frac {a^{3} - b^{3}}{a^{2} + ab + b^{2}} = a - b
and similars we get :
\sum_{cyc}\frac {a^{3}}{a^{2} + ab + b^{2}} = \sum_{cyc}\frac {b^{3}}{a^{2} + ab + b^{2}} = \frac {1}{2}\sum_{cyc}\frac {a^{3...
Now it remains to prove :
\frac {1}{2}\cdot\frac {a^{3} + b^{3}}{a^{2} + ab + b^{2}}\geq\frac {a + b}{6}
which is trivial .
great math wrote:


Problem 32 ( Komal Magazine). Let a,b,c be real numbers. Prove that the following inequality holds
(a^2 + 2)(b^2 + 2)(c^2 + 2) \ge 3(a + b + c)^2


Solution to problem 31
Cauchy-Schwartz gives :
(a^2 + 2)(b^2 + 2) = (a^2 + 1)( 1 + b^2) + a^2 + b^2 + 3 \ge (a + b)^2 + \frac {1}{2}(a + b)^2 + 3 = \frac {3}{2}((a + b)^2 +...
And Cauchy-Schwartz again
(a^2 + 2)(b^2 + 2)(c^2 + 2) \ge \frac {3}{2}((a + b)^2 + 2)(2 + c^2) \ge \frac {3}{2}( \sqrt {2}(a + b) + \sqrt {2}c)^2 = RHS


Problem 32:Let a,b,c\geq 0 and a + b + c = 1 . Prove that :
\frac {a}{\sqrt {b^2 + 3c}} + \frac {b}{\sqrt {c^2 + 3a}} + \frac {c}{\sqrt {a^2 + 3b}}\geq \frac {1}{\sqrt {1 + 3abc}}
_________________
own problems are the best
Last edited by alex2008 on Mon Sep 14, 2009 10:57 pm; edited 2 times in total 
PostPosted: Mon Sep 14, 2009 10:21 pm  Back to top 
  ProfilePMBlog
apratimdefermat
Poincare Conjecture
Poincare Conjecture


Offline
Joined: 13 Jul 2008
Posts: 204
Location: The Dragon's Lair
To rate posts you must be logged in
#56
Solution to Problem 31
By Cauchy Schwarz,
(x^2 + 2)(2 + y^2 ) \geqslant 2(x + y)^2 ......(1)

Again by two applications of AM-GM,
(b^2 + c^2) + (2b^2 c^2 + 2 )\geqslant 2bc + 4\sqrt 2 bc > 6bc
\Leftrightarrow 2(b^2 + c^2 ) + b^2 c^2 + 1 > \frac {3} {2}(b + c)^2
\Leftrightarrow (b^2 + 2)(c^2 + 2) > \frac {3} {2}\{ 2 + (b + c)^2 \} ....(2)

Now , by application of (2) and them (1),
(a^2 + 2)(b^2 + 2)(c^2 + 2) > \frac {3} {2}(a^2 + 2)\{ 2 + (b + c)^2 \} > 3(a + b + c)^2

I wonder if there can be a generalization such as
\prod\limits_{i = 1}^n {(a_i ^2 + 2)} > n\left( {\sum\limits_{i = 1}^n {a_i } } \right)^2


Problem 33(proposed by me)
If a,b,c,d be positive reals then prove that
\frac {{a^{2} + b^{2}}}{{ab + b^{2}}} + \frac {{b^{2} + c^{2}}}{{bc + c^{2}}} + \frac {{c^{2} + d^{2}}}{{cd + d^{2}}}\geq\sqr...
_________________
Sign the "Rehabilitate the slum dwellers of New Gandhinagar!" petition!To sign follow the link in my profile.
Last edited by apratimdefermat on Mon Sep 14, 2009 11:56 pm; edited 1 time in total 
PostPosted: Mon Sep 14, 2009 10:22 pm  Back to top 
  ProfilePMWWWYMBlogAlbum
Agr_94_Math
Yang-Mills Theory
Yang-Mills Theory

Offline
Joined: 17 Feb 2008
Posts: 725
India
To rate posts you must be logged in
#57
Solution to Problem 33

I think the problem was posted in a fellow called Reim or someone {mathlinks username} without any reply.
Here goes my solution:

Write the LHS as \frac {(\frac {a}{b})^2 + 1}{(\frac {a}{b}) + 1} + two other similar terms{feel lazy to write them down}. This is a beautiful application of Jensen's as the function for positve real t such that f(t) = \frac {t^2 + 1}{t + 1} is convex since (\frac {t - 1}{t + 1})^2 \ge 0 .

Thus, we get that LHS \ge \frac {(\frac {(\sum \frac {a}{b} - \frac {d}{a}} {3} )^2 + 1 }{\sum \frac {a}{b} - \frac {d}{a} + 3 }

I would like to write \frac {a}{b} + \frac {b}{c} + \frac {c}{d} = K for my convenience with latexing.

so we have \frac {(\frac {K}{3} ) ^2 + 1 )}{K + 3} = \frac {K + \frac {9}{K}} {1 + \frac {3}{K}} \ge \frac {6}{1 + \frac {3}{K}}

This is from K + \frac {9}{K} \ge 6 by AM GM.

Now K = \sum \frac {a}{b} - \frac {d}{a} \ge 3(\frac {d}{a} )^ {\frac {1}{3}} by AM GM.

Thus, we have \frac {6}{1 + \frac {3}{K}} \ge \frac {6a^{\frac {1}{3}}}{ a^{\frac {1}{3}} + d^{\frac {1}{3}} } \ge \frac {3 a^{\frac {1}{3}...

Very tough to latex than to solvethough.


PS I dont know if I can give nay tough problem for this forum. So someone else post a problem.

PostPosted: Mon Sep 14, 2009 11:49 pm  Back to top 
  ProfilePMBlog
Maths Mechanic
Riemann Hypothesis
Riemann Hypothesis

Offline
Joined: 23 Jan 2009
Posts: 273
Location: New Delhi
India
To rate posts you must be logged in
#58
Problem 34

If a and b are non negative real numbers such that a\ge b.
Prove that a+\frac{1}{b(a-b)}\ge3
_________________
Mathlover
\Re\alpha\delta\Pi\alpha\nu \delta\Re\theta\nu\epsilon\Re

PostPosted: Tue Sep 15, 2009 1:07 am  Back to top 
  ProfilePM
Dimitris X
Yang-Mills Theory
Yang-Mills Theory

Offline
Joined: 17 Sep 2008
Posts: 556
Location: Greece
To rate posts you must be logged in
#59
Maths Mechanic wrote:
Problem 34

If a and b are non negative real numbers such that a\ge b.
Prove that a + \frac {1}{b(a - b)}\ge3


solution to problem 34

If a\ge 3 the problem is obviously true.
Now for a < 3 we have :
a^2b - ab^2 - 3ab + 3b^2 + 1 \ge 0 \Longleftrightarrow ba^2 - (b^2 + 3b)a + 3b^2 + 1 \ge 0
It suffices to prove that D\le0 \Longleftrightarrow b^4 + 6b^3 + 9b^2 - 12b^3 - 4b \Longleftrightarrow b(b - 1)^2(b - 4) \le 0 which is true.


PROBLEM 35
(a^2 - bc)\sqrt {b + c} + (b^2 - ca)\sqrt {c + a} + (c^2 - ab)\sqrt {a + b} \ge 0
_________________
ΠΑΙΡΝΩ ΤΑΜΠΕΛΑ ΚΑΙ ΕΓΩ ΤΟΥ ΕΘΝΙΚΟΥ ΠΡΟΔΟΤΗ ΑΦΙΕΡΩΜΕΝΟ ΚΑΙ ΑΥΤΟ ΣΕ ΚΑΘΕ ΔΟΥΛΟ ΠΑΤΡΙΩΤΗ.....

PostPosted: Tue Sep 15, 2009 2:26 am  Back to top 
  ProfilePM
dgreenb801
Navier-Stokes Equations
Navier-Stokes Equations

Online
Joined: 08 Sep 2007
Posts: 1239
Location: Florida
United States
To rate posts you must be logged in
#60
Another solution to problem 34

a+\frac{1}{b(a-b)}=b+(a-b)+\frac{1}{b(a-b)} \ge 3 by AM-GM

PostPosted: Tue Sep 15, 2009 3:58 am  Back to top 
  ProfilePM
Display posts from previous:   Sort by:   
382 Posts • Page 3 of 20 • Previous 1, 2, 3, 4, 5, ..., 18, 19, 20 Next
Post new topic   Reply to topic View previous topicView next topic
Home » Mathematics » High School Pre-Olympiad (Ages 14+)
 
Jump to:  


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You can download files in this forum
You cannot post calendar events in this forum


© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us