Community

Try our innovative online adaptive learning system, Alcumus.
Over 1100 problems and 60+ video lessons. FREE!
Login Register Memberlist Search AoPS Blogs Contests Galleries Forum Index
The time now is Mon Dec 07, 2009 2:59 am
All times are UTC - 8
 		View posts since last visit
View unanswered posts
Home » Olympiad Section » Inequalities » Inequalities Proposed & Own Problems
 
Am-Gm
Moderators: High School Olympiad Moderators, Arne, blahblahblah, Cezar Lupu, darij grinberg, harazi, Megus, N.T.TUAN, orl, pbornsztein, pvthuan
Post new topic   Reply to topic View previous topicView next topic
23 Posts • Page 1 of 2 • 1, 2 Next
Author Message
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#1
Am-Gm
If x\ge y \ge z\ge 0, then

\frac {(x-z)^2}{4(x+y+z)} \le \frac {x+y+z}{3} -\sqrt[3]{xyz} \le \frac {4(x-z)^2}{3(x+y+z)}.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Wed Nov 04, 2009 6:17 am  Back to top 
  ProfilePM
can_hang2007
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 12 Jan 2008
Posts: 1613
Location: Vietnam
Viet Nam
To rate posts you must be logged in
#2
Very nice inequality, Mr Vasile. Please take a look at my proof at: http://canhang2007.wordpress.com/2009/11/05/inequality-48-v-cirtoaje/
_________________
The love makes us stronger!

V. Q. B. Can

PostPosted: Wed Nov 04, 2009 4:32 pm  Back to top 
  ProfilePMWWW
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#3
Very nice your proof, Can_Hang. Very Happy
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Thu Nov 05, 2009 2:44 pm  Back to top 
  ProfilePM
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#4
What about this ?

If x\ge y \ge z\ge 0, then

\frac {(x - z)^2}{2(x + z)} \le x+y+z - 3\sqrt [3]{xyz} \le \frac {2(x - z)^2}{x+ z}.

Notice that the right inequality is sharper that the one above.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Fri Nov 06, 2009 3:50 pm  Back to top 
  ProfilePM
can_hang2007
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 12 Jan 2008
Posts: 1613
Location: Vietnam
Viet Nam
To rate posts you must be logged in
#5
Vasc wrote:
What about this ?

If x\ge y \ge z\ge 0, then

\frac {(x - z)^2}{2(x + z)} \le x + y + z - 3\sqrt [3]{xyz} \le \frac {2(x - z)^2}{x + z}.

Notice that the right inequality is sharper that the one above.

See my proof here: http://canhang2007.wordpress.com/2009/11/07/inequality-53-v-cirtoaje/
_________________
The love makes us stronger!

V. Q. B. Can

PostPosted: Fri Nov 06, 2009 4:18 pm  Back to top 
  ProfilePMWWW
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#6
Nice solution, Can_hang. Very Happy
Note that for the right inequality there is at least another one nice solution (only with AM-GM inequality).
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Fri Nov 06, 2009 5:11 pm  Back to top 
  ProfilePM
can_hang2007
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 12 Jan 2008
Posts: 1613
Location: Vietnam
Viet Nam
To rate posts you must be logged in
#7
Vasc wrote:
Nice solution, Can_hang. Very Happy
Note that for the right inequality there is at least another one nice solution (only with AM-GM inequality).

Dear Mr Vasile,

Please take a look again the link: http://canhang2007.wordpress.com/2009/11/07/inequality-53-v-cirtoaje/ to see my second proof for (b).
_________________
The love makes us stronger!

V. Q. B. Can

PostPosted: Fri Nov 06, 2009 7:07 pm  Back to top 
  ProfilePMWWW
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#8
can_hang2007 wrote:

Dear Mr Vasile,

Please take a look again the link: http://canhang2007.wordpress.com/2009/11/07/inequality-53-v-cirtoaje/ to see my second proof for (b).

Nice proof, Can_hang. Very Happy
There is still at least another one nice solution.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Sat Nov 07, 2009 1:20 am  Back to top 
  ProfilePM
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#9
Vasc wrote:
can_hang2007 wrote:

Dear Mr Vasile,

Please take a look again the link: http://canhang2007.wordpress.com/2009/11/07/inequality-53-v-cirtoaje/ to see my second proof for (b).

Nice proof, Can_hang. Very Happy
There is still at least another one nice solution.

My solution.
Using twice AM-GM inequality, we show that
x + y + z - 3\sqrt [3]{xyz} \le 2x + z - 3\sqrt [3]{x^2z}\le \frac {2(x - z)^2}{x + z}.
The left inequality reduces to
\sqrt [3]{x^2} + \sqrt [3]{xy} + \sqrt [3]{y^2}\ge 3\sqrt [3]{xz},
and the right to
z^2 + 3\sqrt [3]{x^5z} + \sqrt [3]{x^2z^3}\ge 7xz.

Generalization. If x_1\ge x_2\ge ...\ge x_n\ge 0, then

x_1 + x_2 + ... + x_n - n\sqrt [n]{x_1x_2...x_n}\le \frac {(n - 1)(x_1 - x_n)^2}{x_1 + x_n}.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Sun Nov 08, 2009 6:56 am  Back to top 
  ProfilePM
can_hang2007
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 12 Jan 2008
Posts: 1613
Location: Vietnam
Viet Nam
To rate posts you must be logged in
#10
Vasc wrote:
Vasc wrote:
can_hang2007 wrote:

Dear Mr Vasile,

Please take a look again the link: http://canhang2007.wordpress.com/2009/11/07/inequality-53-v-cirtoaje/ to see my second proof for (b).

Nice proof, Can_hang. Very Happy
There is still at least another one nice solution.

My solution.
Using twice AM-GM inequality, we show that
x + y + z - 3\sqrt [3]{xyz} \le 2x + z - 3\sqrt [3]{x^2z}\le \frac {2(x - z)^2}{x + z}.
The left inequality reduces to
\sqrt [3]{x^2} + \sqrt [3]{xy} + \sqrt [3]{y^2}\ge 3\sqrt [3]{xz},
and the right to
z^2 + 3\sqrt [3]{x^5z} + \sqrt [3]{x^2z^3}\ge 7xz.

Generalization. If x_1\ge x_2\ge ...\ge x_n\ge 0, then

x_1 + x_2 + ... + x_n - n\sqrt [n]{x_1x_2...x_n}\le \frac {(n - 1)(x_1 - x_n)^2}{x_1 + x_n}.

Your solution is really nice, Mr Vasile.

For your generalization, I found now that you have in your book the following problem:

"Let 0 \le a < b and let a_1,a_2, \ldots, a_n \in [a,b]. Prove that
a_1 + a_2 + \cdots + a_n - n\sqrt [n]{a_1a_2\cdots a_n} \le (n - 1)\left( \sqrt {b} - \sqrt {a}\right)^2."

According to the result of this problem, we can see that the above generalization is just a direct consequence of it.
_________________
The love makes us stronger!

V. Q. B. Can

PostPosted: Sun Nov 08, 2009 7:20 am  Back to top 
  ProfilePMWWW
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#11
can_hang2007 wrote:
Your solution is really nice, Mr Vasile.

For your generalization, I found now that you have in your book the following problem:

"Let 0 \le a < b and let a_1,a_2, \ldots, a_n \in [a,b]. Prove that
a_1 + a_2 + \cdots + a_n - n\sqrt [n]{a_1a_2\cdots a_n} \le (n - 1)\left( \sqrt {b} - \sqrt {a}\right)^2."

According to the result of this problem, we can see that the above generalization is just a direct consequence of it.

Indeed, this is true. Very Happy
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Sun Nov 08, 2009 7:49 am  Back to top 
  ProfilePM
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#12
The following inequalities are a little harder.

If x\ge y\ge z\ge 0, then

(a) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \ge \frac {2(y - z)^2}{9(y + z)};

(b) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \ge \frac {(x - y)^2}{5(x + y)}.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Sun Nov 08, 2009 8:26 am  Back to top 
  ProfilePM
can_hang2007
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 12 Jan 2008
Posts: 1613
Location: Vietnam
Viet Nam
To rate posts you must be logged in
#13
Vasc wrote:
The following inequalities are a little harder.

If x\ge y\ge z\ge 0, then

(a) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \ge \frac {2(y - z)^2}{9(y + z)};

(b) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \ge \frac {(x - y)^2}{5(x + y)}.

Please take a look at my proof here, Mr Vasile: http://canhang2007.wordpress.com/2009/11/08/inequality-57-v-cirtoaje/
_________________
The love makes us stronger!

V. Q. B. Can

PostPosted: Sun Nov 08, 2009 9:58 am  Back to top 
  ProfilePMWWW
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#14
can_hang2007 wrote:
Vasc wrote:
The following inequalities are a little harder.

If x\ge y\ge z\ge 0, then

(a) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \ge \frac {2(y - z)^2}{9(y + z)};

(b) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \ge \frac {(x - y)^2}{5(x + y)}.

Please take a look at my proof here, Mr Vasile: http://canhang2007.wordpress.com/2009/11/08/inequality-57-v-cirtoaje/

Nice proofs, Can-Hang. Very Happy
The following inequalities hold, too.

If x\ge y\ge z\ge 0, then

(a) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \le \frac {(x - z)^2}{3z};

(b) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \le \frac {2(x - z)^2}{x + 5z}.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Sun Nov 08, 2009 3:00 pm  Back to top 
  ProfilePM
can_hang2007
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 12 Jan 2008
Posts: 1613
Location: Vietnam
Viet Nam
To rate posts you must be logged in
#15
Vasc wrote:
can_hang2007 wrote:
Vasc wrote:
The following inequalities are a little harder.

If x\ge y\ge z\ge 0, then

(a) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \ge \frac {2(y - z)^2}{9(y + z)};

(b) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \ge \frac {(x - y)^2}{5(x + y)}.

Please take a look at my proof here, Mr Vasile: http://canhang2007.wordpress.com/2009/11/08/inequality-57-v-cirtoaje/

Nice proofs, Can-Hang. Very Happy
The following inequalities hold, too.

If x\ge y\ge z\ge 0, then

(a) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \le \frac {(x - z)^2}{3z};

(b) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \le \frac {2(x - z)^2}{x + 5z}.

I think (a) follows directly from (b). For (b), we can also prove that the stronger inequality holds
x+y+z-3\sqrt[3]{xyz} \le \frac{2(x-z)^2}{x+5z}.

P.s: Actually, I think Mr Vasile meant (b) is my stronger one, but he got somemistakes in his typing.
_________________
The love makes us stronger!

V. Q. B. Can

PostPosted: Mon Nov 09, 2009 2:46 pm  Back to top 
  ProfilePMWWW
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#16
can_hang2007 wrote:
Vasc wrote:

The following inequalities hold, too.

If x\ge y\ge z\ge 0, then

(a) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \le \frac {(x - z)^2}{3z};

(b) \ \ \ \ \ \frac {x + y + z}{3} - \sqrt [3]{xyz} \le \frac {2(x - z)^2}{x + 5z}.

I think (a) follows directly from (b). For (b), we can also prove that the stronger inequality holds
x + y + z - 3\sqrt [3]{xyz} \le \frac {2(x - z)^2}{x + 5z}.
P.s: Actually, I think Mr Vasile meant (b) is my stronger one, but he got somemistakes in his typing.

You are right, Can, I wanted to post
\frac {x + y + z}{3} - \sqrt [3]{xyz} \le \frac {2(x - z)^2}{3(x + 5z)},
but I made a print-mistake.
We can generalize this interesting inequality to n numbers. For instant, if n = 4, then

x + y + z + w - 4\sqrt [4]{xyzw} \le \frac {3(x - w)^2}{x + 5w}

for x\ge y\ge z\ge w\ge 0.

Can someone find the best k_n such that

x_1 + x_2 + ... + x_n - n\sqrt [n]{x_1x_2...x_n}\le \frac {(n - 1)(x_1 - x_n)^2}{x_1 + k_nx_n}

for x_1\ge x_2\ge ...\ge x_n\ge 0?

or equivalently,

If 0\le a < b and x_1, x_2, ..., x_n\in [a,b], find the best k_n such that

x_1 + x_2 + ... + x_n - n\sqrt [n]{x_1x_2...x_n}\le \frac {(n - 1)(b - a)^2}{b + k_na}?

Click to reveal hidden content
k_3 = k_4 = 5; k_5 = 17/2; k_7 = 6.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Tue Nov 10, 2009 5:09 am  Back to top 
  ProfilePM
Potla
Yang-Mills Theory
Yang-Mills Theory


Offline
Joined: 27 Nov 2008
Posts: 517
Location: 22°34' N; 88°30'E
India
To rate posts you must be logged in
#17
can_hang2007 wrote:

x + y + z - 3\sqrt [3]{xyz} \le \frac {2(x - z)^2}{x + 5z}.

EDIT: Flipped a,b,c and x,y,z, sorry. Embarassed
Problem
a + b + c - 3\sqrt [3]{abc} \le \frac {2(a - c)^2}{a + 5c}.
Solution
We replace a = x^3; b = y^3; c = z^3 \Longrightarrow x\geq y\geq z\geq 0, so we are only required to prove that
x^3 + y^3 + z^3 - 3xyz\leq \frac {2(x^3 - z^3)^2}{x^3 + 5z^3}
\Longrightarrow \frac 12 (x + y + z)[(x - y)^2 + (y - z)^2 + (z - x)^2]\leq \frac {2(z - x)^2 (z^2 + x^2 + zx)^2}{x^3 + 5z^3}
Since (x - y)^2 + (y - z)^2\leq (z - x)^2, so it suffices to show that
\frac 12 (x + y + z)(x^3 + 5z^3)\leq (z^2 + zx + x^2)^2
Using x\geq y, this equivalents
\iff (2x + z)(x^3 + 5z^3)\leq 2z^4 + 2z^2x^2 + 2x^4 + 4zx^3 + 4z^3x + 4z^2x^2
\iff 2x^4 + zx^3 + 10z^3x + 5z^4\leq 2x^4 + 2z^4 + 6z^2x^2 + 4zx^3 + 4z^3x
\iff 6z^3x + 3z^4\leq 6z^2x^2 + 3zx^3
\iff 6z^2x(x - z) + 3z(x^3 - z^3)\geq 0
This is perfectly true, since x\geq y\geq z.
(Edited to fix a huge mistake)
_________________
There is no limited age of learning, man can learn anything anytime. Play Ball
The Problem Solver's paradise

PostPosted: Tue Nov 10, 2009 9:51 pm  Back to top 
  ProfilePMBlog
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#18
Nice, Potla. Very Happy
I wait for a solution in the case n=4.
In addition, the best k_n has a nice formula. We have 5\le k_n<7 for n\ge 3.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Wed Nov 11, 2009 2:26 pm  Back to top 
  ProfilePM
can_hang2007
Navier-Stokes Equations
Navier-Stokes Equations


Offline
Joined: 12 Jan 2008
Posts: 1613
Location: Vietnam
Viet Nam
To rate posts you must be logged in
#19
Vasc wrote:
In addition, the best k_n has a nice formula. We have 5\le k_n < 7 for n\ge 3.

In my result, I found that k_n is \frac {2n(n - 1)}{\left\lfloor \frac {n}{2}\right\rfloor \left\lfloor \frac {n + 1}{2}\right\rfloor} - 1. But my proof was too complicated and I am afraid of getting something wrong in my computations. So, I dont wanna post it here. I hope Mr Vasile has an easier proof than mine.

By the way, the case n = 4 is easy and it has a nice proof. I will post it in some days. Now I am a little busy with my school test.
_________________
The love makes us stronger!

V. Q. B. Can

PostPosted: Thu Nov 12, 2009 1:41 am  Back to top 
  ProfilePMWWW
Vasc
Navier-Stokes Equations
Navier-Stokes Equations

Offline
Joined: 17 May 2004
Posts: 1871
Location: PLoiesti
Romania
To rate posts you must be logged in
#20
can_hang2007 wrote:
In my result, I found that k_n is \frac {2n(n - 1)}{\left\lfloor \frac {n}{2}\right\rfloor \left\lfloor \frac {n + 1}{2}\right\rfloor} - 1. But my proof was too complicated and I am afraid of getting something wrong in my computations. So, I dont wanna post it here. I hope Mr Vasile has an easier proof than mine.

Indeed, the best k_n for n\ge 2 is k_{i-1}=k_i=7-\frac 8{i} for even i. My proof uses derivatives, but is not complicated.
_________________
As to inequalities, the simpler and sharper, the more beautiful!

PostPosted: Thu Nov 12, 2009 11:08 am  Back to top 
  ProfilePM
Display posts from previous:   Sort by:   
23 Posts • Page 1 of 2 • 1, 2 Next
Post new topic   Reply to topic View previous topicView next topic
Home » Olympiad Section » Inequalities » Inequalities Proposed & Own Problems
 
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