Inequality, Problem 2

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Inequality, Problem 2

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Kazakhstan National Olympiad 2017, March 14, P2 , matol.kz

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

21 posts

#1 h Mar 15, 2017, 9:47 AM • 5 Y

Y by luofangxiang, isi2003, dalarin01, Adventure10, Mango247

For positive reals $x,y,z\ge \frac{1}{2}$ with $x^2+y^2+z^2=1$ , prove this inequality holds

$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

38 posts

#2 h Mar 15, 2017, 10:59 AM • 1 Y

Y by Adventure10

$$(zy+xz-yz)(yz+xy-xz)$\ge 2$x^2y^2z^2$
then replace $x^2$ by $1-y^2-z^2$
and put $yz=a$ and $y-z=b$
and we have to prove that
$(2a+b^2)(2a^2+b^2)\ge (a^2+b^2)
which is true bu C-S since $a\ge\frac{1}{4}

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3377 posts

mudok

#3 h Mar 15, 2017, 11:19 AM • 2 Y

Y by Adventure10, Mango247

Nice inequality for an olympiad.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4613 posts

#5 h Mar 15, 2017, 12:18 PM • 2 Y

Y by Adventure10, Mango247

Manar-Manth wrote:

$$(zy+xz-yz)(yz+xy-xz)$\ge 2$x^2y^2z^2$
then replace $x^2$ by $1-y^2-z^2$
and put $yz=a$ and $y-z=b$
and we have to prove that
$(2a+b^2)(2a^2+b^2)\ge (a^2+b^2)
which is true bu C-S since $a\ge\frac{1}{4}

Don't understand

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2649 posts

kk108

#7 h Mar 15, 2017, 12:37 PM • 2 Y

Y by Adventure10, Mango247

Manar-Manth wrote:

$(zy+xz-yz)(yz+xy-xz) \ge 2 x^2y^2 z^2$
then replace $x^2$ by $1 -y^2- z^2$
and put $yz=a$ and $y-z=b$
and we have to prove that
$(2a+b^2)(2a^2+b^2) \ge (a^2+b^2)$
which is true bu C-S since $a \ge \frac{1}{4}$

This post has been edited 1 time. Last edited by kk108, Mar 15, 2017, 12:38 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2649 posts

kk108

#8 h Mar 15, 2017, 12:38 PM • 2 Y

Y by Adventure10, Mango247

Is above what you meant?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

38 posts

#9 h Mar 15, 2017, 12:43 PM • 1 Y

Y by Adventure10

yes

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4613 posts

#10 h Mar 15, 2017, 12:55 PM • 2 Y

Y by Adventure10, Mango247

Manar-Manth wrote:

$$(zy+xz-yz)(yz+xy-xz)$\ge 2$x^2y^2z^2$
then replace $x^2$ by $1-y^2-z^2$
and put $yz=a$ and $y-z=b$
and we have to prove that
$(2a+b^2)(2a^2+b^2)\ge (a^2+b^2)
which is true bu C-S since $a\ge\frac{1}{4}

C-S？

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3377 posts

mudok

#11 h Mar 15, 2017, 1:04 PM • 3 Y

Y by MariusStanean, Adventure10, Mango247

http://matol.kz/comments/2926/show

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4613 posts

#12 h Mar 15, 2017, 1:10 PM • 2 Y

Y by Adventure10, Mango247

//cdn.artofproblemsolving.com/images/0/4/a/04a05304b407b66d941f3313eaaca2447e08f8c6.jpg

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3377 posts

mudok

#13 h Mar 15, 2017, 1:20 PM • 2 Y

Y by Adventure10, Mango247

mudok wrote:

http://matol.kz/comments/2926/show

In that link, the first solution is official solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

426 posts

#14 h Mar 15, 2017, 1:22 PM • 1 Y

Y by Adventure10

mudok wrote:

mudok wrote:

http://matol.kz/comments/2926/show

In that link, the first solution is official solution.

But the second is nicer.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4867 posts

#15 h Mar 15, 2017, 2:24 PM • 2 Y

Y by Adventure10, Mango247

Please, check my solution.
We want to prove $\frac{1}{x^2} \geq 2+(\frac{1}{y}-\frac{1}{z})^2$
$\frac{1}{4} \leq yz \leq \frac{1-x^2}{2}<\frac{1}{2}$
$x^2=1-y^2-z^2 \leq 1-2yz$

$\frac{1}{x^2}+\frac{1}{2yz} \geq \frac{1}{1-2yz}+\frac{1}{2yz} \geq 4 \to \frac{1}{x^2} \geq 4-\frac{1}{2yz}$
$2-\frac{1}{2yz} \geq \frac{1}{y^2}+\frac{1}{z^2}-\frac{2}{yz}$
$4y^2z^2+3yz \geq 2y^2+2z^2$
$(4yz-1)(2yz-1)\leq 0 \to 6yz \geq 8y^2z^2+1$
and $(4y^2-1)(4z^2-1) \geq 0 \to 16y^2z^2+1 \geq 4y^2+4z^2$
So
$8y^2z^2+6yz \geq 16y^2z^2+1 \geq 4y^2+4z^2$ - Q.E.D

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4613 posts

#16 h Mar 15, 2017, 2:27 PM • 2 Y

Y by Adventure10, Mango247

//cdn.artofproblemsolving.com/images/2/7/2/272830e80d787476408202839b7564ff12a92551.jpg

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4613 posts

#17 h Mar 15, 2017, 2:46 PM • 2 Y

Y by Adventure10, Mango247

The following inequality is also true
//cdn.artofproblemsolving.com/images/7/9/6/796f485154129674e46b3cb5ac271dbe5449ce8e.jpg

//cdn.artofproblemsolving.com/images/7/9/6/796f485154129674e46b3cb5ac271dbe5449ce8e.jpg

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4613 posts

#18 h Mar 15, 2017, 3:18 PM • 3 Y

Y by isi2003, Adventure10, Mango247

The following inequality is also true?
//cdn.artofproblemsolving.com/images/7/3/9/739a2d03a0768342b200b80bbd1996b28c188c5d.jpg

//cdn.artofproblemsolving.com/images/7/3/9/739a2d03a0768342b200b80bbd1996b28c188c5d.jpg

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41003 posts

sqing

#22 h Jan 30, 2018, 7:50 AM • 2 Y

Y by Adventure10, Mango247

matol.kz wrote:

For positive reals $x,y,z\ge \frac{1}{2}$ with $x^2+y^2+z^2=1$ , prove this inequality holds

$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2$

mudok wrote:

http://matol.kz/comments/2926/show

$(\frac{1}{y^2}-\frac{1}{4})(\frac{1}{z^2}-\frac{1}{4})\ge 0\iff (y-z)^2\le \frac{1}{4}(4yz-1)^2$
$yz\ge \frac{1}{4}, x^2=1-y^2-z^2\le 1-2yz$
$\frac{1}{x^2}-\left(\frac{1}{y}-\frac{1}{z}\right)^2-2=\frac{1}{x^2}-\frac{\left(y-z\right)^2}{y^2z^2}-2\ge\frac{1}{1-2yz}-\frac{\left(4yz-1\right)^2}{4y^2z^2}-2=\frac{\left(4yz-1\right)\left(3y^2z^2+(3yz-1)^2\right)}{4y^2z^2(1-2yz)} \ge 0$
$(\frac{1}{x}+\frac{1}{y}-\frac{1}{z})(\frac{1}{x}-\frac{1}{y}+\frac{1}{z})\ge 2$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a