inequalities

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

inequalities

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

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

LuvThoConBoiRoi

149 posts

LuvThoConBoiRoi

#1 h Aug 1, 2019, 10:27 AM • 1 Y

Y by Adventure10

With $a,b,c$ are all real positive number that: $abc=a+b+c+2$ . Prove that:
$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \leq \frac{3}{4}$

This post has been edited 3 times. Last edited by LuvThoConBoiRoi, Aug 1, 2019, 11:04 AM

Z K Y

The post below has been deleted. Click to close.

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

rctmathadventures

685 posts

rctmathadventures

#2 h Aug 1, 2019, 10:34 AM • 1 Y

Y by Adventure10

LuvThoConBoiRoi wrote:

With $a,b,c$ are all real positive number that: $abc=a+b+c+2$ . Prove that:
$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \ge \frac {3}{4}$

This post has been edited 1 time. Last edited by rctmathadventures, Aug 1, 2019, 10:35 AM

Z K Y

The post below has been deleted. Click to close.

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

rctmathadventures

685 posts

rctmathadventures

#3 h Aug 1, 2019, 10:37 AM • 1 Y

Y by Adventure10

Its 34 or 3/4

Z K Y

The post below has been deleted. Click to close.

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

LuvThoConBoiRoi

149 posts

LuvThoConBoiRoi

#4 h Aug 1, 2019, 10:42 AM • 1 Y

Y by Adventure10

rctmathadventures wrote:

Its 34 or 3/4

It's 3/4, edited

Z K Y

The post below has been deleted. Click to close.

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

6107 posts

#5 h Aug 1, 2019, 10:58 AM • 1 Y

Y by Adventure10

Inequality is incorrect
Try $a=1,b=2,c=5$

Z K Y

The post below has been deleted. Click to close.

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

LuvThoConBoiRoi

149 posts

LuvThoConBoiRoi

#6 h Aug 1, 2019, 11:04 AM • 1 Y

Y by Adventure10

Math-wiz wrote:

Inequality is incorrect
Try $a=1,b=2,c=5$

sorry ^_^

Z K Y

The post below has been deleted. Click to close.

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

41034 posts

sqing

#9 h Yesterday at 1:27 PM

Y by

Let $a,b,c$ be positive numbers such that $abc=a+b+c+2$ . Prove that
$\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}} \leq \frac{3}{2}$ Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y},$ where $x,$ $y$ and $z$ are positives.
Thus, the condition gives $c=\frac{x+y}{z}$ and by AM-GM we obtain:
$\sum_{cyc}\frac{1}{\sqrt{ab}}=\sum_{cyc}\sqrt{\frac{x}{x+z}\cdot\frac{y}{y+z}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{3}{2}.$ The condition $\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2$ is equivalent to $abc= a+b+c+2$
$\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}=2\iff \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1.$ therefore we can make the substitution $a=\frac{y+z}{x}, b=\frac{x+z}{y}$ and $c=\frac{x+y}{z}$ ,

This post has been edited 3 times. Last edited by sqing, 5 hours ago

Z K Y

The post below has been deleted. Click to close.

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

58 posts

#10 h Yesterday at 2:14 PM

Y by

Wow, thank you very much. I only learn about condition $ab+bc+ca +abc$ and $a^2+b^2+c^2+abc$ , i didnt learn about $a+b+c=abc-k$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a