Inequalities Let $abc=1$

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

Inequalities Let $abc=1$

G H J

Reveal topic tags

inequalities proposed

High School Olympiads

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.

23 posts

#1 h Feb 2, 2015, 4:29 AM • 1 Y

Y by Adventure10

Let $a,b,c>0$ and $abc=1$
Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{6}{{a + b +c}} \ge 5$

Z K Y

The post below has been deleted. Click to close.

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

30154 posts

arqady

#2 h Feb 2, 2015, 4:32 AM • 1 Y

Y by Adventure10

Use uvw.

Z K Y

The post below has been deleted. Click to close.

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

23 posts

#3 h Feb 2, 2015, 4:36 AM • 1 Y

Y by Adventure10

arqady wrote:

Use uvw.

Because I'm don't speak English so you can write The solution ..... thanks

Z K Y

The post below has been deleted. Click to close.

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

30154 posts

arqady

#4 h Feb 2, 2015, 4:45 AM • 3 Y

Y by ductiena1k43, duytanvmo, Adventure10

Let $a+b+c=3u$ , $ab+ac+bc=3v^2$ and $abc=w^3$ .
Hence, $\frac{1}{a }+\frac{1}{b }+\frac{1}{c }+\frac{6}{{a+b+c}}\geq5\Leftrightarrow3v^2+\frac{2w^3}{u}\geq5w^2$ .
Since $v^2$ gets a minimal value, when two variables are equal, it remains to check one case only:
$b=a$ and $c=\frac{1}{a^2}$ , which gives $(a-1)^2(2a^4+4a^3-4a^2-a+2)\geq0$ , which is obvious.

Z K Y

The post below has been deleted. Click to close.

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

39 posts

#5 h Feb 2, 2015, 4:57 AM • 1 Y

Y by Adventure10

I'm don't speak..................hahaha

Z K Y

The post below has been deleted. Click to close.

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

41217 posts

sqing

#6 h Feb 2, 2015, 6:24 AM • 1 Y

Y by Adventure10

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=498338

Z K Y

The post below has been deleted. Click to close.

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

41217 posts

sqing

#7 h Mar 22, 2015, 7:25 AM • 2 Y

Y by Adventure10, Mango247

USA 2001:
Let $a,b,c>0$ and $abc=1$
Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{3}{{a + b +c}} \ge 4$

Z K Y

The post below has been deleted. Click to close.

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

1093 posts

#8 h Mar 22, 2015, 8:35 AM • 2 Y

Y by Adventure10, Mango247

This is an old problem!
My solution:
With $p=a+b+c,q=ab+bc+ca,p,q\ge 3$ it returns to:
$q\ge\frac{5p-6}{p}$
We have: (a well-known inequality !!)
$x,y,z\ge 0\Rightarrow x^2+y^2+z^2+2xyz+1\ge 2(xy+yz+zx)\Leftrightarrow$
$\Leftrightarrow x,y,z\ge 0\Rightarrow(x+y+z)^2+2xyz+1\ge 4(xy+yz+zx)$
With $x=bc,y=ca,z=ab$ result $q^2+3\ge 4p\Leftrightarrow q\ge\sqrt{4p-3}$
It is sufficient that $p\ge 3\Rightarrow\sqrt{4p-3}\ge\frac{5p-6}{p}$
$\Leftrightarrow p\ge 3\Rightarrow (p-1)(p-3)^2\ge 0$
which is true!
___
Marin Sandu

Z K Y

The post below has been deleted. Click to close.

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

41217 posts

sqing

#9 h Mar 22, 2015, 9:23 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

USA 2001:
Let $a,b,c>0$ and $abc=1$
Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{3}{{a + b +c}} \ge 4$

$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{3}{{a + b +c}} \geq\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{9}{(\frac{1}{a } + \frac{1}{b } + \frac{1}{c })^2}$
$\geq 4\sqrt[4]{(\frac{1}{3}(\frac{1}{a } + \frac{1}{b } + \frac{1}{c }))^3\cdot \frac{9}{(\frac{1}{a } + \frac{1}{b } + \frac{1}{c })^2} }=4\sqrt[4]{\frac{1}{3}(\frac{1}{a } + \frac{1}{b } + \frac{1}{c })}$
$\geq 4$ .

Z K Y

The post below has been deleted. Click to close.

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

41217 posts

sqing

#10 h Sep 27, 2015, 7:53 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

USA 2001:
Let $a,b,c>0$ and $abc=1$
Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{3}{{a + b +c}} \ge 4$

The inequality is equivalent to
Let $a;b;c$ be positive real numbers satisfying $abc=1$ . Prove that $(a+b)(b+c)(c+a) \geq 4(a+b+c-1).$ MOSP 2001
Find the maximum of $k$ such that $\forall a,b,c>0,abc=1$ the following inequality holds:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{k}{a+b+c}\geq 3+\frac{k}{3}.$ here
$\frac1a+\frac1b+\frac1c+\frac3{a+b+c}=ab+bc+ca+\frac9{3abc(a+b+c)} \geqslant ab+bc+ca+\frac9{(ab+bc+ca)^2}$ $\geqslant1+\frac{ab+bc+ca}3+\frac{ab+bc+ca}3+\frac9{(ab+bc+ca)^2}$ Jiangsu Province,China-2014
MOSP 2001

This post has been edited 1 time. Last edited by sqing, Oct 22, 2021, 2:31 PM

Z K Y

The post below has been deleted. Click to close.

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

41217 posts

sqing

#11 h Jan 11, 2019, 1:44 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

USA 2001:
Let $a,b,c>0$ and $abc=1.$ Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{3}{{a + b +c}} \ge 4$

Let $a,b,c>0$ and $abc=1.$ Prove that: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{7}{a+b+c}\ge\frac{16}{3}$ Let $a,b,c>0$ and $abc=1.$ Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{3}{{a + b +c}} -4\ge \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{7}{a+b+c}-\frac{16}{3}\ge 0.$ Let $a,b,c>0.$ Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{7}{a+b+c}\ge\frac{16}{2+abc}$ Let $a,b,c>0$ and $abc=1 .$ Prove that
$2(a+b+c)+\frac{9}{(ab+bc+ca)^2}\ge7$

This post has been edited 1 time. Last edited by sqing, Feb 3, 2020, 2:52 PM

Z K Y

The post below has been deleted. Click to close.

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

41217 posts

sqing

#12 h Feb 12, 2021, 3:26 PM

Y by

sqing wrote:

USA 2001:
Let $a,b,c>0$ and $abc=1$ Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{3}{{a + b +c}} \ge 4$

Let $a,b,c>0$ and $abc=1.$ Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{66}{{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}} \geq 25$ Can it be strengthened?

This post has been edited 1 time. Last edited by sqing, Feb 12, 2021, 3:28 PM

Z K Y

The post below has been deleted. Click to close.

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

Victoria_Discalceata1

743 posts

Victoria_Discalceata1

#13 h Jul 7, 2021, 2:42 PM • 2 Y

Y by ali3985, Keith50

duytanvmo wrote:

Let $a,b,c>0$ and $abc=1$
Prove that:
$\frac{1}{a } + \frac{1}{b } + \frac{1}{c }+ \frac{6}{{a + b +c}} \ge 5$

With $a=\frac{1}{x},\ b=\frac{1}{y},\ c=\frac{1}{z}$ it is $x+y+z+\frac{6}{xy+yz+zx}\ge 5$ under $xyz=1$ . Let $p=x+y+z$ . By AM-GM $p\ge 3$ . Then by Schur $x+y+z+\frac{6}{xy+yz+zx}-5\ge p+\frac{24p}{p^3+9}-5=\frac{(p-3)\left(\left(\frac{p^3}{2}+\frac{p^3}{2}+4-3p^2\right)+\left(p^2+9-6p\right)+2\right)}{p^3+9}\ge 0$ .

This approach works also for

sqing wrote:

Let $a,b,c>0$ and $abc=1.$ Prove that: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{7}{a+b+c}\ge\frac{16}{3}$

as well as for $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{36}{5(a+b+c)}\ge\frac{27}{5}$ .

This post has been edited 1 time. Last edited by Victoria_Discalceata1, Jul 7, 2021, 9:25 PM
Reason: Additional info.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a