Interesting inequality

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

B

No tags match your search

M

Interesting inequality

G H J

Reveal topic tags

inequalities proposed

L

G H BBookmark kLocked kLocked NReply

Source: Own

The post below has been deleted. Click to close.

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

41250 posts

sqing

#1 h Mar 23, 2025, 3:42 AM

Y by

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=3 .$ Prove that
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{4}{3k}$ Where $k\geq 3.$
$\frac{1}{ab+2c}+\frac{1}{ac+2b} \geq\frac{16}{25}$ $\frac{1}{ab+3c}+\frac{1}{ac+3b} \geq\frac{4}{9}$ $\frac{1}{ab+4c}+\frac{1}{ac+4b} \geq\frac{1}{3}$

Z K Y

The post below has been deleted. Click to close.

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

41250 posts

sqing

#2 h Mar 23, 2025, 3:45 AM

Y by

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=3 .$ Prove that
$\frac{1}{ab+\frac{1}{2} c}+\frac{1}{ac+\frac{1}{2} b} \geq\frac{64}{49}$ $\frac{1}{ab+\frac{3}{2} c}+\frac{1}{ac+\frac{3}{2} b} \geq\frac{64}{81}$ $\frac{1}{ab+\frac{1}{3} c}+\frac{1}{ac+\frac{1}{3} b} \geq\frac{36}{25}$ $\frac{1}{ab+\frac{7}{3} c}+\frac{1}{ac+\frac{7}{3} b} \geq\frac{9}{16}$

Z K Y

The post below has been deleted. Click to close.

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

39 posts

#3 h Mar 23, 2025, 5:49 AM

Y by

sqing wrote:

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=3 .$ Prove that
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{4}{3k}$ Where $k\geq 3.$
$\frac{1}{ab+2c}+\frac{1}{ac+2b} \geq\frac{16}{25}$ $\frac{1}{ab+3c}+\frac{1}{ac+3b} \geq\frac{4}{9}$ $\frac{1}{ab+4c}+\frac{1}{ac+4b} \geq\frac{1}{3}$

use C-S : $(\frac{1}{ab+kc}+\frac{1}{ac+kb})(ab+ac+kb+kc) \geq 4$ $ab+ac+kb+kc=(3-a)(a+k) \leq 3k$ ==>
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{4}{3k}$ where $k \geq 3 .$

This post has been edited 2 times. Last edited by SunnyEvan, Mar 23, 2025, 5:50 AM

Z K Y

The post below has been deleted. Click to close.

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

41250 posts

sqing

#4 h Mar 23, 2025, 1:00 PM

Y by

Nice.Thanks.

sqing wrote:

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=3 .$ Prove that
$\frac{1}{ab+2c}+\frac{1}{ac+2b} \geq\frac{16}{25}$

$\text{LHS} \ge \frac4{(a+2)(b+c)} \ge \frac4{\left( \frac{3+2}2 \right)^2}=\frac{16}{25}$

This post has been edited 2 times. Last edited by sqing, Mar 23, 2025, 1:17 PM

Z K Y

The post below has been deleted. Click to close.

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

41250 posts

sqing

#5 h Mar 23, 2025, 1:13 PM

Y by

sqing wrote:

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=3 .$ Prove that $\frac{1}{ab+\frac{1}{5} c}+\frac{1}{ac+\frac{1}{5} b} \geq\frac{25}{16}$

Solution of DAVROS:
$\text{LHS} \ge \frac4{\left(a+\frac15\right)(b+c)} \ge\frac4{\left( \frac{3+\frac15}2\right)^2}=\frac{25}{16}$ @ $a=\frac75, b=c=\frac45$

sqing wrote:

Let $a,b,c\geq 0$ and $a+b+c=3 .$ Prove that $\frac{1}{ab+2c+1}+\frac{1}{ac+2b+1} \geq\frac{16}{33}$

Solution of DAVROS:
$\text{LHS} \ge \frac4{(a+2)(b+c)+2} \ge \frac4{\left( \frac{3+2}2 \right)^2+2}=\frac{16}{33}$ @ $a=\frac12, b=c=\frac54$

This post has been edited 1 time. Last edited by sqing, Mar 23, 2025, 1:53 PM

Z K Y

The post below has been deleted. Click to close.

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

39 posts

#7 h Mar 23, 2025, 1:30 PM

Y by

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=3 .$ Prove that :
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{16}{(k+3)^2}$ where $k \in [0,3]$

Z K Y

The post below has been deleted. Click to close.

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

41250 posts

sqing

#8 h Mar 23, 2025, 1:37 PM

Y by

SunnyEvan wrote:

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=3 .$ Prove that :
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{16}{(k+3)^2}$ where $k \in [0,3]$

$\frac{1}{ab+kc}+\frac{1}{ac+kb}\geq \frac4{\left(a+k\right)(b+c) } \ge\frac4{\left( \frac{3+k}2\right)^2 }=\frac{16}{(k+3)^2}$

This post has been edited 1 time. Last edited by sqing, Mar 23, 2025, 1:43 PM

Z K Y

The post below has been deleted. Click to close.

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

41250 posts

sqing

#9 h Mar 23, 2025, 1:43 PM

Y by

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=3 .$ Prove that
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{16}{(k+3)^2}$ Where $0\leq k\leq 3.$
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{4}{3k}$ Where $k\geq 3.$

Z K Y

The post below has been deleted. Click to close.

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

39 posts

#10 h Mar 23, 2025, 1:56 PM

Y by

Let $a,b,c\geq 0,(ab+c)(ac+b)\neq 0$ and $a+b+c=s=constant .$ Prove that :
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{16}{(k+s)^2}$ Where $0\leq k\leq s.$
$\frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{4}{ks}$ Where $k\geq s.$

Z K Y

The post below has been deleted. Click to close.

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

41250 posts

sqing

#11 h Monday at 12:18 PM

Y by

Good.Thanks.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a