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

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.

41195 posts

sqing

#1 h Yesterday at 3:45 AM

Y by

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$ Where $4\leq k\in N^+.$
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$ Where $16\geq k>0 .$

This post has been edited 3 times. Last edited by sqing, Yesterday at 1:52 PM

Z K Y

The post below has been deleted. Click to close.

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

949 posts

#2 h Yesterday at 3:59 AM

Y by

sqing wrote:

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{2k}{1+k^2 a^2b^2}$ Where $5\leq k\in N^+.$

$\frac{1}{a}+\frac{1}{b}-\frac{2k}{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}}=\frac{a+b}{ab}-\frac{2k}{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}}=\frac{1+{{k}^{2}}{{a}^{2}}{{b}^{2}}-2kab}{ab\left( 1+{{k}^{2}}{{a}^{2}}{{b}^{2}} \right)}=\frac{{{\left( kab-1 \right)}^{2}}}{ab\left( 1+{{k}^{2}}{{a}^{2}}{{b}^{2}} \right)}\ge 0,\forall k\in \mathbb{R}$

Z K Y

The post below has been deleted. Click to close.

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

41195 posts

sqing

#3 h Yesterday at 1:14 PM

Y by

Nice.Thanks.

This post has been edited 1 time. Last edited by sqing, Yesterday at 1:21 PM

Z K Y

The post below has been deleted. Click to close.

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

41195 posts

sqing

#4 h Yesterday at 1:19 PM

Y by

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$ Where $16\geq k>0 .$
$\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$

This post has been edited 3 times. Last edited by sqing, Yesterday at 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.

35 posts

#5 h Yesterday at 1:29 PM

Y by

sqing wrote:

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$

$a+b=1$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$ <===> $a^2b^2- \frac{17}{4}ab+1 \geq 0$ <===> $ab \leq \frac{1}{4}$

Z K Y

The post below has been deleted. Click to close.

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

41195 posts

sqing

#6 h Yesterday at 1:48 PM

Y by

Nice.Thanks.

Z K Y

The post below has been deleted. Click to close.

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

35 posts

#8 h Yesterday at 1:54 PM

Y by

sqing wrote:

Let $a,b> 0$ and $a+b=1 .$ Prove that
$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$ Where $16\geq k>0 .$
$\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{17}{4}}{1+ a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{5}{1+4 a^2b^2}$ $\frac{1}{a}+\frac{1}{b}\geq \frac{\frac{25}{4}}{1+9a^2b^2}$

$\frac{1}{a}+\frac{1}{b}\geq \frac{4+\frac{k}{4}}{1+ ka^2b^2}$ $(4+ \frac{k}{4})ab \leq 1+ka^2b^2$ <===> $(kab-4)(4ab-1) \geq 0$ <===> $ab \leq \frac{1}{4}$

Z K Y

The post below has been deleted. Click to close.

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

41195 posts

sqing

#9 h Yesterday at 2:46 PM

Y by

Nice.Thanks.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a