Interesting inequalities

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 inequalities

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.

41825 posts

sqing

#1 h Apr 15, 2025, 8:32 AM

Y by

Let $a,b$ be reals such that $a^2+ab+b^2 =1$ . Prove that
$\frac{8}{ 5 }> \frac{1}{ a^2+1 }+ \frac{1}{ b^2+1 } \geq 1$ $\frac{9}{ 5 }\geq\frac{1}{ a^4+1 }+ \frac{1}{ b^4+1 } \geq 1$ $\frac{27}{ 14 }\geq \frac{1}{ a^6+1 }+ \frac{1}{ b^6+1 } \geq 1$ Let $a,b$ be reals such that $a^2+ab+b^2 =3$ . Prove that
$\frac{13}{ 10 }> \frac{1}{ a^2+1 }+ \frac{1}{ b^2+1 } \geq \frac{1}{ 2 }$ $\frac{6}{ 5 }>\frac{1}{ a^4+1 }+ \frac{1}{ b^4+1 } \geq \frac{1}{ 5 }$ $\frac{1}{ a^6+1 }+ \frac{1}{ b^6+1 } \geq \frac{1}{ 14 }$

This post has been edited 3 times. Last edited by sqing, Apr 15, 2025, 8:53 AM

Z K Y

The post below has been deleted. Click to close.

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

552 posts

#2 h Apr 15, 2025, 10:55 AM

Y by

sqing wrote:

Let $a,b$ be reals such that $a^2+ab+b^2 =1$ . Prove that
$\frac{8}{ 5 }> \frac{1}{ a^2+1 }+ \frac{1}{ b^2+1 } \geq 1$

Since $a^2+b^2=2(a^2+ab+b^2)-(a+b)^2\le 2,$ it follows that

$\frac{1}{a^2+1}+\frac{1}{b^2+1}\ge \frac{4}{a^2+1+b^2+1}\ge 1.$
Let $c=-a-b.$ Then

$a+b+c=0,\quad ab+bc+ca=-1.$
Thus $a^2b^2 + b^2c^2 + c^2a^2 = (ab+bc+ca)^2 - 2(a+b+c)abc = 1$ and ,

$\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}=\frac{8}{a^2b^2c^2+4}\le2.$
Moreover, since it is easy to see that $c^2=(a+b)^2\le \frac{4}{3}(a^2+ab+b^2) = \frac 43,$ we have

$\frac{1}{a^2+1}+\frac{1}{b^2+1}\le 2-\frac{1}{c^2+1}\le\frac{11}{7}<\frac{8}{5}.$

This post has been edited 3 times. Last edited by lbh_qys, Apr 16, 2025, 2:00 AM

Z K Y

The post below has been deleted. Click to close.

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

41825 posts

sqing

#3 h Apr 15, 2025, 11:37 AM

Y by

Very very nice.Thank lbh_qys:
Let $a,b$ be reals such that $a^2+ab+b^2 =1$ . Prove that
$\frac{11}{ 7 }> \frac{1}{ a^2+1 }+ \frac{1}{ b^2+1 } \geq 1$

Z K Y

The post below has been deleted. Click to close.

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

552 posts

#4 h Apr 16, 2025, 2:23 AM

Y by

sqing wrote:

Let $a,b$ be reals such that $a^2+ab+b^2 =1$ . Prove that
$\frac{9}{ 5 }\geq\frac{1}{ a^4+1 }+ \frac{1}{ b^4+1 } \geq 1$

Let $c = -a - b$ . Then,
$a^2 + b^2 + c^2 = a^4 + b^4 + c^4 = 2,$ and it is evident that
$a^2,\ b^2,\ c^2 \leq \frac{4}{3}.$
Thus,

$\frac{1}{a^4+1}+\frac{1}{b^4+1} \geq \frac{4}{a^4+1+b^4+1} = \frac{4}{4-c^4} \geq 1,$
$\frac{1}{a^4+1}+\frac{1}{b^4+1}+\frac{1}{c^4+1} = \frac{1}{50}\sum \left( 27(2-a^2)-\frac{(4-3a^2)(3a^2-1)^2}{a^4+1} \right) \leq \frac{27}{50}\sum (2-a^2)=\frac{54}{25}.$
Therefore,
$\frac{1}{a^4+1}+\frac{1}{b^4+1} \leq \frac{54}{25}-\frac{1}{c^4+1}\leq \frac{9}{5}.$

Z K Y

The post below has been deleted. Click to close.

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

41825 posts

sqing

#5 h Apr 16, 2025, 2:34 AM

Y by

Very very nice.Thank lbh_qys.

Z K Y

The post below has been deleted. Click to close.

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

41825 posts

sqing

#6 h Apr 17, 2025, 8:09 AM

Y by

sqing wrote:

Let $a,b$ be reals such that $a^2+ab+b^2 =3$ . Prove that $\frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$

Solution of DAVROS:
Let $v=ab, f= \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab$ then $a^4+b^4=(3-v)^2-2v^2=9-6v-v^2$

$v\ge0 \implies 2v+v\le3 \implies v\le1$ and $v\le0 \implies -2v+v\le3 \implies v\ge-3$

$f \ge \frac4{a^4+b^4+6}+ab = \frac4{24-(v+3)^2}+v \ge \frac4{24}-3 = -\frac{17}6$ at $v=-3, a=-b=\pm \sqrt3$

$f = -\frac13\left(\frac{a^4}{a^4+3} + \frac{b^4}{b^4+3}-3ab-2\right)\le -\frac13\left( \frac{(a^2+b^2)^2}{a^4+b^4+6}-3ab-2\right)$

$f \le \frac13\left(\frac{2 (3 - a^2 b^2)}{6 + a^4 + b^4}+3ab+1\right) = \frac13\left(\frac{2 (3 - v^2)}{15-6v-v^2}+3v+1\right) = \frac{4(v-2)}{24-(v+3)^2}+v+1$

$f \le \frac{4(1-2)}{24-4^2}+1+1=\frac32$ at $v=1, a=b=\pm1$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a