a+b+c=2 ine

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#1 h Jun 25, 2024, 4:00 PM • 3 Y

Y by jokehim, Zuyong, TNKT

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{a}{\sqrt{2a+3bc}}+\frac{b}{\sqrt{2b+3ca}}+\frac{c}{\sqrt{2c+3ab}} \le \sqrt{\frac{2}{ab+bc+ca}}. }$

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#2 h Jun 26, 2024, 8:34 AM

Y by

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{a}{\sqrt{2a+3bc}}+\frac{b}{\sqrt{2b+3ca}}+\frac{c}{\sqrt{2c+3ab}} \le \sqrt{\frac{2}{ab+bc+ca}}. }$

By C-S $\sum_{cyc}\frac{a}{\sqrt{2a+3bc}}\leq\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{2a+3bc}}\leq \sqrt{\frac{2}{ab+bc+ca}},$ where the last inequality is nice, but easy.

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#3 h Jun 26, 2024, 8:37 AM • 4 Y

Y by arqady, jokehim, Zuyong, TNKT

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{a}{\sqrt{2a+3bc}}+\frac{b}{\sqrt{2b+3ca}}+\frac{c}{\sqrt{2c+3ab}} \le \sqrt{\frac{2}{ab+bc+ca}}. }$

$\frac{a}{\sqrt{7a+3bc}}+\frac{b}{\sqrt{7b+3ca}}+\frac{c}{\sqrt{7c+3ab}}\ge \sqrt{\frac{2}{7}\left(ab+bc+ca+1\right)}.$

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#5 h Jul 3, 2024, 9:19 PM • 3 Y

Y by jokehim, Zuyong, TNKT

#3 is very hard.
(Wrong!) Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{red}{\frac{bc+1}{8-3a^2}+\frac{ca+1}{8-3b^2}+\frac{ab+1}{8-3c^2}\ge \frac{13}{20}.}$

This post has been edited 1 time. Last edited by KhuongTrang, Aug 1, 2024, 8:26 AM

Z K Y

The post below has been deleted. Click to close.

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

NO_SQUARES

1135 posts

NO_SQUARES

#6 h Jul 3, 2024, 10:05 PM

Y by

KhuongTrang wrote:

#3 is very hard.
Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{bc+1}{8-3a^2}+\frac{ca+1}{8-3b^2}+\frac{ab+1}{8-3c^2}\ge \frac{13}{20}.}$

Isn't this inequality wrong for $a=1.99$ , $b=0.01$ and $c=0$ or I'm mistaken?

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#7 h Jul 4, 2024, 3:37 AM • 3 Y

Y by jokehim, Zuyong, TNKT

NO_SQUARES wrote:

Isn't this inequality wrong for $a=1.99$ , $b=0.01$ and $c=0$ or I'm mistaken?

Thank you, you're right.
#3 is very hard.
Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{bc}{\sqrt{5a+4}}+\frac{ca}{\sqrt{5b+4}}+\frac{ab}{\sqrt{5c+4}}\le \frac{1}{2}.}$

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#8 h Jul 4, 2024, 4:45 AM

Y by

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{bc}{\sqrt{5a+4}}+\frac{ca}{\sqrt{5b+4}}+\frac{ab}{\sqrt{5c+4}}\le \frac{1}{2}.}$

$\sum_{cyc}\frac{bc}{\sqrt{5a+4}}\leq\sqrt{\sum_{cyc}bc\sum_{cyc}\frac{bc}{5a+4}}\leq\frac{1}{2}.$

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#10 h Aug 1, 2024, 8:25 AM • 3 Y

Y by jokehim, Zuyong, TNKT

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{a}{\sqrt{2a+3bc}}+\frac{b}{\sqrt{2b+3ca}}+\frac{c}{\sqrt{2c+3ab}} \le \sqrt{\frac{2}{ab+bc+ca}}. }$

$\frac{a}{\sqrt{7a+3bc}}+\frac{b}{\sqrt{7b+3ca}}+\frac{c}{\sqrt{7c+3ab}}\ge \sqrt{\frac{2}{7}\left(ab+bc+ca+1\right)}.$

Have you tried Holder, arqady?

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#11 h Aug 1, 2024, 9:57 AM

Y by

KhuongTrang wrote:

Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that
$\frac{a}{\sqrt{7a+3bc}}+\frac{b}{\sqrt{7b+3ca}}+\frac{c}{\sqrt{7c+3ab}}\ge \sqrt{\frac{2}{7}\left(ab+bc+ca+1\right)}.$

Have you tried Holder, arqady?

I tried. We need non-trivial Holder here. Nice inequality!

This post has been edited 1 time. Last edited by arqady, Aug 1, 2024, 9:57 AM

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#13 h Aug 20, 2024, 3:10 AM • 3 Y

Y by jokehim, Zuyong, TNKT

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{3a-2}{\sqrt{a+bc}}+\frac{3b-2}{\sqrt{b+ca}}+\frac{3c-2}{\sqrt{c+ab}}\le 0. }$

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#14 h Aug 20, 2024, 4:17 AM

Y by

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{3a-2}{\sqrt{a+bc}}+\frac{3b-2}{\sqrt{b+ca}}+\frac{3c-2}{\sqrt{c+ab}}\le 0. }$

SOS, of course.

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#17 h Sep 2, 2024, 2:08 PM • 3 Y

Y by jokehim, Zuyong, TNKT

arqady wrote:

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{3a-2}{\sqrt{a+bc}}+\frac{3b-2}{\sqrt{b+ca}}+\frac{3c-2}{\sqrt{c+ab}}\le 0. }$

SOS, of course.

It is $\color{blue}{\frac{b+c-2a}{\sqrt{a^2+ab+2bc+ac}}+\frac{c+a-2b}{\sqrt{b^2+bc+2ca+ab}}+\frac{a+b-2c}{\sqrt{c^2+ca+2ab+bc}}\ge 0.}$ Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{1}{2ab+1}+\frac{1}{2bc+1}+\frac{1}{2ca+1}\ge \frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}. }$ Equality holds iff $a=b=c=\dfrac{2}{3}$ or $a=b=\dfrac{1}{2},c=1$ and any cyclic permutations.

This post has been edited 1 time. Last edited by KhuongTrang, Sep 6, 2024, 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.

arqady

30261 posts

arqady

#18 h Sep 2, 2024, 4:55 PM

Y by

KhuongTrang\ wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{1}{2ab+1}+\frac{1}{2bc+1}+\frac{1}{2ca+1}\ge \frac{1}{2a^2+1}+\frac{1}{2b^2+1}+\frac{1}{2c^2+1}. }$

We need to prove that:
$\sum_{cyc}\left(\frac{1}{2bc+1}-\frac{1}{2a^2+1}\right)\geq0$ or
$\sum_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{(2bc+1)(2a^2+1)}\geq0$ or
$\sum_{cyc}(a-b)\left(\frac{a+c}{(2bc+1)(2a^2+1)}-\frac{b+c}{(2ac+1)(2b^2+1)}\right)\geq0$ or
$\sum_{cyc}(a-b)^2(2c^2+1)^2(2ab+1)(1-2ab)\geq0$ and the rest is smooth.

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#24 h Sep 9, 2024, 12:27 PM • 3 Y

Y by jokehim, Zuyong, TNKT

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{a\sqrt{b+bc+c}}{b+c}+\frac{b\sqrt{c+ca+a}}{c+a}+\frac{c\sqrt{a+ab+b}}{a+b}\ge 2.}$

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#25 h Sep 9, 2024, 4:32 PM

Y by

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{a\sqrt{b+bc+c}}{b+c}+\frac{b\sqrt{c+ca+a}}{c+a}+\frac{c\sqrt{a+ab+b}}{a+b}\ge 2.}$

By Holder $\left(\sum_{cyc}\frac{a\sqrt{b+bc+c}}{b+c}\right)^2\sum_{cyc}\frac{a(b+c)^2}{b+bc+c}\geq(a+b+c)^3$ and it's enough to prove that:
$a+b+c\geq2\sum_{cyc}\frac{a(b+c)^2}{(b+c)(a+b+c)+2bc}$ or
$\sum_{sym}(a^6b+3a^5b^2-4a^4b^3+4a^5bc+9a^4b^2c-10a^3b^3c-3a^3b^2c^2)\geq0,$ which is true by Muirhead.

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#27 h Sep 17, 2024, 11:33 AM • 3 Y

Y by jokehim, Zuyong, TNKT

#25
Nice again, arqady.
Problem. Given non-negative real numbers satisfying $a+b+c=2.$ Prove that $\color{blue}{\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge 3\sqrt{ab+bc+ca-\frac{3}{4}abc}.}$

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#28 h Sep 17, 2024, 5:27 PM • 1 Y

Y by ehuseyinyigit

KhuongTrang wrote:

Problem. Given non-negative real numbers satisfying $a+b+c=2.$ Prove that $\color{blue}{\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge 3\sqrt{ab+bc+ca-\frac{3}{4}abc}.}$

Easiest Holder helps:
$\left(\sum_{cyc}\sqrt{a+bc}\right)^2\sum_{cyc}\frac{1}{a+bc}\geq27.$

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#29 h Sep 29, 2024, 4:38 AM • 4 Y

Y by ehuseyinyigit, jokehim, Zuyong, TNKT

#28, it turns out obvious by AM-GM
$\color{blue}{(a+bc)(b+ca)(c+ab)\ge \left(ab+bc+ca-\frac{3}{4}abc\right)^3.}$

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\frac{3}{2}\ge \sqrt{\frac{a}{2a^2+2a+bc}}+\sqrt{\frac{b}{2b^2+2b+ca}}+\sqrt{\frac{c}{2c^2+2c+ab}}\ge 1.$

Z K Y

The post below has been deleted. Click to close.

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

nikiiiita

100 posts

nikiiiita

#30 h Sep 29, 2024, 6:51 AM

Y by

KhuongTrang wrote:

#28, it turns out obvious by AM-GM
$\color{blue}{(a+bc)(b+ca)(c+ab)\ge \left(ab+bc+ca-\frac{3}{4}abc\right)^3.}$

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\frac{3}{2}\ge \sqrt{\frac{a}{2a^2+2a+bc}}+\sqrt{\frac{b}{2b^2+2b+ca}}+\sqrt{\frac{c}{2c^2+2c+ab}}\ge 1.$

For RHS, its clearly that:
$16 \sum_{cyc} \frac{ab}{(2a^{2}+2a+bc)(2b^2+2b+ca)}+297\frac{abc}{(2a^{2}+2a+bc)(2b^2+2b+ca)(2c^2+2c+ab)} \ge 1$

Z K Y

The post below has been deleted. Click to close.

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

jokehim

1028 posts

jokehim

#31 h Sep 29, 2024, 9:25 AM

Y by

nikiiiita, what do you mean?

Z K Y

The post below has been deleted. Click to close.

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

nikiiiita

100 posts

nikiiiita

#32 h Sep 29, 2024, 10:14 AM

Y by

jokehim wrote:

nikiiiita, what do you mean?

Lemma: For all $x,y,z$ are non-negative reals number such that:
$\boxed{16x^2y^2+16y^2z^2+16z^2x^2+297x^2y^2z^2 \ge 1}$
then we will have $x+y+z \ge 1$ as a consequence.

This lemma can be easily proved by uvw or MV

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#33 h Oct 25, 2024, 6:28 PM

Y by

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\frac{3}{2}\ge \sqrt{\frac{a}{2a^2+2a+bc}}+\sqrt{\frac{b}{2b^2+2b+ca}}+\sqrt{\frac{c}{2c^2+2c+ab}}$

By Jensen $\sum_{cyc}\sqrt{\frac{a}{2a^2+2a+bc}}\leq\sqrt{3\sum_{cyc}\frac{a}{2a^2+2a+bc}}$ and it's enough to prove that:
$\sum_{cyc}\frac{a}{3a^2+ab+ac+bc}\leq\frac{3}{2(a+b+c)}$ or
$\sum_{sym}(a^4b^2+a^4bc+7a^3b^3-10a^3b^2c+a^2b^2c^2)\geq0,$ which is true by Schur and Murhead.

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#34 h Oct 25, 2024, 7:09 PM

Y by

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\sqrt{\frac{a}{2a^2+2a+bc}}+\sqrt{\frac{b}{2b^2+2b+ca}}+\sqrt{\frac{c}{2c^2+2c+ab}}\ge 1.$

Also Holder with $\sum_{cyc}a^2(3a^2+ab+ac+bc)(b+c)^3$ helps.

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#36 h Oct 29, 2024, 9:17 AM • 3 Y

Y by jokehim, Zuyong, TNKT

Problem. Given positive real number satisfying $a+b+c=2.$ Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{4abc}\ge 4(a^{2}+b^{2}+c^{2}).$

Z K Y

The post below has been deleted. Click to close.

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

Nguyenhuyen_AG

3331 posts

Nguyenhuyen_AG

#37 h Oct 29, 2024, 10:07 AM

Y by

KhuongTrang wrote:

Problem. Given positive real number satisfying $a+b+c=2.$ Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{4abc}\ge 4(a^{2}+b^{2}+c^{2}).$

It's equivalent to
$\sum (b+c)(7a^2-b^2-c^2-10ab+14bc-10ca)^2 \geqslant 0.$

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#38 h Feb 21, 2025, 12:20 PM • 3 Y

Y by jokehim, Zuyong, TNKT

Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\sqrt{3a^{2}\left(b+c\right)+1}+\sqrt{3b^{2}\left(c+a\right)+1}+\sqrt{3c^{2}\left(a+b\right)+1}\le 5.}$

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#39 h Feb 21, 2025, 12:29 PM

Y by

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\sqrt{3a^{2}\left(b+c\right)+1}+\sqrt{3b^{2}\left(c+a\right)+1}+\sqrt{3c^{2}\left(a+b\right)+1}\le 5.}$

The Vasc's LCF Theorem helps.

Z K Y

The post below has been deleted. Click to close.

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

KhuongTrang

731 posts

KhuongTrang

#40 h Feb 21, 2025, 12:32 PM • 3 Y

Y by jokehim, Zuyong, TNKT

arqady wrote:

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\sqrt{3a^{2}\left(b+c\right)+1}+\sqrt{3b^{2}\left(c+a\right)+1}+\sqrt{3c^{2}\left(a+b\right)+1}\le 5.}$

The Vasc's LCF Theorem helps.

C-S with $a+1$ also helps.
The following inequality is very nice :-D

Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3} \ge 5.}$

This post has been edited 1 time. Last edited by KhuongTrang, Feb 21, 2025, 12:32 PM

Z K Y

The post below has been deleted. Click to close.

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

arqady

30261 posts

arqady

#41 h Feb 21, 2025, 2:51 PM

Y by

KhuongTrang wrote:

The following inequality is very nice :-D

Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\sqrt{a-2bc+3}+\sqrt{b-2ca+3}+\sqrt{c-2ab+3} \ge 5.}$