Stronger inequality than an old result

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KhuongTrang

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KhuongTrang

#1 h Aug 1, 2024, 3:13 PM • 4 Y

Y by kiyoras_2001, NguyenVanHoa29, Zuyong, TNKT

Problem. Find the best constant $k$ satisfying $(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$ holds for all $a,b,c\ge 0: ab+bc+ca>0.$

This post has been edited 2 times. Last edited by KhuongTrang, Aug 2, 2024, 6:43 AM

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ehuseyinyigit

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ehuseyinyigit

#2 h Aug 1, 2024, 5:55 PM

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Even the 1996 Iran's inequality was strong enough, can't think about this problem :lol:

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arqady

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arqady

#3 h Aug 1, 2024, 8:41 PM

Y by

KhuongTrang wrote:

Problem. Find the best constant $k$ satisfying $(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$ holds for all $a,b,c\ge 0: ab+bc+ca>0.$

$k_{max}=\frac{27}{8}.$

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KhuongTrang

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KhuongTrang

#7 h Aug 4, 2024, 5:18 AM • 3 Y

Y by NguyenVanHoa29, Zuyong, TNKT

Problem. Find the best constant $k$ satisfying $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k\cdot\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\ge 3$ holds for all $a,b,c>0.$

This post has been edited 3 times. Last edited by KhuongTrang, Aug 6, 2024, 3:16 AM

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Nguyenhuyen_AG

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Nguyenhuyen_AG

#9 h Aug 4, 2024, 12:09 PM

Y by

KhuongTrang wrote:

Problem. Find the best constant $k$ satisfying $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k\cdot\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\ge 3$ holds for all $a,b,c>0.$

It's old $k = 8.$

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KhuongTrang

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KhuongTrang

#11 h Aug 5, 2024, 2:54 PM • 3 Y

Y by NguyenVanHoa29, Zuyong, TNKT

Nguyenhuyen_AG wrote:

KhuongTrang wrote:

Problem. Find the best constant $k$ satisfying $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+k\cdot\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}\ge 3$ holds for all $a,b,c>0.$

It's old $k = 8.$

Are you sure?

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Nguyenhuyen_AG

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Nguyenhuyen_AG

#12 h Aug 5, 2024, 3:02 PM

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KhuongTrang wrote:

Are you sure?

Not sure =)) I made a mistake $k_{\max} = 15.50669170$ :sleeping:

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KhuongTrang

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KhuongTrang

#14 h Aug 7, 2024, 5:05 AM • 4 Y

Y by DipoleOfMonorak, NguyenVanHoa29, Zuyong, TNKT

@above, it's same as mine.
Problem. Find the best constant $k$ satisfying $\frac{ab}{(a+b)^{2}}+\frac{bc}{(b+c)^{2}}+\frac{ca}{(c+a)^{2}}\le \frac{177}{500}+k\cdot\frac{ab+bc+ca}{a^2+b^2+c^2}$ holds for $\forall a,b,c \in R :(a+b)(b+c)(c+a) \neq 0.$

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DipoleOfMonorak

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DipoleOfMonorak

#15 h Aug 7, 2024, 9:36 AM

Y by

KhuongTrang wrote:

@above, it's same as mine.
Problem. Find the best constant $k$ satisfying $\frac{ab}{(a+b)^{2}}+\frac{bc}{(b+c)^{2}}+\frac{ca}{(c+a)^{2}}\le \frac{177}{500}+k\cdot\frac{ab+bc+ca}{a^2+b^2+c^2}$ holds for $\forall a,b,c \in R :(a+b)(b+c)(c+a) \neq 0.$

$\frac{177}{500} + \frac{54}{125}\cdot\frac{ab+bc+ca}{a^2+b^2+c^2} - \frac{ab}{(a+b)^{2}}-\frac{bc}{(b+c)^{2}}-\frac{ca}{(c+a)^{2}}=$ $=\sum\limits_{cyc}{\frac{\Bigl(232(c+b)a^3-(639c^2+578bc+639b^2)a^2-(c+b)(521c^2-810bc+521b^2)a+bc(289c^2+1278bc+289b^2)\Bigr)^2}{2088000\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(a^2+b^2+c^2\right)}} +$ $+\frac{\Bigl(7\left(a^3(b+c)+b^3(c+a)+c^3(a+b)\right) -6\left(a^2b^2+b^2c^2+c^2a^2\right)-32abc\left(a+b+c\right)\Bigr)^2}{750\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(a^2+b^2+c^2\right)}+$ $+\frac{431\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2\left(a^2+b^2+c^2-ab-bc-ca\right)}{4640\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(a^2+b^2+c^2\right)}$
maxima code

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KhuongTrang

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KhuongTrang

#16 h Aug 7, 2024, 9:55 AM • 3 Y

Y by NguyenVanHoa29, Zuyong, TNKT

@above, thank you. Your result is same as mine.
Problem. Find the best constant $k$ satisfying $k\left(a+b+c-3\right)abc+6\le 2(a+b+c).$ holds for $\forall a,b,c \ge 0: ab+bc+ca+abc=4.$

This post has been edited 1 time. Last edited by KhuongTrang, Aug 7, 2024, 1:02 PM

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DipoleOfMonorak

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DipoleOfMonorak

#17 h Aug 7, 2024, 11:42 AM

Y by

KhuongTrang wrote:

@above, thank you. Your result is same as mine.
Problem. Find the best constant $k$ satisfying $k\left(a+b+c-3\right)abc+6\ge 2(a+b+c).$ holds for $\forall a,b,c \ge 0: ab+bc+ca+abc=4.$

May be $k\left(a+b+c-3\right)abc+6\le 2(a+b+c)?$
Let $F(a,b,c)=2(a+b+c)-\Bigl(2\left(a+b+c-3\right)abc+6\Bigr),$ then

$F\left(\frac{2a}{b+c},\frac{2b}{c+a},\frac{2c}{a+b}\right)=\frac{2(bc^2+ac^2+b^2c+a^2c+ab^2+a^2b-6abc)(2a^3+2b^3+2c^3-bc^2-ac^2-b^2c-a^2c-ab^2-a^2b)}{(a+b)^2(c+a)^2(c+b)^2}=$

$=\sum\limits_{cyc}{\frac{2(a-b)^2(a-c)^2(5c^2+5b^2+2bc+6ac+6ab)}{3(a+b)^2(c+a)^2(c+b)^2}}$

maxima code

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KhuongTrang

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KhuongTrang

#22 h Aug 16, 2024, 2:19 AM • 3 Y

Y by NguyenVanHoa29, Zuyong, TNKT

@above, it's same as mine.
Problem. (Pham Le Van) Find the best constant $k$ satisfying $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}+k\cdot\left(\frac{a-b}{a+b}\right)^{2}\left(\frac{b-c}{b+c}\right)^{2}\left(\frac{c-a}{c+a}\right)^{2}$ holds for all $a,b,c\ge 0: ab+bc+ca>0.$

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Victoria_Discalceata1

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Victoria_Discalceata1

#23 h Aug 18, 2024, 6:49 AM • 1 Y

Y by kiyoras_2001

arqady wrote:

KhuongTrang wrote:

Problem. Find the best constant $k$ satisfying $(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$ holds for all $a,b,c\ge 0: ab+bc+ca>0.$

$k_{max}=\frac{27}{8}.$

@DipoleOfMonorak, can you please use your kung fu on this one, too?

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DipoleOfMonorak

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DipoleOfMonorak

#25 h Aug 18, 2024, 1:20 PM • 1 Y

Y by Victoria_Discalceata1

Victoria_Discalceata1 wrote:

arqady wrote:

KhuongTrang wrote:

Problem. Find the best constant $k$ satisfying $(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$ holds for all $a,b,c\ge 0: ab+bc+ca>0.$

$k_{max}=\frac{27}{8}.$

@DipoleOfMonorak, can you please use your kung fu on this one, too?

$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]- \frac{9}{4}-\frac{27}{8}\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}=$

$=\sum\limits_{cyc}\frac{bc\left(b-c\right)^4(b+c)\left(5\,b^2+11\,bc+5\,c^2\right)}{8\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(a+b+c\right)^3} + \sum\limits_{cyc}\frac{a\left(a-b\right)^2\left(a-c\right)^2\Bigl( 3\,(b+c)a^3+\left(5\,b^2+10\,bc+5\,c^2\right)a^2+4\,(b+c)\left(b^2+bc+c^2\right)a+5\,bc\left(b^2+c^2\right)\Bigr) }{8\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(a+b+c\right)^3}$

maxima code

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Victoria_Discalceata1

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Victoria_Discalceata1

#26 h Aug 18, 2024, 2:10 PM • 2 Y

Y by DipoleOfMonorak, ehuseyinyigit

Thanks, here is my best shot at it ( $p=a+b+c,\ q=ab+bc+ca,\ r=abc)$ :
$\Bigl(8pq\bigl(p^3-4pq+9r\bigr)+(3pq+5r)(pq-9r)\Bigr)\bigl(p^3-4pq+9r\bigr)+2r(pq-9r)^2\ge 0.$

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DipoleOfMonorak

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DipoleOfMonorak

#27 h Aug 18, 2024, 2:21 PM

Y by

Victoria_Discalceata1 wrote:

Thanks, here is my best shot at it ( $p=a+b+c,\ q=ab+bc+ca,\ r=abc)$ :
$\Bigl(8pq\bigl(p^3-4pq+9r\bigr)+(3pq+5r)(pq-9r)\Bigr)\bigl(p^3-4pq+9r\bigr)+2r(pq-9r)^2\ge 0.$

It's very nice

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KhuongTrang

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KhuongTrang

#32 h Oct 23, 2024, 1:18 PM • 3 Y

Y by NguyenVanHoa29, Zuyong, TNKT

Problem. Let $a,b,c$ be non-negative real numbers with $ab+bc+ca>0.$
Find the largest constant $k$ such that the following inequality holds $(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge k\cdot\frac{a^2+b^2+c^2-ab-bc-ca}{(a+b+c)^2}.$ When does equality occur?

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Victoria_Discalceata1

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Victoria_Discalceata1

#33 h Oct 23, 2024, 3:54 PM

Y by

KhuongTrang wrote:

Problem. Let $a,b,c$ be non-negative real numbers with $ab+bc+ca>0.$
Find the largest constant $k$ such that the following inequality holds $(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge k\cdot\frac{a^2+b^2+c^2-ab-bc-ca}{(a+b+c)^2}.$ When does equality occur?

Take $b=1,\ c=0$ to obtain $k\le 8+\frac{\left(a^2-3a+1\right)^2}{a\left(a^2-a+1\right)}$ . That means that taking $a=\frac{1}{2}\left(3+\sqrt{5}\right)$ we obtain $k\le 8$ . We will show that the given inequality holds with $k=8$ .
In terms of $p=a+b+c,\ q=ab+bc+ca,\ r=abc$ we are to prove $\frac{q\left(p^4-2p^2q+4pr+q^2\right)}{(pq-r)^2}\ge \frac{8\left(p^2-3q\right)}{p^2}$ which is $\left(p^2-3q\right)(pq-9r)r+\left(p^2-3q\right)r^2+pq\left(4p^2+15\left(p^2-3q\right)\right)r+p^2q\left(p^2-5q\right)^2\ge 0$ Equality occurs IFF $(a,b,c)$ is proportional to a permutation of $\left(0,\sqrt{5}-1,\sqrt{5}+1\right)$ .

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KhuongTrang

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KhuongTrang

#34 h Oct 26, 2024, 3:37 PM • 3 Y

Y by NguyenVanHoa29, Zuyong, TNKT

KhuongTrang wrote:

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c+abc=4.$ Prove that
$\color{blue}{\frac{1}{2ab+2bc+ca}+\frac{1}{2bc+2ca+ab}+\frac{1}{2ca+2ab+bc}\ge \frac{1}{a+b+3c}+\frac{1}{c+b+3a}+\frac{1}{a+c+3b}}.$ Equality holds iff $a=b=c=1$ or $a=b=2,c=0$ and any permutations.

Dear Victoria_Discalceata1, does your kung fu work for this problem?

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shilad11

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shilad11

#35 h Oct 26, 2024, 4:13 PM

Y by

Why is the font so small, dude?

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KhuongTrang

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KhuongTrang

#36 h Apr 22, 2025, 9:09 AM • 3 Y

Y by NguyenVanHoa29, Zuyong, TNKT

Problem. Let $a,b,c$ be three real numbers. Prove that $\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right)\ge 3\left(a+b+c\right)^{2}+\frac{8}{9}\left(ab+bc+ca-3\right)^{2}.$

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arqady

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arqady

#37 h Apr 22, 2025, 12:50 PM

Y by

KhuongTrang wrote:

Problem. Let $a,b,c$ be three real numbers. Prove that $\left(a^{2}+2\right)\left(b^{2}+2\right)\left(c^{2}+2\right)\ge 3\left(a+b+c\right)^{2}+\frac{8}{9}\left(ab+bc+ca-3\right)^{2}.$

Because $\prod_{cyc}(a^2+2)-3(a+b+c)^2-\frac{8}{9}(ab+ac+bc-3)^2=\frac{1}{9}\sum_{cyc}(3(a-b)^2+5c^2(a-b)^2+3c^2(ab-1)^2)\geq0.$

This post has been edited 2 times. Last edited by arqady, Apr 22, 2025, 12:51 PM

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