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70 posts

#1 h Apr 21, 2025, 1:51 PM • 1 Y

Y by RainbowJessa

Let $a,b,c>0$ and $a+b+c=3$ . Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

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#2 h Apr 21, 2025, 2:21 PM • 1 Y

Y by RainbowJessa

Cobedangiu wrote:

Let $a,b,c>0$ and $a+b+c=3$ . Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

Tangent line :whistling:

:whistling:

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148 posts

#3 h Apr 21, 2025, 2:38 PM • 1 Y

Y by RainbowJessa

giangtruong13 wrote:

Bài này giống với bài BĐT trong đề thi HSG Thái Bình năm 2024-2025

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#4 h Apr 21, 2025, 3:17 PM • 1 Y

Y by RainbowJessa

use the constants

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70 posts

#5 h Apr 21, 2025, 4:00 PM • 1 Y

Y by RainbowJessa

giangtruong13 wrote:

giangtruong13 wrote:

Bài này giống với bài BĐT trong đề thi HSG Thái Bình năm 2024-2025

nó có thể giải đc chỉ với Schwarz .-.

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30256 posts

arqady

#6 h Apr 22, 2025, 8:20 AM • 1 Y

Y by RainbowJessa

Cobedangiu wrote:

Let $a,b,c>0$ and $a+b+c=3$ . Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

It's just Popoviciu for $f(x)=\frac{1}{x}$ on $(0,+\infty).$

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209 posts

#7 h Apr 23, 2025, 3:45 PM • 1 Y

Y by RainbowJessa

m4thbl3nd3r wrote:

Cobedangiu wrote:

Let $a,b,c>0$ and $a+b+c=3$ . Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

Tangent line :whistling:

:whistling:

mhmm,Tangent Line
I like

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30256 posts

arqady

#8 h Apr 23, 2025, 5:25 PM • 1 Y

Y by RainbowJessa

IceyCold wrote:

mhmm,Tangent Line
I like

Did you try? I think, it does not help here.

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ReticulatedPython

713 posts

ReticulatedPython

#9 h Apr 24, 2025, 3:03 PM • 1 Y

Y by RainbowJessa

Interesting problem. I suspect that AM-GM might be applicable here, since equality is achieved at $a=b=c=1$ (which is the AM-GM equality condition).

This post has been edited 1 time. Last edited by ReticulatedPython, Apr 24, 2025, 3:04 PM

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209 posts

#10 h Apr 28, 2025, 12:37 PM • 1 Y

Y by RainbowJessa

arqady wrote:

IceyCold wrote:

mhmm,Tangent Line
I like

Did you try? I think, it does not help here.

It was one of our test.The graders marked my method correct,so I hope darn well I am right lol-

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70 posts

#11 h Apr 28, 2025, 4:16 PM • 1 Y

Y by RainbowJessa

IceyCold wrote:

arqady wrote:

IceyCold wrote:

mhmm,Tangent Line
I like

Did you try? I think, it does not help here.

It was one of our test.The graders marked my method correct,so I hope darn well I am right lol-

Can you write your method?

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ReticulatedPython

713 posts

ReticulatedPython

#12 h Apr 28, 2025, 7:10 PM • 1 Y

Y by RainbowJessa

IceyCold wrote:

arqady wrote:

IceyCold wrote:

mhmm,Tangent Line
I like

Did you try? I think, it does not help here.

It was one of our test.The graders marked my method correct,so I hope darn well I am right lol-

Yeah can you share the method with us?

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131 posts

#13 h Apr 28, 2025, 7:43 PM

Y by

Cobedangiu wrote:

Let $a,b,c>0$ and $a+b+c=3$ . Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$

$a=\frac{3x}{x+y+z},...$
$\sum_{sym} x^4y^2-x^4yz+x^3y^3-x^2y^2z^2\ge0.$

This post has been edited 1 time. Last edited by Edward_Tur, Apr 28, 2025, 7:44 PM

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#14 h Apr 30, 2025, 1:21 AM

Y by

Cobedangiu wrote:

Can you write your method?

Fakesolve

This post has been edited 3 times. Last edited by IceyCold, Apr 30, 2025, 1:36 AM

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#15 h Apr 30, 2025, 1:28 AM

Y by

IceyCold wrote:

Cobedangiu wrote:

Can you write your method?

Show that $\frac{4}{3-c} \le \frac{1}{c} -2c + 3$ .

This is equivalent to $\frac{(c-1)^2(2c+3)}{c(c-3)} \le 0$ ,trivially true.

Never mind,I see the issue now.
Sorry for a waste of time-

This post has been edited 1 time. Last edited by IceyCold, Apr 30, 2025, 1:34 AM

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arqady

#17 h May 1, 2025, 3:37 AM

Y by

ReticulatedPython wrote:

Yeah can you share the method with us?

See here

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