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lgx57   4
N 4 minutes ago by sqing
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Find min of $\dfrac{a^2}{ab+1}+\dfrac{b^2+2}{a+b}$
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lgx57
Yesterday at 3:01 PM
sqing
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an exponential inequality with two variables
teresafang   1
N 13 minutes ago by CHESSR1DER
x and y are positive real numbers.prove that [(x^y)/y]^(1/2)+[(y^x)/x]^(1/2)>=2.
sorry.I’m not good at English.Also I don’t know how to use Letax.
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teresafang
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CHESSR1DER
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hard problem
Cobedangiu   15
N May 1, 2025 by arqady
Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a} \le \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3$
15 replies
Cobedangiu
Apr 21, 2025
arqady
May 1, 2025
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hard problem
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Cobedangiu
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Cobedangiu
#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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m4thbl3nd3r
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m4thbl3nd3r
#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:
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giangtruong13
140 posts
giangtruong13
#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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Jackson0423
66 posts
Jackson0423
#4 h  Apr 21, 2025, 3:17 PM • 1 Y
Y by RainbowJessa
use the constants
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Cobedangiu
66 posts
Cobedangiu
#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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arqady
30231 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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IceyCold
208 posts
IceyCold
#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:

mhmm,Tangent Line
I like
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arqady
30231 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
617 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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IceyCold
208 posts
IceyCold
#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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Cobedangiu
66 posts
Cobedangiu
#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
617 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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Edward_Tur
127 posts
Edward_Tur
#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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IceyCold
208 posts
IceyCold
#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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IceyCold
208 posts
IceyCold
#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
30231 posts
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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