Old problem

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chien than

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chien than

#1 h Sep 30, 2007, 2:55 AM • 3 Y

Y by Adventure10, HWenslawski, Mango247

Let $a;b;c$ be positive real numbers satisfying $abc=1$
Prove that:
$(a+b)(b+c)(c+a) \geq 4(a+b+c-1)$

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Svejk

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Svejk

#2 h Sep 30, 2007, 10:50 AM • 2 Y

Y by Adventure10, HWenslawski

Here is a stronger one: http://www.mathlinks.ro/viewtopic.php?t=128529
I think the one above could be solved by the same technique

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Mather

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Mather

#3 h Sep 30, 2007, 12:31 PM • 2 Y

Y by Adventure10, Mango247

chien than wrote:

Let $a;b;c$ be positive real numbers satisfying $abc = 1$
Prove that:
$(a + b)(b + c)(c + a) \geq 4(a + b + c - 1)$

$f(a,b,c) = LHS - RHS;t = \sqrt {ab}$
$\Rightarrow t^2c = 1$
$f(a,b,c) - f(t,t,c) = (\sqrt {a} - \sqrt {b})^2(ab + c^2 + c(\sqrt {a} + \sqrt {b}) - 4)$
$ab + c^2 + c\sqrt {a} + c\sqrt {b}\ge 4\sqrt [4]{a^3b^3c^3} = 4$
$\Rightarrow f(a,b,c)\ge f(t,t,c)$
$f(t,t,c) = 2t(t + \frac {1}{t^2})^2 - 4(2t + \frac {1}{t^2} - 1) = \frac {2(t - 1)^2(t^4 + 2t^3 - t^2 + 1)}{t^3}\ge 0$
So $f(a,b,c)\ge 0$

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arqady

30253 posts

arqady

#4 h Sep 30, 2007, 2:15 PM • 1 Y

Y by Adventure10

chien than wrote:

Let $a;b;c$ be positive real numbers satisfying $abc = 1$
Prove that:
$(a + b)(b + c)(c + a) \geq 4(a + b + c - 1)$

$(a + b)(b + c)(c + a) + 3k - 8\geq k(a + b + c)$ is true too
for all non-negative $a,$ $b$ and $c$ such that $abc = 1,$ where
$k_{best} = \min_{t\in(0,1)}\frac {2(t^2 + t + 1)^2}{t(2t + 1)} = 5.7028809...$

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FOURRIER

727 posts

FOURRIER

#5 h Sep 30, 2007, 8:17 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

$k_{best} = \min_{t\in(0,1)}\frac {2(t^2 + t + 1)^2}{t(2t + 1)} = 5.7028809...$

Could you tell me what is $k$ best and why its equal to this?

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arqady

30253 posts

arqady

#6 h Sep 30, 2007, 9:03 PM • 2 Y

Y by NaPrai, Adventure10

FOURRIER wrote:

arqady wrote:

$k_{best} = \min_{t\in(0,1)}\frac {2(t^2 + t + 1)^2}{t(2t + 1)} = 5.7028809...$

Could you tell me what is $k$ best and why its equal to this?

Let $f(a,b,c)=(a + b)(b + c)(c + a) + 3k - 8-k(a + b + c)$ and $a=\max\{a,b,c\}.$
Then $a\geq1$ and $f(a,b,c)-f\left(a,\sqrt{bc},\sqrt{bc}\right)=$
$=\left(\sqrt b-\sqrt c\right)^2\left(a^2+bc+a\left(\sqrt b+\sqrt c\right)^2-k\right)\geq\left(\sqrt b-\sqrt c\right)^2\left(6a\sqrt{bc}-k\right)=$
$=\left(\sqrt b-\sqrt c\right)^2\left(6\sqrt{a}-k\right)\geq0$ for $k\leq6.$
Thus, it remains to check out when $f(a,b,b)\geq0,$ where $a=\frac{1}{b^2}$ and $b\in(0,1].$
We obtain $f(a,b,b)\geq0\Leftrightarrow2b^6-2kb^4+(3k-4)b^3-kb+2\geq0\Leftrightarrow$
$\Leftrightarrow2(b^3-1)^2-kb(2b^3-3b^2+1)\geq0\Leftrightarrow$
$\Leftrightarrow(b-1)^2\left(2(b^2+b+1)^2-(2b^2+b)k\right)\geq0.$
Id est, $k\leq\frac{2(b^2+b+1)^2}{2b^2+b}.$ But $\min\frac{2(b^2+b+1)^2}{2b^2+b}<6.$
Thus, $k_{best}=\min\frac{2(b^2+b+1)^2}{2b^2+b}.$
If $k>k_{best}$ then $f(a,b,b)$ can be negative.
If $k\leq k_{best}$ then $f(a,b,c)\geq f\left(a,\sqrt{bc},\sqrt{bc}\right)$ and $f(a,b,b)\geq0.$
Hence, $f(a,b,c)\geq0$ is true.

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FOURRIER

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FOURRIER

#7 h Oct 1, 2007, 8:06 PM • 1 Y

Y by Adventure10

arqady wrote:

Thus, it remains to check out when $f(a,b,b)\geq0,$ where $a = \frac {1}{b^2}$ and $b\in(0,1].$

First, thank you soooo much for your explanation!!

I think it remains to check $f(a,\sqrt{ab},\sqrt{ab})\geq0,$
Why $a = \frac {1}{b^2}$ ?

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FOURRIER

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FOURRIER

#8 h Oct 1, 2007, 9:07 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Mather wrote:

$ab + c^2 + c\sqrt {a} + c\sqrt {b}\ge 4\sqrt [4]{a^3b^3c^3} = 4$

If you used $AM$ $GM$ here then Its false

If you used something else please tell me about it

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arqady

30253 posts

arqady

#9 h Oct 2, 2007, 5:23 AM • 2 Y

Y by Adventure10, Mango247

FOURRIER wrote:

arqady wrote:

Thus, it remains to check out when $f(a,b,b)\geq0,$ where $a = \frac {1}{b^2}$ and $b\in(0,1].$

Why $a = \frac {1}{b^2}$ ?

Because $b = c$ and $abc = 1.$

FOURRIER wrote:

I think it remains to check $f(a,\sqrt {ab},\sqrt {ab})\geq0,$

We have proved that $f(a,b,c)\geq f(a,\sqrt {bc},\sqrt {bc}).$
Hence, it remains to ptove that $f(a,\sqrt {bc},\sqrt {bc})\geq0,$ where $abc=1.$
Thus, it remains to ptove that $f(a,b,b)\geq0,$ where $abb=1$ since, $a\cdot\sqrt{bc}\cdot\sqrt{bc}=abc=1.$

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FOURRIER

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FOURRIER

#10 h Oct 2, 2007, 11:30 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Hence, it remains to ptove that $f(a,\sqrt {bc},\sqrt {bc})\geq0,$ where $abc = 1.$
Thus, it remains to ptove that $f(a,b,b)\geq0,$ where $abb = 1$ since, $a\cdot\sqrt {bc}\cdot\sqrt {bc} = abc = 1.$

I think we put $t=\sqrt{bc}$
It remains to prove $f(a,t,t)\geq0$ where $at^2=1$

Right?

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Mather

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Mather

#11 h Oct 2, 2007, 11:35 AM • 3 Y

Y by Adventure10, Mango247, Mango247

FOURRIER wrote:

Mather wrote:

$ab + c^2 + c\sqrt {a} + c\sqrt {b}\ge 4\sqrt [4]{a^3b^3c^3} = 4$

If you used $AM$ $GM$ here then Its false

If you used something else please tell me about it

I'm sorry

Suppose $c = Max(a,b,c)$
so $1 = abc\le c^3 \Rightarrow c\ge 1$
$ab + c^2 + c\sqrt {a} + c\sqrt {b}\ge 4\sqrt [4]{\sqrt [2]{ab}abc^4}\ge 4\sqrt [12]{abc} = 4$

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arqady

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arqady

#12 h Oct 2, 2007, 12:27 PM • 2 Y

Y by Adventure10, Mango247

FOURRIER wrote:

arqady wrote:

Hence, it remains to ptove that $f(a,\sqrt {bc},\sqrt {bc})\geq0,$ where $abc = 1.$
Thus, it remains to ptove that $f(a,b,b)\geq0,$ where $abb = 1$ since, $a\cdot\sqrt {bc}\cdot\sqrt {bc} = abc = 1.$

I think we put $t = \sqrt {bc}$
It remains to prove $f(a,t,t)\geq0$ where $at^2 = 1$

Right?

Yes!

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FOURRIER

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FOURRIER

#13 h Oct 2, 2007, 1:00 PM • 2 Y

Y by Adventure10, Mango247

Mather wrote:

$f(a,b,c) - f(t,t,c) = (\sqrt {a} - \sqrt {b})^2(ab + c^2 + c(\sqrt {a} + \sqrt {b}) - 4)$

This is FALSE

I Expanded it many many times

Just give: $a=4$ $b=9$ $c=\frac{1}{36}$ It will give you $\frac{41653}{36^2}=\frac{42373}{36^2}$

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Mather

42 posts

Mather

#14 h Oct 2, 2007, 1:33 PM • 1 Y

Y by Adventure10

FOURRIER wrote:

Mather wrote:

$f(a,b,c) - f(t,t,c) = (\sqrt {a} - \sqrt {b})^2(ab + c^2 + c(\sqrt {a} + \sqrt {b}) - 4)$

This is FALSE

I Expanded it many many times

Just give: $a = 4$ $b = 9$ $c = \frac {1}{36}$ It will give you $\frac {41653}{36^2} = \frac {42373}{36^2}$

Sorry
$f(a,b,c) - f(t,t,c) = (\sqrt {a} - \sqrt {b})^2(ab + c^2 + c(\sqrt {a} + \sqrt {b})^2 - 4)$

Done !

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nguyenvuthanhha

482 posts

nguyenvuthanhha

#15 h Feb 11, 2010, 4:56 AM • 4 Y

Y by AB2016, wrc_, Adventure10, Mango247

This problem is very easy :

Note that : $(ab + bc + ca )(a+b+c) = (a+b)(b+c)(c+a) + abc = (a+b)(b+c)(c+a) +1$

so the initial ineq is equivalent to :

$(ab + bc + ca )(a+b+c) + 3 \ge 4 (a+b+c)$

so we need to prove : $\frac{3}{a+b+c} + (ab+bc+ca) \ge 4$

from the well-known lemma : $(ab+bc+ca)^2 \ge 3abc(a+b+c)$ and the hypothesis $abc=1$

we deduce that : $(ab+bc+ca)^2 \ge 3(a+b+c)$ and it's really trivial that : $ab +bc + ca \ge 3$

Now , applying $AM-GM$ for $4$ positive real numbers

$\frac{3}{a+b+c} + (ab+bc+ca) = \frac{3}{a+b+c} + \frac{ab+bc+ca}{3} + \frac{ab+bc+ca}{3} + \frac{ab+bc+ca}{3} \ge 4 \sqrt[4]{ \frac{ (ab+bc+ca)^3}{9(a+b+c)}} = 4 \sqrt[4]{ \frac{ (ab+bc+ca)^2}{3(a+b+c)} \cdot \frac{(ab+bc+ca)}{3}} \ge 4$ , Done

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SBM

780 posts

SBM

#35 h Nov 13, 2020, 9:36 AM

Y by

sqing wrote:

Ji Chen (2019-09-21):
Let $a,b,c$ be positive real numbers .Prove that $c(a^3+b^3+c^3-3-4c)> 7(abc-ab-c).$ (p/6262708495)

We have:
$c(a^3+b^3+c^3-3-4c)-7(abc-ab-c)$ $= -\left[ 3ac+3bc+7c-7 \right] ab+c \left[ \left( a +b \right) ^{3}+\left({c}^{3}-4c+4\right) \right]$ $= \left[ {\frac{1}{4}{\mkern 1mu} {{\left( {a - b} \right)}^2} + \frac{{{{\left( {3{\mkern 1mu} c\left( {a + b} \right) - 14{\mkern 1mu} c + 14} \right)}^2}}}{{108{\mkern 1mu} {c^2}}}} \right]\left( {3{\mkern 1mu} ac + 3{\mkern 1mu} bc + 7{\mkern 1mu} c - 7} \right) + \frac{{27{\mkern 1mu} {c^6} - 108{\mkern 1mu} {c^4} - 235{\mkern 1mu} {c^3} + 1029{\mkern 1mu} {c^2} - 1029{\mkern 1mu} c + 343}}{{27{c^2}}}$ Now $,$ easy prove.

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sqing

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sqing

#36 h Nov 13, 2020, 9:48 AM

Y by

Thanks.
Let $a,b,c$ be positive real numbers satisfying $a^2+b^2+c^2=3.$ Prove that $4(a+b+c-2)< (a+b)(b+c)(c+a) \leq 4(a+b+c-1)$

This post has been edited 1 time. Last edited by sqing, Aug 8, 2021, 7:10 AM

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sqing

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sqing

#38 h Oct 22, 2021, 3:01 PM

Y by

Let $a,b,c$ be positive real numbers satisfying $abc=1.$ Prove that
$(a+b)(b+c)(c+a) \geq 8(a+b+c-2)$ Let $a,b,c$ be non-negative numbers satisfying $a(b+1)c=1.$ Prove that
$(a+b)(b+c)(c+a) \geq 2(a+b+c-1)$ h

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ZJ42

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ZJ42

#40 h Oct 22, 2021, 4:44 PM

Y by

why is this thread still active from 2007? and more importantly why does sqing bump it every few years?

This post has been edited 1 time. Last edited by ZJ42, Oct 22, 2021, 4:45 PM

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Jwenslawski

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Jwenslawski

#41 h Oct 22, 2021, 4:51 PM

Y by

Why does sqing post a bunch of problems though on other peoples threads?

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franzliszt

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franzliszt

#42 h Oct 22, 2021, 4:58 PM • 1 Y

Y by Jwenslawski

I believe it is because sqing posts problems related to the original, so that people can try to solve them.

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Jwenslawski

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Jwenslawski

#43 h Oct 22, 2021, 7:49 PM

Y by

franzliszt wrote:

I believe it is because sqing posts problems related to the original, so that people can try to solve them.

But why doesn't he make a thread of his own problems?

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franzliszt

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franzliszt

#44 h Oct 22, 2021, 8:29 PM • 1 Y

Y by Jwenslawski

Because then there would be too many threads

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jasperE3

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jasperE3

#45 h Oct 22, 2021, 8:34 PM • 1 Y

Y by megarnie

You can get postbanned for creating too many threads.

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#46 h Jun 18, 2022, 9:24 PM

Y by

chien than wrote:

Let $a;b;c$ be positive real numbers satisfying $abc=1$
Prove that:
$(a+b)(b+c)(c+a) \geq 4(a+b+c-1)$

$\left ( a+b \right )\left ( b+c \right )\left ( c+a \right )\geq 4\left ( a+b+c-1 \right )$

$\Longleftrightarrow (a+b+c)(ab+bc+ca)-abc\geq 4\left ( a+b+c-1 \right )$

$\Longleftrightarrow (a+b+c)(ab+bc+ca)+3\geq 4(a+b+c).$

Proof :

$ab+bc+ca+\frac{3}{a+b+c}\geq \sqrt{3abc(a+b+c)}+\frac{3}{a+b+c}$

$=\sqrt{3}x+\frac{3}{x^2}$ where $x=\sqrt{a+b+c}\geq \sqrt{3}$ by AM-GM

$=\frac{(x-\sqrt{3})\{(x-\sqrt{3})(\sqrt{3}x+2)+\sqrt{3}\}}{x^2}+4\geq 4.$

Equality : $x = \sqrt{3}\Longleftrightarrow a = b = c = 1.$

This post has been edited 3 times. Last edited by Kunihiko_Chikaya, Jun 18, 2022, 9:46 PM

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sqing

42260 posts

sqing

#47 h Jun 18, 2022, 11:59 PM

Y by

$ab+bc+ca +\frac{3}{a+b+c}\geq 4\sqrt[4]{\left(\frac{ab+bc+ca}{3}\right)^3\cdot \frac{3}{a+b+c} }\geq 4\sqrt[4]{\frac{ ab+bc+ca }{3 }}\geq 4$

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sqing

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sqing

#48 h Aug 13, 2022, 8:21 AM

Y by

Let $a,b,c$ be positive real numbers satisfying $a^2+b^2+c^2+abc=4$ .Show that $\frac{a}{1+2b}+\frac{b}{1+2c}+\frac{c}{1+2a}\geq1$ $\frac{a}{1+b+c}+\frac{b}{1+c+a}+\frac{c}{1+a+b}\geq1$

sqing wrote:

Let $a,b,c$ be positive real numbers satisfying $a^2+b^2+c^2+abc=4$ .Show that $(b+c)(c+a)(a+b)\geq 4(a+b+c+abc-2) .$

Solution of mudok:
Let $a+b+c=x, abc=z$ . Then $ab+bc+ca=\frac{1}{2}(x^2+z-4)$

We need to prove: $x^3-12x+16\ge z(10-x)$

After using this: $\frac{(x-2)^2}{4-x}\ge z$

it is sufficient to prove:

$(x^3-12x+16)(4-x)\ge (x-2)^2(10-x)$ $\iff (3-x)(x-2)^2(x+2)\ge 0$

This post has been edited 1 time. Last edited by sqing, Aug 14, 2022, 2:07 PM

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mihaig

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mihaig

#49 h Aug 13, 2022, 9:28 AM

Y by

chien than wrote:

Let $a;b;c$ be positive real numbers satisfying $abc=1$
Prove that:
$(a+b)(b+c)(c+a) \geq 4(a+b+c-1)$

Very beautiful

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sqing

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sqing

#50 h Aug 15, 2024, 1:47 AM

Y by

Let $a,b,c>0$ and $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ .Prove that
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c}\geq 2$ Let $a,b,c>0$ and $\frac{a+b}c+1=\frac{b+c}a+\frac{c+a}b.$ Prove that
$\frac{a+b}c\geq \frac12+\frac{\sqrt{17}}2$

szl6208 wrote:

Let $a;b;c;d$ be positive real numbers satisfying $abcd=1$ Prove that:
$(a+b)(b+c)(c+d)(d+a) \geq \frac{16}{3}(a+b+c+d-1)$

https://artofproblemsolving.com/community/c6h422665p2389576:
Let $a,b,c,d>0$ and $abcd=1$ . Prove that
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{16}{3(a+b+c+d)} \ge \frac{16}{3}$