a²+b²+c²+abc=4

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younesmath2012maroc

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younesmath2012maroc

#1 h Nov 9, 2012, 10:03 PM • 2 Y

Y by Adventure10, Mango247

If $a , b, c \ge 0$ such that $\boxed{ a^2+b^2+c^2+abc = 4}$ prove that $\boxed{a+b+c \geq ab+bc+ca}$ .

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#2 h Nov 9, 2012, 11:20 PM • 3 Y

Y by Olemissmath, Adventure10, Mango247

It is easy to check that:

$a^2+b^2+c^2+abc = 4 \Leftrightarrow \exists_{x,y,z > 0} \ a=\frac{2x}{y+z}, b=\frac{2y}{z+x}, c=\frac{2z}{x+y}$ .

Our inequality is equivalent to:

$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge \frac{2xy}{(z+y)(z+x)}+\frac{2yz}{(x+z)(x+y)}+\frac{2zx}{(x+y)(y+z)}$

But this is:

$x(x-y)(x-z)+y(y-z)(y-x)+z(z-x)(z-y) \ge 0$

true by Schur inequality.

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arqady

#3 h Nov 10, 2012, 1:39 AM • 4 Y

Y by sabkx, Adventure10, Mango247, and 1 other user

Marcinek665 wrote:

It is easy to check that:

$a^2+b^2+c^2+abc = 4 \Leftrightarrow \exists_{x,y,z > 0} \ a=\frac{2x}{y+z}, b=\frac{2y}{z+x}, c=\frac{2z}{x+y}$ .

Are you sure?

younesmath2012maroc wrote:

$a , b, c \ge 0$ such that $\boxed{ a^2+b^2+c^2+abc = 4}$ prove that : $\boxed{ a+b+c \ge ab+bc+ca}$

Let $a+b+c<ab+ac+bc$ and $a=kx$ , $b=ky$ , $c=kz$ , where $k>0$ and $x+y+z=xy+xz+yz$ .
Hence, $k(x+y+z)<k^2(xy+xz+yz)$ , which gives $k>1$ .
Now $4=a^2+b^2+c^2+abc=k^2(x^2+y^2+z^2)+k^3xyz>x^2+y^2+z^2+xyz$ ,
which is contradiction because we'll prove that $4\leq x^2+y^2+z^2+xyz$ .
We obtain: $4\leq x^2+y^2+z^2+xyz\Leftrightarrow$
$\Leftrightarrow\frac{4(xy+xz+yz)^3}{(x+y+z)^3}\leq\frac{(xy+xz+yz)(x^2+y^2+z^2)}{x+y+z}+xyz$ ,
which is easy by $SOS$ or by $uvw$ , which you don't like.

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#4 h Nov 10, 2012, 2:24 AM • 2 Y

Y by Adventure10, Mango247

Oh, sorry. My substitution works correctly only if $ab+bc+ca+abc=4$

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younesmath2012maroc

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younesmath2012maroc

#5 h Nov 10, 2012, 7:54 AM • 2 Y

Y by Adventure10, Mango247

thus the exercise is not solved !!!
ho can try again !!!

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224 posts

#6 h Nov 10, 2012, 10:07 AM • 3 Y

Y by Adventure10, Mango247, kiyoras_2001

younesmath2012maroc wrote:

thus the exercise is not solved !!!
ho can try again !!!

You can let $\sqrt{ab}=x, \sqrt{bc}=y, \sqrt{ca}=z$ . Then let $x=2cosA, y=2cosB, z=2cosC$ , ok now

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Lawasu

#7 h Nov 10, 2012, 10:16 AM • 2 Y

Y by Adventure10, Mango247

The Method of Lagrange Multipliers kills it immediately.

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#8 h Nov 10, 2012, 11:13 AM • 2 Y

Y by Adventure10, Mango247

Lawasu wrote:

The Method of Lagrange Multipliers kills it immediately.

You are right! More idea from this inquality? You can see it in the book "Problem from the book 1"

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mudok

#9 h Nov 10, 2012, 12:10 PM • 2 Y

Y by Adventure10, Mango247

Let's prove $a+b+c\le 3$ .

Two of $(a-1),(b-1),(c-1)$ have the same sign. Assume that
$(a-1)(b-1)\ge 0 \ \ \ \iff \ \ \ a+b+c\le 1+c+ab$ . If $c+ab\le 2$ then it is done. If $c+ab>2$ then $4=(a^2+b^2) + (c+ab)c\ge 2ab+(c+ab)c>2ab+2c=2(ab+c)>4$ Contradiction. So $a+b+c\le 3$ is proved. Then use only $(a+b+c)^2\ge 3(ab+bc+ca)$

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Sayan

#10 h Nov 10, 2012, 12:59 PM • 4 Y

Y by sabkx, Adventure10, Mango247, and 1 other user

In fact the following inequality also holds:
$a+b+c \ge ab+bc+ca+\frac15\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$
Dear mudok, can you give a similar proof to this one?

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mudok

#11 h Nov 11, 2012, 1:50 PM • 4 Y

Y by younesmath2012maroc, Adventure10, Mango247, and 1 other user

The following inequality also holds:
$a+b+c \ge ab+bc+ca+\frac{1}{\sqrt{24}}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$
there exist a similar solution

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younesmath2012maroc

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younesmath2012maroc

#12 h Nov 11, 2012, 5:11 PM • 2 Y

Y by Adventure10, Mango247

ho has another solution please !!!

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Kunihiko_Chikaya

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Kunihiko_Chikaya

#13 h Nov 14, 2012, 5:02 AM • 4 Y

Y by hctb00, bitrak, Adventure10, Mango247

$(a-b)^2=a^2+b^2-2ab=4-abc-c^2-2ab=(c+2)(2-c-ab)\geq 0$ for $a,\ b,\ c\geq 0$ , yielding $ab+c\leq 2.$

Analogously, $bc+a\leq 2,\ ca+b\leq 2$ , thus we have $ab+bc+ca+a+b+c\leq 6$ ,

combine with $a+b+c\geq \sqrt{3}(ab+bc+ca)$ ,

we get $ab+bc+ca+\sqrt{3}\sqrt{ab+bc+ca}-6\leq 0$

$\Longleftrightarrow (\sqrt{ab+bc+ca}-\sqrt{3})(\sqrt{ab+bc+ca}+2\sqrt{3})\leq 0.$

Since $a,\ b,\ c\geq 0$ , yielding $0\leq ab+bc+ca\leq 3$ , then

$(a+b+c)^2-(ab+bc+ca)^2\geq 3(ab+bc+ca)-(ab+bc+ca)^2$

$=(ab+bc+ca)(3-ab-bc-ca)\geq 0\ (a,\ b,\ c\geq 0)$ , yielding $a+b+c\geq ab+bc+ca$ , as desired.

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#14 h Dec 15, 2012, 5:16 AM • 2 Y

Y by Adventure10, Mango247

hieu1411997 wrote:

Lawasu wrote:

The Method of Lagrange Multipliers kills it immediately.

You are right! More idea from this inquality? You can see it in the book "Problem from the book 1"

where i can find this book?

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sqing

#15 h Feb 4, 2013, 1:15 AM • 2 Y

Y by Adventure10, Mango247

If $a , b, c > 0$ such that $\boxed{\sum \frac{1}{a+b+1}\ge 1}$ prove that $\boxed{a+b+c \geq ab+bc+ca}$ .
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=1422711

$\boxed{ a+b+c+abc = 4}$ $\Rightarrow \boxed{a+b+c \geq ab+bc+ca}$ $\Rightarrow \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \geq \frac{a+b+c}{\sqrt{2}}$

Poland Olympiad 2007:
If $a , b, c > 0$ such that $a+b+c+abc = 4$ ,prove that $\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \geq \frac{a+b+c}{\sqrt{2}}$

If $a , b, c > 0$ , prove that $\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \geq \sqrt{\frac{3}{2}(a+b+c)}$

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younesmath2012maroc

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younesmath2012maroc

#16 h Feb 4, 2013, 8:47 AM • 2 Y

Y by Adventure10, Mango247

younesmath2012maroc wrote:

If $a , b, c \ge 0$ such that $\boxed{ a^2+b^2+c^2+abc = 4}$ prove that $\boxed{a+b+c \geq ab+bc+ca}$ .

$my~solution~is~:~(putting~t=a+b+c~and~B=ab+bc+ca~)\\ \bullet we~have~t=a+2\times 1\times \frac{b+c}{2}\leq a+1+(\frac{b+c}{2})^2\leq 1+2=3 \\ (because~we~know~that~a+(\frac{b+c}{2})^2\leq 2~*~)\\ \bullet we~have~3t\geq t^2(t\leq 3)\geq 3B\Rightarrow t\geq B$
* see here http://www.artofproblemsolving.com/blog/78923

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hctb00

#17 h Feb 4, 2013, 12:06 PM • 2 Y

Y by Adventure10, Mango247

hieu1411997 wrote:

Lawasu wrote:

The Method of Lagrange Multipliers kills it immediately.

You are right! More idea from this inquality? You can see it in the book "Problem from the book 1"

please , where i can download this book?

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1158 posts

#18 h Feb 4, 2013, 12:27 PM • 2 Y

Y by Adventure10, Mango247

Denote by :
$a=2cosA,b=2cosB,c=2cosC$ and $\sum cosA=\frac{R+r}{R},\sum cosAcosB=\frac{s^{2}+r^{2}-4R^{2}}{4R^{2}}$ , then we :
$\sum a^{2}+abc=4\Leftrightarrow \sum cos^{2}A+2\prod cosA=1,\forall A,B,C\in (0,\frac{\pi }{2})$
We:
$a+b+c\geq ab+bc+ca\Leftrightarrow 2\sum cosA\geq 4\sum cosAcosB\Leftrightarrow \frac{R+r}{R}\geq \frac{2(s^{2}+r^{2}-4R^{2})}{4R^{2}}\Leftrightarrow s^{2}\leq 6R^{2}+2rR-r^{2}$ ,
using Gerresten's, we have : $s^{2}\leq 4R^{2}+4Rr+3r^{2}$ , or,

$4R^{2}+4Rr+3r^{2}\leq 6R^{2}+2rR-r^{2}\Leftrightarrow R^{2}-Rr-2r^{2}\geq 0\Leftrightarrow (R-2r)(R+r)\geq 0\Leftrightarrow R\geq 2r (Euler), or, R+r\geq 0, it's true$

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#19 h Feb 4, 2013, 1:25 PM • 4 Y

Y by younesmath2012maroc, hctb00, Adventure10, Mango247

My proof
$(a,b,c >0,a^2+b^2+c^2+abc=4)\Leftrightarrow\sum\frac{a}{2a+bc}=\ 1\Leftrightarrow\sum\frac{bc}{2a+bc}=\ 1$
Applying Cachy-schwarz get:
$1=\sum\frac{b^2c^2}{2abc+b^2c^2}\geq\frac{(bc+ca+ab)^2}{6abc+\sum{b^2c^2}}=\frac{2abc(a+b+c)+\sum{b^2c^2}}{6abc+\sum{b^2c^2}}\Rightarrow{6abc+\sum{b^2c^2}\geq 2abc(a+b+c)+\sum b^2c^2\Leftrightarrow 6abc\geq 2abc(a+b+c)\Leftrightarrow 3\geq a+b+c }$
Derive the relationship:
1) $3\geq a+b+c$
Taking into account the relation (1), we obtain:
$3(a+b+c)\geq (a+b+c)^2\geq 3(ab+bc+ca)\Rightarrow a+b+c\geq ab+bc+ca$
The demonstration is over here!
______
Sandu Marin

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41782 posts

sqing

#20 h Feb 4, 2013, 11:14 PM • 2 Y

Y by Adventure10, Mango247

If $a , b, c \ge 0$ such that $\boxed{ a^2+b^2+c^2+abc = 4}$ prove that $\boxed{a+b+c \geq ab+bc+ca}$ .
Vietnam MO96
Let $a, b, c$ three positive reals such that $\frac{1}{a+b+1}+\frac{1}{b+c+1}+\frac{1}{c+a+1}\geq 1.$ Show that $a+b+c\geq ab+bc+ca.$ (Romania TST 2007)
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=261698
Positive real numbers $a,b,c$ satisfy $\frac{1}{2+a} + \frac{1}{2+b} + \frac{1}{2+c} = 1$ Prove that $a+b+c \geq ab+bc+ca$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2137595
If $a , b, c >0$ such that $\boxed{ ab+bc+ca+abc = 4}$ prove that $\boxed{a+b+c \geq ab+bc+ca}$ .
Let $a,b,c>0,$ and $bc+ca+ab+abc=4$ , show that $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq\frac{3}{4}+\frac{1}{4}\sqrt{3(a^2+b^2+c^2)}\geq\frac{3}{2}.$
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=518647

This post has been edited 1 time. Last edited by sqing, Oct 7, 2014, 1:29 AM

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younesmath2012maroc

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younesmath2012maroc

#21 h Sep 21, 2014, 9:37 AM • 1 Y

Y by Adventure10

mudok wrote:

The following inequality also holds:
$a+b+c \ge ab+bc+ca+\frac{1}{\sqrt{24}}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$
there exist a similar solution

what is the similar solution please dear "mudok"

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#22 h Sep 21, 2014, 11:27 AM • 2 Y

Y by Adventure10, Mango247

Proof: $(a+b+c)^2\ge 3(ab+bc+ca)\ge (a+b+c)(ab+bc+ca)\implies a+b+c\ge ab+bc+ca$

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154 posts

#23 h Sep 21, 2014, 11:28 AM • 2 Y

Y by Adventure10, Mango247

Proof: $(a+b+c)^2\ge 3(ab+bc+ca)\ge (a+b+c)(ab+bc+ca)\implies a+b+c\ge ab+bc+ca$

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#24 h Sep 21, 2014, 6:58 PM • 2 Y

Y by Adventure10, Mango247

younesmath2012maroc wrote:

If $a , b, c \ge 0$ such that $\boxed{ a^2+b^2+c^2+abc = 4}$ prove that $\boxed{a+b+c \geq ab+bc+ca}$ .

we have $9(\sum a)^2=9(4-abc+2\sum ab)=\\=(\sum ab+6)^2+(\sum ab)(3-\sum ab)+3(\sum ab-3abc) \ge (\sum ab+6)^2$
because $3\ge \sum ab \ge 3abc$
therefore $a+b+c \ge \frac{ab+ac+bc}{3}+2$
which is stronger.

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#25 h Sep 21, 2014, 7:05 PM • 2 Y

Y by Adventure10, Mango247

with the same conditions prove that $\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq a^2+b^2+c^2$

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arqady

#26 h Sep 21, 2014, 7:23 PM • 1 Y

Y by Adventure10

elmrini wrote:

with the same conditions prove that $\frac{ab}{c}+\frac{ac}{b}+\frac{bc}{a}\geq a^2+b^2+c^2$

It's Schur!

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younesmath2012maroc

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younesmath2012maroc

#27 h Sep 22, 2014, 6:26 AM • 1 Y

Y by Adventure10

younesmath2012maroc wrote:

mudok wrote:

The following inequality also holds:
$a+b+c \ge ab+bc+ca+\frac{1}{\sqrt{24}}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$
there exist a similar solution

what about this one ?

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30210 posts

arqady

#28 h Sep 22, 2014, 2:49 PM • 2 Y

Y by Adventure10, Mango247

Maybe it should be the following?

Let $a$ , $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2+abc=4$ . Prove that:
$a+b+c\geq ab+ac+bc+(\sqrt2-1)(a^2+b^2+c^2-ab-ac-bc)$

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younesmath2012maroc

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younesmath2012maroc

#29 h Sep 23, 2014, 6:57 PM • 1 Y

Y by Adventure10

arqady wrote:

Maybe it should be the following?

Let $a$ , $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2+abc=4$ . Prove that:
$a+b+c\geq ab+ac+bc+(\sqrt2-1)(a^2+b^2+c^2-ab-ac-bc)$

dear "arqady" can you prove it please !!!

i'm traying to cherche the best constante k such that $a+b+c\geq ab+ac+bc+k(a^2+b^2+c^2-ab-ac-bc)$
mudok find $k=\frac{1}{\sqrt{6}}$
"arqady"find $k=\sqrt{2}-1$

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arqady

#30 h Sep 23, 2014, 7:38 PM • 5 Y

Y by younesmath2012maroc, centslordm, sabkx, Adventure10, Mango247

I have no proof of this inequality.

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younesmath2012maroc

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younesmath2012maroc

#31 h Sep 23, 2014, 7:57 PM • 2 Y

Y by Adventure10, Mango247

dear ''mudok'' have you a solution please !!!

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mudok

#32 h Oct 1, 2014, 3:40 PM • 1 Y

Y by Adventure10

My solution is by homogenization, so, my inequality must be very weak !

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30210 posts

arqady

#33 h Aug 10, 2024, 5:28 PM

Y by

younesmath2012maroc wrote:

arqady wrote:

Maybe it should be the following? :maybe:

:maybe:

Let $a$ , $b$ and $c$ be non-negative numbers such that $a^2+b^2+c^2+abc=4$ . Prove that:
$a+b+c\geq ab+ac+bc+(\sqrt2-1)(a^2+b^2+c^2-ab-ac-bc)$

dear "arqady" can you prove it please !!!

Try a Contradiction method and $uvw$ .

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