Inequality with xyz=1

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Inequality with xyz=1

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MI6

#1 h Jun 9, 2012, 6:55 AM • 1 Y

Y by Adventure10

Let x,y,z be positive numbers such that $xyz=1$ .Prove that:
$x^{2}+y^{2}+z^{2}+3\geq 2\left ( xy+yz+zx \right )$

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2130 posts

Sayan

#2 h Jun 9, 2012, 8:38 AM • 4 Y

Y by MI6, integrated_JRC, Adventure10, Mango247

I will use a technique that i just learned today: Mixing Variables

let $f(x,y,z)=x^2+y^2+z^2+3-2(xy+yz+zx)$
WLOG, assume $x=\text{min}(x,y,z)$

$f(x,\sqrt{yz},\sqrt{yz})=x^2+2yz+3-2(2x\sqrt{yz}+yz)=x^2+3-4x\sqrt{yz}$
so,
$f(x,y,z)-f(x,\sqrt{yz},\sqrt{yz})=(\sqrt{y}-\sqrt{z})^2((\sqrt{y}+\sqrt{z})^2-2x)$
now,
$(\sqrt{y}+\sqrt{z})^2 \ge 4\sqrt{yz}=\frac{4}{\sqrt{x}}\ge 4\ge 2x$ since $x=\text{min}(x,y,z)$

so, $f(x,y,z) \ge f(x,\sqrt{yz},\sqrt{yz})$
so it remains to prove

$f(x,\sqrt{yz},\sqrt{yz})\ge 0 \implies x^2+3 \ge 4\sqrt{x}$
but,
$x^2+3=x^2+1+1+1 \stackrel{AM-GM}{\ge} 4\sqrt{x}$

and equality holds when $x=y=z=1$

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120 posts

#3 h Jun 9, 2012, 10:51 AM • 2 Y

Y by Adventure10, Mango247

MI6 wrote:

Let x,y,z be positive numbers such that $xyz=1$ .Prove that:
$x^{2}+y^{2}+z^{2}+3\geq 2\left ( xy+yz+zx \right )$

After Homogenizing we should prove:
$x^2+y^2+z^2+3\sqrt[3]{(x^2y^2z^2)}\geq 2(xy+yz+zx)$
It's a symmetric inequality of degree $2$ ,So we just need to check it when $x=y$ and $x=0$ .
Case 1. $x=y$ :
$2x^2+z^2+3\sqrt[3]{x^4z^2}\geq 2x^2+4xz\Leftrightarrow z^2+3\sqrt[3]{x^4z^2}\geq 4xz\rightarrow \text{AM-GM}:z^2+3\sqrt[3]{x^4z^2}\geq 4\sqrt[4]{(z^2)(\sqrt[3]{x^4z^2})^3}=4xz$
Case 2. $x=0$ :
$y^2+z^2\geq 2yz$
Which is obvious by AM-GM.
So Done.

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1479 posts

#4 h Jun 9, 2012, 11:12 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Put $x=a/b,y=b/c,z=c/a$
so we have to prove $a^4.c^2+b^4.a^2+c^4.b^2+3(abc)^2>=2abc(a^2.b+b^2.c+c^2.a)$
it's direct from Scur's.

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215 posts

#5 h Jun 9, 2012, 12:31 PM • 2 Y

Y by Adventure10, Mango247

subham1729 wrote:

Put $x=a/b,y=b/c,z=c/a$
so we have to prove $a^4.c^2+b^4.a^2+c^4.b^2+3(abc)^2>=2abc(a^2.b+b^2.c+c^2.a)$
it's direct from Scur's.

How Schur's inequality works here subham

.
Also see here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138

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327 posts

#6 h Jun 9, 2012, 12:39 PM • 2 Y

Y by Adventure10, Mango247

subham1729 wrote:

Put $x=a/b,y=b/c,z=c/a$
so we have to prove $a^4.c^2+b^4.a^2+c^4.b^2+3(abc)^2>=2abc(a^2.b+b^2.c+c^2.a)$
it's direct from Scur's.

Nice soln!

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1479 posts

#7 h Jun 9, 2012, 1:34 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

vanthanhlhp wrote:

subham1729 wrote:

Put $x=a/b,y=b/c,z=c/a$
so we have to prove $a^4.c^2+b^4.a^2+c^4.b^2+3(abc)^2>=2abc(a^2.b+b^2.c+c^2.a)$
it's direct from Scur's.

How Schur's inequality works here subham

.
Also see here http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=359138

Using $x^3+y^3+z^3+3xyz>=2[(xy)^{3/2}+(yz)^{3/2}+(xz)^{3/2}]$

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560 posts

#8 h Jun 10, 2012, 2:15 AM • 1 Y

Y by Adventure10

See here http://dxdy.ru/topic59038.html

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318 posts

#9 h Jun 11, 2012, 12:47 PM • 1 Y

Y by Adventure10

You can use of schur inequality.

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15142 posts

#10 h Jun 11, 2012, 2:07 PM • 4 Y

Y by Adventure10 and 3 other users

Lagrange Multipliers method. Define $E(x,y,z) = x^{2}+y^{2}+z^{2} - 2\left ( xy+yz+zx \right ) + 3$ and $L(x,y,z;\lambda) = E(x,y,z) - \lambda(xyz-1)$ . Then $\frac {\partial L} {\partial x} = 2(x-y-z) - \lambda yz$ et. al. Equaling the partial derivatives to zero, by convenient manipulations we get $(x-y)((x+y+z) - 2z) = 0$ et. al. So

1. $x=y=z=1$ , for which $E(1,1,1) = 0$ ;
2. $x=y=a\neq z$ , when we need $z=0$ , impossible;
3. $x,y,z$ pairwise distinct, again impossible.

It remains to check what happens on the border of the domain of $L$ . Since $\lim_{x\to 0} E(x,y,z) = (y-z)^2 + 3 \geq 3$ , it follows that $E(x,y,z) \geq 0$ , with equality if and only if $x=y=z=1$ .

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389 posts

Djurre

#11 h Jun 24, 2012, 11:39 PM • 2 Y

Y by Adventure10, Mango247

threehandsnal wrote:

After Homogenizing we should prove:
$x^2+y^2+z^2+3\sqrt[3]{(x^2y^2z^2)}\geq 2(xy+yz+zx)$
It's a symmetric inequality of degree $2$ ,So we just need to check it when $x=y$ and $x=0$ .

Why do you only have to prove it for $x=y$ and $x=0$ ?

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2322 posts

oty

#12 h Jun 25, 2012, 3:11 AM • 2 Y

Y by Ronald Widjojo, Adventure10

it is : $x^2+y^2+z^2+2xyz+1\geq{2(xy+yz+xz)}$ and $LHS-RHS=(x-y)^2+2z(x-1)(y-1)+(z-1)^2\geq{0}$

.

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#13 h Jun 18, 2014, 9:07 AM • 2 Y

Y by Adventure10, Mango247

Because it is known that:
$x,y,z\ge 0\Rightarrow x^2+y^2+z^2+2xyz+1\ge 2(xy+yz+zx)$
____
Marin Sandu

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42590 posts

sqing

#14 h Jun 19, 2014, 11:16 PM • 2 Y

Y by Adventure10, Mango247

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76676
http://www.artofproblemsolving.com/viewtopic.php?t=24093
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=50745

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42590 posts

sqing

#15 h Jun 25, 2014, 2:16 PM • 2 Y

Y by Adventure10, Mango247

MI6 wrote:

Let x,y,z be positive numbers such that $xyz=1$ .Prove that:
$x^{2}+y^{2}+z^{2}+3\geq 2\left ( xy+yz+zx \right )$

$x^{2}+y^{2}+z^{2}+3\geq 2\left ( xy+yz+zx \right )\iff (y+z-x)^2+3-4yz\geq 0.$
we should prove:
If $3-4yz<0,$ then $(y+z-x)^2+3-4yz\geq 0.$
If $3-4yz<0,$ have $yz>\frac{3}{4},a<\frac{4}{3},$
$\Rightarrow y+z\geq 2\sqrt{yz}>\sqrt{3}\geq\frac{4}{3}>a,$
$\Rightarrow y+z-x\geq 2\sqrt{yz}-x> 0,$
$\Rightarrow (y+z-x)^2+3-4yz\geq (2\sqrt{yz}-x)^2+3-4yz$
$=x^2-4\sqrt{x}+3=(\sqrt{x}-1)^2(x+2\sqrt{x}+3)\geq 0.$

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#16 h Jun 26, 2014, 8:49 AM • 1 Y

Y by Adventure10

$consequence\; Schur\; inequality$
$let\; x,y,z \; non-negative \; real\; numbers \; and \; a\; positive\; number\; t\in ( 0,3 ]$
$it \; always\; true$
$(3-t)+txyz^{\frac{2}{t}}+x^{2}+y^{2}+z^{2}\geq 2(xy+yz+zx)$

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42590 posts

sqing

#17 h Jul 1, 2014, 11:37 PM • 1 Y

Y by Adventure10

samariddin wrote:

$consequence\; Schur\; inequality$
$let\; x,y,z \; non-negative \; real\; numbers \; and \; a\; positive\; number\; t\in ( 0,3 ]$
$it \; always\; true$
$(3-t)+txyz^{\frac{2}{t}}+x^{2}+y^{2}+z^{2}\geq 2(xy+yz+zx)$

Let $x^2=a^3,y^2=b^3,z^2=c^3,$ hence
$(3-t)+t(xyz)^{\frac{2}{t}}+x^{2}+y^{2}+z^{2}\geq 2(xy+yz+zx)\iff$
$(3-t)+t(abc)^{\frac{3}{t}}+a^{3}+b^{3}+c^{3}\geq 2((ab)^{\frac{3}{2}}+(bc)^{\frac{3}{2}}+(ca)^{\frac{3}{2}}):$
$(3-t)+t(abc)^{\frac{3}{t}}+a^{3}+b^{3}+c^{3}\geq 3abc+a^{3}+b^{3}+c^{3}$
$\geq a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\geq 2((ab)^{\frac{3}{2}}+(bc)^{\frac{3}{2}}+(ca)^{\frac{3}{2}}).$

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695 posts

#18 h Jul 7, 2014, 7:54 PM • 2 Y

Y by Adventure10, Mango247

MI6 wrote:

Let x,y,z be positive numbers such that $xyz=1$ .Prove that:
$x^{2}+y^{2}+z^{2}+3\geq 2\left ( xy+yz+zx \right )$

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42590 posts

sqing

#19 h Jul 29, 2014, 2:16 PM • 2 Y

Y by Adventure10, Mango247

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559652
Let $a,b,c>0$ and $n \geq 1$ . Prove that $(2n-1)(a^2+b^2+c^2)+2abc+1 \geq 2n(ab+bc+ca)$

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Dr Sonnhard Graubner

16100 posts

Dr Sonnhard Graubner

#20 h Jul 29, 2014, 5:01 PM • 2 Y

Y by Adventure10 and 1 other user

hello, but the proof is here
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559652
Sonnhard.

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42590 posts

sqing

#21 h Sep 9, 2014, 6:24 AM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559652
Let $a,b,c>0$ and $n \geq 1$ . Prove that $(2n-1)(a^2+b^2+c^2)+2abc+1 \geq 2n(ab+bc+ca)$

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=605419&p3597172#p3597172

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sqing

#22 h Dec 3, 2017, 9:37 AM • 1 Y

Y by Adventure10

MI6 wrote:

Let x,y,z be positive numbers such that $xyz=1$ .Prove that:
$x^{2}+y^{2}+z^{2}+3\geq 2\left ( xy+yz+zx \right )$

Let $a$ , $b$ , $c$ be positive real numbers such that $abc=1.$ Prove that $a^4+b^4+c^4+9\geq 4\left(ab+bc+ca\right)$

oty wrote:

it is : $x^2+y^2+z^2+2xyz+1\geq{2(xy+yz+xz)}$ and $LHS-RHS=(x-y)^2+2z(x-1)(y-1)+(z-1)^2\geq{0}$

.

https://artofproblemsolving.com/community/c6h1519087p9064612

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This post has been edited 2 times. Last edited by sqing, Jan 29, 2021, 12:52 PM

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#23 h Dec 3, 2017, 1:39 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=3559652
Let $a,b,c>0$ and $n \geq 1$ . Prove that $(2n-1)(a^2+b^2+c^2)+2abc+1 \geq 2n(ab+bc+ca)$

For $n=1$ the inequality is true.
Assume the inequality is true for some $n \in Z_+$ . Then adding to it obvious inequality $2(a^2+b^2+c^2)\ge2(ab+bc+ca)$ , we get
$(2(n+1)-1)(a^2+b^2+c^2)+2abc+1 \geq 2(n+1)(ab+bc+ca)$ as desired. The proof is finished by induction.

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