xyz(x+y+z)=4

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#1 h Feb 3, 2019, 9:11 AM • 1 Y

Y by Adventure10

Let $x, y, z$ be positive real numbers such that $xyz(x + y + z) = 4$ . Prove that
$(x+y)^2+3(z+y)^2+(x+z)^2 \geq 8\sqrt7$

This post has been edited 1 time. Last edited by RnstTrjyn, Feb 3, 2019, 3:19 PM

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pco

#2 h Feb 3, 2019, 9:55 AM • 1 Y

Y by Adventure10

RnstTrjyn wrote:

Let $x, y, z$ be positive real numbers such that $xyz(x + y + z) = 4$ . Prove that
$(x+y)^2+(z+y)^2+(x+z)^2 \geq 8\sqrt7$

Wrong.
Choose as counter-example $(x,y,z)=(1,1,\sqrt 5-1)$

In fact, just writing :
$(x+y)^2+(y+z)^2+(z+x)^2$ $=2\frac{xyz(x+y+z)-4}{xy}+2(x-y)^2+\left(\frac{2\sqrt 2}{\sqrt{xy}}-\sqrt{6xy}\right)^2+8\sqrt 3$

We get, when $xyz(x+y+z)-4=0$ : $(x+y)^2+(y+z)^2+(z+x)^2\boxed{\ge 8\sqrt 3}$ (and not $8\sqrt 7$ )

And it is easy to check that value may be reached with $x=y=z=\sqrt{\frac 2{\sqrt 3}}$ .

This post has been edited 1 time. Last edited by pco, Feb 3, 2019, 10:39 AM
Reason: Added best limit

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#3 h Feb 3, 2019, 3:19 PM • 2 Y

Y by Adventure10, Mango247

pco wrote:

RnstTrjyn wrote:

Let $x, y, z$ be positive real numbers such that $xyz(x + y + z) = 4$ . Prove that
$(x+y)^2+(z+y)^2+(x+z)^2 \geq 8\sqrt7$

Wrong.
Choose as counter-example $(x,y,z)=(1,1,\sqrt 5-1)$

In fact, just writing :
$(x+y)^2+(y+z)^2+(z+x)^2$ $=2\frac{xyz(x+y+z)-4}{xy}+2(x-y)^2+\left(\frac{2\sqrt 2}{\sqrt{xy}}-\sqrt{6xy}\right)^2+8\sqrt 3$

We get, when $xyz(x+y+z)-4=0$ : $(x+y)^2+(y+z)^2+(z+x)^2\boxed{\ge 8\sqrt 3}$ (and not $8\sqrt 7$ )

And it is easy to check that value may be reached with $x=y=z=\sqrt{\frac 2{\sqrt 3}}$ .

Sorry, I've corrected.There was 3.

This post has been edited 1 time. Last edited by RnstTrjyn, Feb 3, 2019, 3:20 PM

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pco

#5 h Feb 3, 2019, 3:57 PM • 3 Y

Y by RnstTrjyn, Adventure10, Mango247

RnstTrjyn wrote:

Let $x, y, z$ be positive real numbers such that $xyz(x + y + z) = 4$ . Prove that
$(x+y)^2+3(z+y)^2+(x+z)^2 \geq 8\sqrt7$

That's OK now.

Let $s=y+z$ and $p=yz$

$xyz(x+y+z)=4$ is $x^2+sx=\frac 4p$

$(x+y)^2+3(y+z)^2+(z+x)^2$ $=2x^2+2x(y+z)+y^2+z^2+3(y+z)^2$

$=2(x^2+sx)+4s^2-2p$

$=\frac 8p+4s^2-2p\ge\frac 8p+16p-2p=\frac 8p+14p$

And $\frac 8p+14p=\left(\sqrt{\frac 8p}-\sqrt{14p}\right)^2+8\sqrt 7$

Q.E.D.

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#6 h Feb 3, 2019, 4:03 PM • 1 Y

Y by Adventure10

pco wrote:

RnstTrjyn wrote:

Let $x, y, z$ be positive real numbers such that $xyz(x + y + z) = 4$ . Prove that
$(x+y)^2+3(z+y)^2+(x+z)^2 \geq 8\sqrt7$

That's OK now.

Let $s=y+z$ and $p=yz$

$xyz(x+y+z)=4$ is $x^2+sx=\frac 4p$

$(x+y)^2+3(y+z)^2+(z+x)^2$ $=2x^2+2x(y+z)+y^2+z^2+3(y+z)^2$

$=2(x^2+sx)+4s^2-2p$

$=\frac 8p+4s^2-2p\ge\frac 8p+16p-2p=\frac 8p+14p$

And $\frac 8p+14p=\left(\sqrt{\frac 8p}-\sqrt{14p}\right)^2+8\sqrt 7$

Q.E.D.

Thanks Sir Pco. Nice solution.

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#8 h Thursday at 9:02 AM

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RnstTrjyn wrote:

Let $x, y, z$ be positive real numbers such that $xyz(x + y + z) = 4$ . Prove that
$(x+y)^2+3(z+y)^2+(x+z)^2 \geq 8\sqrt7$

Let $a,b,c$ be positive real number such that $abc(a+b+c)=4.$ Prove that
$(a+b)^2+(b+c)^2+ k(c+a)^2\geq 8\sqrt{2k+1}$ Where $k\in N^+.$

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