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109 posts

#1 h May 14, 2005, 7:34 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Let x,y,z are positive reals.
Prove that
$x^6+y^6+z^6+15(x^3y^3+y^3z^3+z^3x^3)\geq 16xyz(x^3 +y^3 +z^3)$

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#2 h May 14, 2005, 9:30 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

WLOG $xyz=1$ .
Take, $x=\sqrt[3]{a}$ , etc., so $abc=1$ .
So we get:
$a^2+b^2+c^2+15(ab+bc+ca) \geq 16(a+b +c)$
Can anyone continue from here? Or give another solution maybe

This post has been edited 4 times. Last edited by perfect_radio, May 15, 2005, 8:26 PM

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#3 h May 15, 2005, 7:58 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

You can show using mixing variables the last inequality. Or at least, you can show it reduces to the case when $a=b$ and then it's just huge computation.

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2607 posts

#4 h May 15, 2005, 8:27 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

What do you mean by mixing variables?
Do you mean the following:

Take $f(a,b,c)=a^2+b^2+c^2+15(ab+bc+ca)-16(a+b +c)$ .
Then prove that $f(a,b,c) \geq f(\sqrt{ab},\sqrt{ab},c)$

?
If yes, i don't know whether it is true, nor how to prove it

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Soarer

#5 h May 15, 2005, 8:32 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

perfect_radio wrote:

What do you mean by mixing variables?
Do you mean the following:

Take $f(a,b,c)=a^2+b^2+c^2+15(ab+bc+ca)-16(a+b +c)$ .
Then prove that $f(a,b,c) \geq f(\sqrt{ab},\sqrt{ab},c)$

?
If yes, i don't know whether it is true, nor how to prove it

it is true if c is the largest among a,b,c.
You can try to prove it.

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2607 posts

#6 h May 15, 2005, 9:42 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

$f(a,b,c)=a^2+b^2+c^2+15(ab+bc+ca)-16(a+b +c)$ .

take $c \geq a \geq b$

Then
$T=f(a,b,c) - f(\sqrt{ab},\sqrt{ab},c)=a^2+b^2+c^2+15(ab+bc+ca)-16(a+b +c) - \left ( 2ab+c^2+15(ab+2c \sqrt{ab})-16(2 \cdot \sqrt{ab}+c) \right )$
$=a^2+b^2+15(bc+ca)-16(a+b) -2ab-15(2c \sqrt{ab})+16(2 \cdot \sqrt{ab})$
$=(a-b)^2-16(\sqrt{a}-\sqrt{b})^2+15(\sqrt{bc}-\sqrt{ca})^2$
$=(a-b)^2-16(\sqrt{a}-\sqrt{b})^2+15c(\sqrt{b}-\sqrt{a})^2$
$=(a-b)^2+(\sqrt{a}-\sqrt{b})^2 \cdot (15c-16)$

If $15c-16 \geq 0$ then $T \geq 0$
Else we have $1 \leq c < \frac{16}{15}$ .
$T>(a-b)^2-(\sqrt{a}-\sqrt{b})^2$
$(a-b)^2-(\sqrt{a}-\sqrt{b})^2 \geq 0$ is equivalent with $(a-b)(a+b-1)+2 \sqrt{ab}(\sqrt{a}-\sqrt{b}) \geq 0$
But by AM-GM $a+b \geq 2 \sqrt{ab} = 2 \sqrt{\frac{1}{c}} >1$
So everything is positive and $T \geq 0$ in all cases.

Therefore it is sufficient to prove the ineq for $a=b \leq c$ .
$2a^2+c^2+15(a^2+2ac) \geq 16 (2a+c)$
$17a^2+c^2+30ac \geq 32a+16c$
use the fact $a^2c=1$
$17a^6-32a^5+30a^3-16a^2+1 \geq 0$
$(a-1)^2(17a^4+2a^3-13a^2+2a+1) \geq 0$
By using derivatives we get $17a^4+2a^3-13a^2+2a+1 \geq 0$ for $a > 0$ $\blacksquare$
Hope it's correct...

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109 posts

#7 h May 16, 2005, 3:19 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

when $y=z=1,$
$x^6 +2+15(2x^3 + 1) \geq 16x(x^3 + 2)$
$x^6 -16x^4 + 30x^3 - 32x + 17 \geq 0$
$(x-1)^2 ( x^4 +2x^3 - 3x^2 + 2x +17) = (x-1)^2 (x^4 + 2x(x-1)^2 + x^2 + 17 ) \geq 0$
OK?!

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#8 h May 16, 2005, 3:37 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Parkdoosung wrote:

$(x-1)^2 ( x^4 +2x^3 - 3x^2 + 2x +17)$
OK?!

$3x^2$ or $13x^2$ ?

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#9 h May 16, 2005, 9:00 PM • 5 Y

Y by Adventure10, Mango247, and 3 other users

siuhochung wrote:

Parkdoosung wrote:

$(x-1)^{2}( x^{4}+2x^{3}-3x^{2}+2x+17)$
OK?!

$3x^{2}$ or $13x^{2}$ ?

Clearly it is $(x-1)^{2}( x^{4}+2x^{3}-13x^{2}+2x+17)$

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#10 h May 17, 2005, 9:11 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

I used derivatives to solve this problem.
I could not find triky method.

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#11 h May 17, 2005, 9:38 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

then why did you say it is not hard but tricky?

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#12 h May 18, 2005, 10:28 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

We can also solve this ineq by n-1 Equal Variable Principle.
We write it in the form
$15(a+b+c)^2-32(a+b+c) \ge 13(a^2+b^2+c^2)$ ,
where $abc=1$ . For $abc=1$ and fixed $a+b+c$ , the sum $a^2 +b^2+c^2$ is maximal for
$a=b \le c$ .

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827 posts

#13 h Apr 28, 2007, 5:28 PM • 5 Y

Y by Nathanisme, Adventure10, Mango247, and 2 other users

Parkdoosung wrote:

Let $x,y,z$ are nonnegative real numbers. Prove that
$x^{6}+y^{6}+z^{6}+15(y^{3}z^{3}+z^{3}x^{3}+x^{3}y^{3})\geq 16xyz(x^{3}+y^{3}+z^{3})$ .

Let we show the stronger inequality

$x^{6}+y^{6}+z^{6}+(5+6\sqrt{3}\cos{\frac{\pi}{18}})(y^{3}z^{3}+z^{3}x^{3}+x^{3}y^{3})$

$\geq(6+6\sqrt{3}\cos{\frac{\pi}{18}})xyz(x^{3}+y^{3}+z^{3})$ ,

with equality if $y=z=\frac{x}{2}\sec{\frac{\pi}{9}}$ , where

$5+6\sqrt{3}\cos{\frac{\pi}{18}}=15.234422383429318515539\cdots$ ;

$2\cos{\frac{\pi}{9}}=1.8793852415718167681082\cdots$ .

Proof. $\sum{x^{6}}+(5+6\sqrt{3}\cos{\frac{\pi}{18}})\sum{y^{3}z^{3}}-(6+6\sqrt{3}\cos{\frac{\pi}{18}})xyz\sum{x^{3}}$

$=\frac{(2x-y-z)^{2}}{4}[x-(y+z)\cos{\frac{\pi}{9}}]^{2}[x^{2}+(1+2\cos{\frac{\pi}{9}})x(y+z)+(\frac{3}{4}+\frac{1}{2}\cos{\frac{\pi}{9}})(y+z)^{2}]$

$+(\frac{3}{4}+\frac{3\sqrt{3}}{4}\cos{\frac{\pi}{18}})(y-z)^{2}\{(2x-y-z)[x^{3}+(\frac{9}{4}-\sqrt{3}\sin{\frac{\pi}{9}})x(y+z)(2x+y+z)$

$+(\frac{5}{8}-\frac{\sqrt{3}}{2}\sin{\frac{\pi}{9}})(y+z)^{3}]+(\frac{9\sqrt{3}}{4}\sin{\frac{\pi}{9}}-\frac{9}{8})(y+z)^{4}\}$

$+(\frac{9}{8}+\frac{9\sqrt{3}}{8}\cos{\frac{\pi}{18}})(y-z)^{4}[x(y+z)+(\frac{5}{\sqrt{3}}\sin{\frac{\pi}{9}}-\frac{2}{3})(y^{2}+z^{2})+(\frac{8}{\sqrt{3}}\sin{\frac{\pi}{9}}-\frac{2}{3})yz]$

$\geq0$ ,

which is clearly true for $x=\max\{x,y,z\}$ .

Remark.

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