Inequality with two conditions

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Inequality with two conditions

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inequalities proposed

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#1 h Jul 25, 2011, 3:55 PM • 3 Y

Y by Adventure10, Mango247, kiyoras_2001

$x,y,z\in\Bbb{R},\;x+y+z=0, \;x^2+y^2+z^2=6$ . Find $\max\{x^2y+y^2z+z^2x\}$ .

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#2 h Jul 25, 2011, 4:44 PM • 3 Y

Y by MariusStanean, Adventure10, Mango247

MariusStanean wrote:

$x,y,z\in\Bbb{R},\;x+y+z=0, \;x^2+y^2+z^2=6$ . Find $\max\{x^2y+y^2z+z^2x\}$ .

Let $3u = x+y+z, 3v^2 = xy+yz+zx, w^3 = xyz$ , so $u=0, v^2 = -1$ .
And $2(x^2y+y^2z+z^2x) = (x-y)(x-z)(y-z)+9uv^2-3w^3 = (x-y)(x-z)(y-z)-3w^3$ . Assume wlog $(x-y)(x-z)(y-z) \ge 0$ .
$(x-y)^2(x-z)^2(y-z)^2 = 27( -(w^3-3uv^2+2u^3)^2+4(u^2-v^2)^3)= 27(4-w^6)$ . So:
$x^2y+y^2z+z^2x = \frac{1}{2} \left ( (x-y)(x-z)(y-z)-3w^3 \right ) = \frac{1}{2} \left ( \sqrt{27(4-w^6)} - 3w^3 \right )$
Let $g(t) = \sqrt{3(4-t^2)}-t$ , then $x^2y+y^2z+z^2x = \frac{3}{2} g(w^3)$ . And $g'(t) = \frac{-\sqrt{3}t}{\sqrt{4-t^2}}-1$ , so if $g'(t) = 0$ we have $t < 0$ , and: $(-\sqrt{3}t)^2 =4-t^2$ , so $t = -1$ , and it is seen that $g'(-1)=0$ . And since $g'(0)=-1<0$ and $g'\left(-\sqrt{2}\right)=\sqrt{3}-1>0$ , $g$ has max at $t=-1$ . So:
$x^2y+y^2z+z^2x \le \frac{3}{2}g(-1) = 6$
The maximum is $6$ , and this is reached when $w^3=-1$ , i.e. $x,y,z$ being the roots of with the roots of $t^3-3t+1=0$ . (They are real since the roots of $t^3-3ut^2+3v^2t-w^3=0$ are real iff $-(w^3-3uv^2+2u^3)^2+4(u^2-v^2)^3 \ge 0$ , and in this case $-(w^3-3uv^2+2u^3)^2+4(u^2-v^2)^3 = 3$ )

This post has been edited 2 times. Last edited by Mathias_DK, Jul 25, 2011, 9:09 PM

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arqady

#3 h Jul 25, 2011, 5:09 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Mathias_DK, try $x=\sqrt3\sin80^{\circ}-\cos80^{\circ}$ , $y=2\cos80^{\circ}$ and $z=-\sqrt3\sin80^{\circ}-\cos80^{\circ}$ .

By the way, the minimum is equal to $-6$ .

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#4 h Jul 25, 2011, 5:19 PM • 2 Y

Y by Adventure10, Mango247

or $x=-\sqrt{3},\,y=\sqrt{3},\,z=0\Rightarrow x^2y+y^2z+z^2x=3\sqrt{3}$ . I think this is the maximum.

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#5 h Jul 25, 2011, 6:01 PM • 2 Y

Y by Adventure10, Mango247

MariusStanean, see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2373078#p2373078
You can get something grater.

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#6 h Jul 25, 2011, 9:04 PM • 1 Y

Y by Adventure10

arqady wrote:

Mathias_DK, try $x=\sqrt3\sin80^{\circ}-\cos80^{\circ}$ , $y=2\cos80^{\circ}$ and $z=-\sqrt3\sin80^{\circ}-\cos80^{\circ}$ .

By the way, the minimum is equal to $-6$ .

Thank you! I edited my post

How did you find $x,y,z$ explicitly?

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arqady

#7 h Jul 26, 2011, 3:46 AM • 1 Y

Y by Adventure10

Mathias_DK wrote:

How did you find $x,y,z$ explicitly?

It follows from my solution.
Since $x^2+y^2+z^2=6$ and $x+y+z=0$ , we obtain $x^2+xy+y^2=3$ , which is $\left(\frac{2x+y}{2\sqrt3}\right)^2+\left(\frac{y}{2}\right)^2=1$ .
Hence, we can assume $\frac{2x+y}{2\sqrt3}=\sin\alpha$ and $\frac{y}{2}=\cos\alpha$ .
From here we obtain $x=\sqrt3\sin\alpha-\cos\alpha$ and $y=2\cos\alpha$ .
Thus, $x^2y+y^2z+z^2x=x^3+3x^2y-y^3=f\left(\alpha\right)$ and we can use a calculus.
I think, your solution is better. $uvw$ forever!

After these solutions we see that
$|x^3+3x^2y-y^3|\leq6\left(\frac{x^2+xy+y^2}{3}\right)^{\frac{3}{2}}$ is true because
$|x^3+3x^2y-y^3|\leq6\left(\frac{x^2+xy+y^2}{3}\right)^{\frac{3}{2}}\Leftrightarrow(x^3-3x^2y-6xy^2-y^3)^2\geq0$ .

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#8 h Jul 26, 2011, 4:46 AM • 2 Y

Y by Adventure10, Mango247

I remember a unbelievable solution saying that:
$12-2(x^2y+y^2z+z^2x)=(x-xy-1)^2+(y-yz-1)+(z-zx-1)^2\geq 0$
so the max is $6$ .

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sqing

#9 h Mar 14, 2020, 7:57 AM

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Let $x,y,z\in\Bbb{R},\;x+y+z=0, \;x^2+y^2+z^2=6$ . Prove that $|(x-y)(y-z)(z-x)|\leq 6\sqrt 3.$

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#11 h Mar 14, 2020, 1:31 PM

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sqing wrote:

Let $x,y,z\in\Bbb{R},\;x+y+z=0, \;x^2+y^2+z^2=6$ . Prove that $|(x-y)(y-z)(z-x)|\leq 6\sqrt 3.$

And your nice solution!

sqing
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#7
VPM Nov 30, 2013, 8:37 AM • 1 Y
WLOG $x\ge y\ge z$ , hence $x-z\le \sqrt{2(x^2+y^2+z^2)}=2\sqrt{3}$ ,

$|(x-y)(y-z)(z-x) |=(x-y)(y-z)(x-z)$

$\le (\frac{x-y+y-z}{2})^2(x-z)=\frac{1}{4}(x-z)^3 \le 6\sqrt{3}$ .

when $x=\sqrt{3},y=0,z=-\sqrt{3}$ ,the maximum of $|(x-y)(y-z)(z-x) |$ is $6\sqrt{3}$ .

This post has been edited 1 time. Last edited by teomihai, Mar 14, 2020, 1:31 PM

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sqing

#12 h Mar 16, 2020, 1:17 AM

Y by

Nice. Thanks.
Another
$((x-y)(y-z)(z-x))^2 =108-27x^2y^2z^2.$

This post has been edited 2 times. Last edited by sqing, Mar 16, 2020, 1:24 AM

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sqing

#13 h Aug 26, 2020, 9:21 AM • 1 Y

Y by teomihai

sqing wrote:

Let $x,y,z\in\Bbb{R},\;x+y+z=0, \;x^2+y^2+z^2=6$ . Prove that $|(x-y)(y-z)(z-x)|\leq 6\sqrt 3.$

If $a,\,b,\,c$ are real numbers and $t \geqslant 0$ such that $a+b+c=0,$ $a^2+b^2+c^2 = 6t^2.$ Prove that
$\left|(a-b)(b-c)(c-a)\right| \leqslant 6\sqrt{3}t^3.$ Let $a, b, c\ge 0$ prove that
$6\sqrt{3}(a-b)(b-c)(c-a)\le (a+b+c)^3.$ Let $a,b,c > 0$ such that $a+b+c=3$ .Prove:
$(a-b)(b-c)(c-a) \le \frac{4}{\sqrt{27abc}}$ Let $a,b,c \in [0,1].$ Prove that $|(a-b)(b-c)(c-a)|\leq \frac{1}{4}.$ h
Let $a,b,c$ be non-negative real numbers such that $a+b+c=6$ and $a^3+b^3+c^3=54.$ .Prove that $ab+bc+ca\geq 9.$

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#14 h Apr 24, 2023, 1:47 PM • 1 Y

Y by teomihai

Let $a,b,c$ be real number satisfy $a+b+c=6$ and $a^2+b^2+c^2=14$ .Prove that $(a-b)(b-c)(c-a) \leq 2$

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#17 h Nov 14, 2024, 11:36 PM

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sqing wrote:

Let $a,b,c$ be real number satisfy $a+b+c=6$ and $a^2+b^2+c^2=14$ .Prove that $(a-b)(b-c)(c-a) \leq 2$

NOTE THIS IS UNFINISHED I HAD TO GO DO SOMETHING ELSE
ILL TRY MY BEST TO AT LEAST FINISH IS TOMMOROW
uhh ok let see
$(-b^2+ab+bc-ac)(c-a)=(-b^2c$

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sqing

#18 h Apr 14, 2025, 2:53 AM

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Let $a,b,c$ be real numbers such that $a+b+c=0,a^2+b^2+c^2=6$ . Prove that $a^3b+b^3c+c^3a\leq -9$

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