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Source: Vietnam National Olympiad 2014

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34 posts

#1 h Jan 4, 2014, 4:54 AM • 2 Y

Y by lenhathoang1998, Adventure10

Find the maximum of
$P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$
where $x,y,z$ are positive real numbers.

This post has been edited 2 times. Last edited by vutuanhien, Jan 4, 2014, 3:58 PM

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#2 h Jan 4, 2014, 2:42 PM • 4 Y

Y by lenhathoang1998, lehoangphuc, Adventure10, Mango247

No one has idea?
This is my solution for this problem
Let $\frac{x}{y}=a, \frac{y}{z}=b, \frac{z}{x}=c\Rightarrow abc=1$
$P=\frac{1}{(a^4+1)(b+c)^3}+\frac{1}{(b^4+1)(c+a)^3}+\frac{1}{(c^4+1)(a+b)^3}$ By AM-GM inequality:
$(b+c)^3\geq 8bc\sqrt{bc}=\frac{8}{a\sqrt{a}}\geq \frac{16}{a^2+a}$ $\Rightarrow 16P\leq \frac{a^2+a}{a^4+1}+\frac{b^2+b}{b^4+1}+\frac{c^2+c}{c^4+1} (1)$ We have:
$\frac{a^2+a}{a^4+1}\leq \frac{3(a+1)}{2(a^2+a+1)}\Leftrightarrow (a-1)^2(a+1)(3a^2+4a+3)\geq 0$ (true)
$\Rightarrow \dfrac{a^2+a}{a^4+1}+\dfrac{b^2+b}{b^4+1}+\dfrac{c^2+c}{c^4+1}\leq \dfrac{3}{2}\sum \dfrac{a+1}{a^2+a+1}\leq 3 (2)$ (This is Vasc inequality:If $abc=1$ then $\sum \frac{a+1}{a^2+a+1}\leq 2$ )
From (1) and (2) we have $P\leq \frac{3}{16}$
Equality occurs if and only if $a=b=c=1$ , or $x=y=z$

This post has been edited 5 times. Last edited by vutuanhien, Oct 18, 2015, 5:03 AM
Reason: latex error

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723 posts

codyj

#3 h Jan 4, 2014, 3:03 PM • 2 Y

Y by Adventure10, Mango247

vutuanhien wrote:

$P=\frac{1}{(a^4+1)(b+c)^3}+\frac{1}{(b^4+1)(c+a)^3}+\frac{1}{(c^4+1)(a+b)^3}$

This is wrong given your definitions of $a$ , $b$ , and $c$ .

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#4 h Jan 4, 2014, 3:08 PM • 2 Y

Y by Adventure10, Mango247

codyj wrote:

vutuanhien wrote:

$P=\frac{1}{(a^4+1)(b+c)^3}+\frac{1}{(b^4+1)(c+a)^3}+\frac{1}{(c^4+1)(a+b)^3}$

This is wrong given your definitions of $a$ , $b$ , and $c$ .

Sorry I posted wrong problem

. But my solution is right

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3428 posts

#5 h Jan 4, 2014, 3:20 PM • 2 Y

Y by Adventure10, Mango247

vutuanhien wrote:

No one has idea?
This is my solution for this problem
Let $\frac{x}{y}=a, \frac{y}{z}=b, \frac{z}{x}=c\Rightarrow abc=1$
$P=\frac{1}{(a^4+1)(b+c)^3}+\frac{1}{(b^4+1)(c+a)^3}+\frac{1}{(c^4+1)(a+b)^3}$
By AM-GM inequality:
$(b+c)^3\geq 8bc\sqrt{bc}=\frac{8}{a\sqrt{a}}\geq \frac{16}{a^2+a}$
$\Rightarrow 16T\leq \frac{a^2+a}{a^4+1}+\frac{b^2+b}{b^4+1}+\frac{c^2+c}{c^4+1} (1)$
We have:
$\frac{a^2+a}{a^4+1}\leq \frac{3(a+1)}{2(a^2+a+1)}\Leftrightarrow (a-1)^2(a+1)(3a^2+4a+3)\geq 0$ (right)

\[\Rightarrow \[\frac{a^2+a}{a^4+1}+\frac{b^2+b}{b^4+1}+\frac{c^2+c}{c^4+1}\leq \frac{3}{2}\sum \frac{a+1}{a^2+a+1}\leq 3 (2)\]

(This is Vasc inequality:If $abc=1$ then $\sum \frac{a+1}{a^2+a+1}\leq 2$ )
From (1) and (2) we have $P\leq \frac{3}{16}$
Equality occurs if and only if $a=b=c=1$ , or $x=y=z$

It's wrong. Try $x=\frac{1}{3}, y=\frac{1}{4}, z=1$

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34 posts

#6 h Jan 4, 2014, 3:28 PM • 2 Y

Y by Adventure10, Mango247

oldbeginner wrote:

vutuanhien wrote:

No one has idea?
This is my solution for this problem
Let $\frac{x}{y}=a, \frac{y}{z}=b, \frac{z}{x}=c\Rightarrow abc=1$
$P=\frac{1}{(a^4+1)(b+c)^3}+\frac{1}{(b^4+1)(c+a)^3}+\frac{1}{(c^4+1)(a+b)^3}$
By AM-GM inequality:
$(b+c)^3\geq 8bc\sqrt{bc}=\frac{8}{a\sqrt{a}}\geq \frac{16}{a^2+a}$
$\Rightarrow 16T\leq \frac{a^2+a}{a^4+1}+\frac{b^2+b}{b^4+1}+\frac{c^2+c}{c^4+1} (1)$
We have:
$\frac{a^2+a}{a^4+1}\leq \frac{3(a+1)}{2(a^2+a+1)}\Leftrightarrow (a-1)^2(a+1)(3a^2+4a+3)\geq 0$ (right)

\[\Rightarrow \[\frac{a^2+a}{a^4+1}+\frac{b^2+b}{b^4+1}+\frac{c^2+c}{c^4+1}\leq \frac{3}{2}\sum \frac{a+1}{a^2+a+1}\leq 3 (2)\]

(This is Vasc inequality:If $abc=1$ then $\sum \frac{a+1}{a^2+a+1}\leq 2$ )
From (1) and (2) we have $P\leq \frac{3}{16}$
Equality occurs if and only if $a=b=c=1$ , or $x=y=z$

It's wrong. Try $x=\frac{1}{3}, y=\frac{1}{4}, z=1$

Where's wrong?

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2130 posts

Sayan

#7 h Jan 5, 2014, 3:33 AM • 4 Y

Y by Kunihiko_Chikaya, Ashutoshmaths, hctb00, Adventure10

vutuanhien wrote:

Find the maximum of
$P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$
where $x,y,z$ are real numbers, $x,y,z> 0$

Note that $xy+z^2 \ge 2z\sqrt{xy}$ and $x^4+y^4 \ge xy\sqrt{xy}\sqrt{2(x^2+y^2)}$ . Hence
$8P\le \sum_{cyc} \frac{xy^2\sqrt{xy}}{x^4+y^4} \le \sum_{cyc}\frac{y}{\sqrt{2(x^2+y^2)}}\le \frac32$
The last result follows from this well known inequality. So,
$P \le \frac3{16}$

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3428 posts

#8 h Jan 5, 2014, 10:15 AM • 2 Y

Y by Adventure10, Mango247

vutuanhien wrote:

oldbeginner wrote:

vutuanhien wrote:

No one has idea?
This is my solution for this problem
Let $\frac{x}{y}=a, \frac{y}{z}=b, \frac{z}{x}=c\Rightarrow abc=1$
$P=\frac{1}{(a^4+1)(b+c)^3}+\frac{1}{(b^4+1)(c+a)^3}+\frac{1}{(c^4+1)(a+b)^3}$
By AM-GM inequality:
$(b+c)^3\geq 8bc\sqrt{bc}=\frac{8}{a\sqrt{a}}\geq \frac{16}{a^2+a}$
$\Rightarrow 16T\leq \frac{a^2+a}{a^4+1}+\frac{b^2+b}{b^4+1}+\frac{c^2+c}{c^4+1} (1)$
We have:
$\frac{a^2+a}{a^4+1}\leq \frac{3(a+1)}{2(a^2+a+1)}\Leftrightarrow (a-1)^2(a+1)(3a^2+4a+3)\geq 0$ (right)

\[\Rightarrow \[\frac{a^2+a}{a^4+1}+\frac{b^2+b}{b^4+1}+\frac{c^2+c}{c^4+1}\leq \frac{3}{2}\sum \frac{a+1}{a^2+a+1}\leq 3 (2)\]

(This is Vasc inequality:If $abc=1$ then $\sum \frac{a+1}{a^2+a+1}\leq 2$ )
From (1) and (2) we have $P\leq \frac{3}{16}$
Equality occurs if and only if $a=b=c=1$ , or $x=y=z$

It's wrong. Try $x=\frac{1}{3}, y=\frac{1}{4}, z=1$

Where's wrong?

It is very bad that I no longer recognize that initially you posted the wrong version of a seeming inequality, but that's your problem.

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#9 h Jan 5, 2014, 2:38 PM • 5 Y

Y by lenhathoang1998, NYY, hwksltz, lehoangphuc, Adventure10

Here's another solution
We have these inequalities:
$x^4+y^4\geq \frac{2}{3}xy(x^2+y^2+xy)$
$(x+y)^3\geq 4xy(x+y)$
This implies that
$\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}\leq \frac{3}{8}.\frac{x^3y^4z^3}{xy(x^2+xy+y^2)xyz^2(xy+z^2)}$
$\leq \frac{3}{8}.\frac{xy^2z}{(x^2+y^2+xy)(xy+z^2)}$
$=\frac{3}{8}.\frac{xy^2z}{(x^2y^2+y^2z^2+z^2x^2)+xy(x^2+y^2+z^2)}$
$\leq \frac{3xy^2z}{32}.(\frac{1}{x^2y^2+y^2z^2+z^2x^2}+\frac{1}{xy(x^2+y^2+z^2)})$
$=\frac{3}{32}.(\frac{xyz.y}{x^2y^2+y^2z^2+z^2x^2}+\frac{xy}{x^2+y^2+z^2})$
Therefore
$P\leq \frac{3}{32}.(\frac{xyz(x+y+z)}{x^2y^2+y^2z^2+z^2x^2}+\frac{xy+yz+zx}{x^2+y^2+z^2})\leq \frac{3}{16}$

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41897 posts

sqing

#10 h Apr 15, 2014, 4:58 AM • 2 Y

Y by Adventure10, Mango247

vutuanhien wrote:

Here's another solution
We have these inequalities:
$x^4+y^4\geq \frac{2}{3}xy(x^2+y^2+xy)$
$(x+y)^3\geq 4xy(x+y)$
This implies that
$\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}\leq \frac{3}{8}.\frac{x^3y^4z^3}{xy(x^2+xy+y^2)xyz^2(xy+z^2)}$
$\leq \frac{3}{8}.\frac{xy^2z}{(x^2+y^2+xy)(xy+z^2)}$
$=\frac{3}{8}.\frac{xy^2z}{(x^2y^2+y^2z^2+z^2x^2)+xy(x^2+y^2+z^2)}$
$\leq \frac{3xy^2z}{32}.(\frac{1}{x^2y^2+y^2z^2+z^2x^2}+\frac{1}{xy(x^2+y^2+z^2)})$
$=\frac{3}{32}.(\frac{xyz.y}{x^2y^2+y^2z^2+z^2x^2}+\frac{xy}{x^2+y^2+z^2})$
Therefore
$P\leq \frac{3}{32}.(\frac{xyz(x+y+z)}{x^2y^2+y^2z^2+z^2x^2}+\frac{xy+yz+zx}{x^2+y^2+z^2})\leq \frac{3}{16}$

$=\frac{3}{32}.(\frac{xyz\cdot y}{x^2y^2+y^2z^2+z^2x^2}+\frac{yz}{x^2+y^2+z^2})$

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30219 posts

arqady

#11 h Apr 15, 2014, 6:56 AM • 2 Y

Y by Adventure10, Mango247

sqing, we take $\sum_{cyc}$ of this.

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41897 posts

sqing

#12 h Apr 15, 2014, 7:02 AM • 2 Y

Y by Adventure10, Mango247

I know. Proof to be correct . But
$\frac{3xy^2z}{32}.(\frac{1}{x^2y^2+y^2z^2+z^2x^2}+\frac{1}{xy(x^2+y^2+z^2)})$
$\neq \frac{3}{32}.(\frac{xyz.y}{x^2y^2+y^2z^2+z^2x^2}+\frac{xy}{x^2+y^2+z^2})$

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sayantanchakraborty

505 posts

sayantanchakraborty

#13 h Apr 22, 2014, 3:18 PM • 2 Y

Y by Adventure10, Mango247

Sayan wrote:

vutuanhien wrote:

Find the maximum of
$P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$
where $x,y,z$ are real numbers, $x,y,z> 0$

Note that $xy+z^2 \ge 2z\sqrt{xy}$ and $x^4+y^4 \ge xy\sqrt{xy}\sqrt{2(x^2+y^2)}$ . Hence
$8P\le \sum_{cyc} \frac{xy^2\sqrt{xy}}{x^4+y^4} \le \sum_{cyc}\frac{y}{\sqrt{2(x^2+y^2)}}\le \frac32$
The last result follows from this well known inequality. So,
$P \le \frac3{16}$

Sayan,your proof is nice,but the problem you used as a 'theorem' is very strong.

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#14 h May 1, 2014, 4:03 AM • 3 Y

Y by lehoangphuc, Adventure10, Mango247

sqing wrote:

I know. Proof to be correct . But
$\frac{3xy^2z}{32}.(\frac{1}{x^2y^2+y^2z^2+z^2x^2}+\frac{1}{xy(x^2+y^2+z^2)})$
$\neq \frac{3}{32}.(\frac{xyz.y}{x^2y^2+y^2z^2+z^2x^2}+\frac{xy}{x^2+y^2+z^2})$

Oh sorry. Thanks for your correction. It must be $\frac{yz}{x^2+y^2+z^2}$ . But I think my proof is true

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41897 posts

sqing

#15 h Sep 18, 2016, 11:39 AM • 1 Y

Y by Adventure10

vutuanhien wrote:

Find the maximum of
$P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$ where $x,y,z$ are positive real numbers.

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1618 posts

#16 h May 21, 2019, 9:36 AM • 2 Y

Y by Adventure10, Mango247

Here is my solution for this problem
Solution
We have: $P = \sum_{cyc} \dfrac{x^3y^4z^3}{(x^4 + y^4)(xy + z^2)^3} \le \sum_{cyc} \dfrac{x^3y^4z^3}{xy(x^2 + y^2) . 4xyz^2(xy + z^2)} = \sum_{cyc} \dfrac{xy^2z}{4(x^2 + y^2)(xy + z^2)}$ $=$ $\sum_{cyc} \dfrac{xy^2z}{4xy(x^2 + y^2) + 4y^2z^2 + 4x^2z^2} \le \sum_{cyc} \dfrac{xy^2z}{8x^2y^2 + 4y^2z^2 + 4x^2z^2}$
Let $xy = a$ , $yz = b$ , $zx = c$ then: $P \le \sum_{cyc} \dfrac{ab}{8a^2 + 4b^2 + 4c^2}$
Suppose $a \ge b \ge c$ so: $\dfrac{1}{2c^2 + b^2 + a^2} \ge \dfrac{1}{2b^2 + c^2 + a^2} \ge \dfrac{1}{2a^2 + b^2 + c^2}$ and $bc \le ca \le ab$
Hence: $P \le \dfrac{ab}{8c^2 + 4b^2 + 4a^2} + \dfrac{ca}{8b^2 + 4c^2 + 4a^2} + \dfrac{bc}{8a^2 + 4b^2 + 4c^2}$ $\le$ $\dfrac{(a + b)^2}{32c^2 + 16b^2 + 16a^2} + \dfrac{(c + a)^2}{32b^2 + 16c^2 + 16a^2} + \dfrac{(b + c)^2}{32a^2 + 16b^2 + 16c^2}$ $\le$ $\dfrac{1}{16} \left(\dfrac{a^2}{a^2 + c^2} + \dfrac{b^2}{b^2 + c^2} + \dfrac{a^2}{a^2 + b^2} + \dfrac{c^2}{b^2 + c^2} + \dfrac{b^2}{a^2 + b^2} + \dfrac{c^2}{a^2 + c^2} \right) = \dfrac{3}{16}$
Equality holds when: $a = b = c > 0$ or $x = y = z > 0$

This post has been edited 3 times. Last edited by khanhnx, May 21, 2019, 9:39 AM

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