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Nice problem from leviethai

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2948 posts

#1 h Dec 15, 2010, 12:14 PM • 2 Y

Y by Adventure10, Mango247

Let $a,$ $b,$ $c$ be nonnegative real numbers such that $a^2+b^2+c^2+ab+bc+ca=6.$ Determine the maximum value of
$P=a^2b+b^2c+c^2a.$

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512 posts

diks94

#2 h Dec 15, 2010, 1:36 PM • 2 Y

Y by Adventure10, Mango247

according to me answer to the above question will be 3.
As By simply Putting a=b=c=1
else if we put either of a or b to be >1
Then if We Find the value of c by quadratic formulae...c will be either negative or will be complex ..hence max values of a=b=c=1
therefore answer =3

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3377 posts

mudok

#3 h Sep 16, 2016, 6:52 AM • 2 Y

Y by Adventure10, Mango247

unsolved?

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4613 posts

#4 h Sep 16, 2016, 7:08 AM • 4 Y

Y by bel.jad5, lminsl, Adventure10, Mango247

The maximum is not 3

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zdyzhj

#5 h Sep 16, 2016, 9:16 AM • 3 Y

Y by bel.jad5, Adventure10, Mango247

luofangxiang wrote:

The maximum is not 3

kmax= $\frac{368}{3}-\frac{176\sqrt{33}}{9}$

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#7 h Sep 16, 2016, 9:54 AM • 2 Y

Y by Adventure10, Mango247

Why??? Can you post your solution?

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30150 posts

arqady

#8 h Sep 16, 2016, 10:24 AM • 2 Y

Y by Adventure10, Mango247

We can solve this problem by $uvw$ 's technique, but it's a lot of work.

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CEH

#9 h Sep 16, 2016, 10:33 AM • 1 Y

Y by Adventure10

How,arqady? Can you just post some outline please?

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30150 posts

arqady

#10 h Sep 16, 2016, 11:05 AM • 3 Y

Y by Akatsuki1010, Adventure10, Mango247

Let $k$ is a maximum, $a+b+c=3u$ , $ab+ac+bc=3v^2$ , $abc=w^3$ .
Hence, we need to prove that $a^2b+b^2c+c^2a\leq k\left(\frac{9u^2-6v^2+3v^2}{6}\right)^{\frac{3}{2}}$ or
$\sum_{cyc}(a^2b-a^2c)\leq 2k\left(\frac{3u^2-v^2}{2}\right)^{\frac{3}{2}}-\sum_{cyc}(a^2b+a^2c)$ or
$(a-b)(a-c)(b-c)\leq2k\left(\frac{3u^2-v^2}{2}\right)^{\frac{3}{2}}-9uv^2+3w^3$ .
We can show that for all $k\geq3$ the right side is non-negative and it remains to prove that
$\prod_{cyc}(a-b)^2\leq\left(2k\left(\frac{3u^2-v^2}{2}\right)^{\frac{3}{2}}-9uv^2+3w^3\right)^2$ or
$27(u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\leq\left(2k\left(\frac{3u^2-v^2}{2}\right)^{\frac{3}{2}}-9uv^2+3w^3\right)^2$ ,
which is a quadratic inequality of $w^3$ . The rest is smooth, but a lot of work.

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#11 h Sep 16, 2016, 4:35 PM • 2 Y

Y by Adventure10, Mango247

when the equality occur ??

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fox04

#12 h Sep 16, 2016, 4:40 PM • 2 Y

Y by Adventure10, Mango247

Why was this thing revived from nearly 6 years ago?

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830 posts

#13 h Sep 16, 2016, 4:42 PM • 2 Y

Y by Adventure10, Mango247

fox04 wrote:

Why was this thing revived from nearly 6 years ago?

we shouldn't let any problem be unsolved

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981 posts

zdyzhj

#14 h Sep 17, 2016, 1:57 AM • 2 Y

Y by Adventure10, Mango247

thuanz123 wrote:

fox04 wrote:

Why was this thing revived from nearly 6 years ago?

we shouldn't let any problem be unsolved

but why don't you help me with f(a,b,c)?

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#15 h Sep 17, 2016, 3:01 AM • 2 Y

Y by Adventure10, Mango247

problem seems not that efficient...

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4655 posts

#16 h Sep 17, 2016, 5:57 AM • 1 Y

Y by Adventure10

It' efficiency will be determined by it's solution to which arqady made a nice breakthrough .

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dm_edogawasonichi

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dm_edogawasonichi

#18 h Sep 17, 2016, 3:20 PM • 1 Y

Y by Adventure10

I think C-S helps here. Who can find it?

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#19 h Sep 17, 2016, 10:26 PM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

can_hang2007 wrote:

Let $a,$ $b,$ $c$ be nonnegative real numbers such that $a^2+b^2+c^2+ab+bc+ca=6.$ Determine the maximum value of
$P=a^2b+b^2c+c^2a.$

This problem is an easy application of Lagrange multiplier method. The critical point lead to 3 and we only need to check the borders which lead to a maximum : $\frac{4}{3}\sqrt{69-11\sqrt{33}}>3$

This post has been edited 5 times. Last edited by bel.jad5, Sep 18, 2016, 8:21 AM

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#20 h Sep 18, 2016, 4:01 AM • 2 Y

Y by Adventure10, Mango247

denote $x=\sum{a^2}$ and $\sum{ab}=y$ we have $x+y=6$
and we have $\sum{(c^2+2ab)^2}\sum{(2a+b+c)^2} \ge \left(\sum{(c^2+2ab)(2a+b+c)}\right)^2$
$\Leftrightarrow (x^2+2y^2)(6x+10y) \ge \left(\sqrt{(x+2y)^3}+3P \right)^2$
use $x+y=6$ we can get the answer

This post has been edited 1 time. Last edited by thuanz123, Sep 18, 2016, 4:02 AM

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#21 h Sep 18, 2016, 6:45 AM • 2 Y

Y by Adventure10, Mango247

zdyzhj wrote:

luofangxiang wrote:

The maximum is not 3

kmax= $\frac{368}{3}-\frac{176\sqrt{33}}{9}$

I am sure you mean: $\sqrt{\frac{368}{3}-\frac{176\sqrt{33}}{9}} =\frac{4}{3}\sqrt{69-11\sqrt{33}}>3$

Which is what I have found as well using Lagrange.

This post has been edited 3 times. Last edited by bel.jad5, Sep 18, 2016, 6:49 AM

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#22 h Sep 18, 2016, 10:13 AM • 1 Y

Y by Adventure10

another proof
WLOG we assume that $c=min\{a;b;c\}$
denote $f(a,b,c)=\frac{\left(\sum{a^2b}\right)^2}{\left(\sum{a^2}+\sum{ab}\right)^3}$
we prove $f(a,b,c) \le f(a,b,0)$

This post has been edited 2 times. Last edited by thuanz123, Sep 18, 2016, 1:51 PM

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#23 h Sep 18, 2016, 10:33 AM • 1 Y

Y by Adventure10

thuanz123 wrote:

another proof
WLOG we assume that $c=min\{a;b;c\}$
denote $f(a,bc)=\frac{\left(\sum{a^2b}\right)^2}{\left(\sum{a^2}+\sum{ab}\right)^3}$
we prove $f(a,b,c) \le f(a,b,0)$

Show your proof

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3750 posts

#24 h Sep 20, 2016, 6:47 AM • 1 Y

Y by Adventure10

To solve the critical points system:
http://www.artofproblemsolving.com/community/u21755h1307293p6991918

To solve the problem on the border:
http://www.artofproblemsolving.com/community/c6t243f6h1307419_inequality_with_x_y

This post has been edited 1 time. Last edited by bel.jad5, Sep 20, 2016, 6:48 AM

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#25 h Sep 20, 2016, 7:45 AM • 2 Y

Y by Adventure10, Mango247

But the most important is prove that $P$ will maximum value when $a = 0$ or $b = 0$ or $c=0.$

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#26 h Sep 20, 2016, 2:20 PM • 2 Y

Y by Adventure10, Mango247

Nguyenhuyen_AG wrote:

But the most important is prove that $P$ will maximum value when $a = 0$ or $b = 0$ or $c=0.$

I did it using Lagrange Multipliers. It is definitely tedious but the critical points lead to a maximum of 3, so you just have to check the borders.

This post has been edited 2 times. Last edited by bel.jad5, Sep 20, 2016, 2:20 PM

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