Trig problem

treegoner

Let ABC be an acute triangle. Prove that
$a^4b^2+b^4a^2+c^4a^2+a^4c^2+b^4c^2+c^4b^2 -3a^2b^2c^2\geq (a^2+b^2+c^2)^2R^2$
where R is it's circumcircle radius.

**Posted:** Thu Jul 01, 2004 9:53 am

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Life doesn't always follow what I have planned or expected. No matter what way my life leads me, I am still thankful.

ma_go

what about setting a=x+y, b=y+z, c=z+x and (sadicly) expand all?
maybe muirheads (bunching?) inequality may solve Smile

(i don't really want to do it Mr. Green

)

**Posted:** Thu Jul 01, 2004 10:16 am

treegoner

Dear ma-go, you should do it. You may face difficulty when solving it. I have a very beautiful solution to it ( because I don't like heavy computation Confused

).

**Posted:** Thu Jul 01, 2004 10:33 am

_________________
Life doesn't always follow what I have planned or expected. No matter what way my life leads me, I am still thankful.

darij grinberg

**Posted:** Thu Jul 01, 2004 12:19 pm

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darij grinberg

Using my substitution $a^2=y+z$ , $b^2=z+x$ , $c^2=x+y$ , your inequality simplifies to

$2\left( yz^{2}+zy^{2}+zx^{2}+xz^{2}+xy^{2}+yx^{2}+x^{3}+y^{3}+z^{3}+3xyz\right) \geq \left( x+y+z\right) ^{2}\frac{\left( y+z...$ ,

or, equivalently,

$8\left( \left( x+y+z\right) \left( x^{2}+y^{2}+z^{2}\right) +3xyz\right) \left( yz+zx+xy\right) \geq \left( x+y+z\right) ^{2}...$ .

(Now Mr. Hyde speaking)

But this is trivial, since

$8\left( \left( x+y+z\right) \left( x^{2}+y^{2}+z^{2}\right) +3xyz\right) \left( yz+zx+xy\right) -\left( x+y+z\right) ^{2}\lef...$ ,

what is surely $\geq 0$ .

(Dr. Jekyll again)

Any simpler proof?

Darij

**Posted:** Fri Jul 02, 2004 2:50 am

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Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

thus the inequality isn't sharp at all from your proof(or did I see something wrong?).

but setting $a=b=c$ , thus $R=\frac{1}{\sqrt{3}}a$ , we get equality...I didn't check your computations, but there must be something wrong i think.

Peter

**Posted:** Fri Jul 02, 2004 3:05 am

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

btw darij, in the original inequality, we had $blabla...-3a^2b^2c^2$ not $blabla...+3a^2b^2c^2$ Mr. Green

peter

**Posted:** Fri Jul 02, 2004 3:07 am

darij grinberg

Thanks, Peter, I was wrong indeed, there should be equality for a = b = c. The error was not the $blabla...-3a^2b^2c^2$ (in fact, the minus sign was there but it was killed by the substitution), but because I forgot a "4" before the $\left( x+y+z\right)^2$ . So the correct inequality should be

$2\left( yz^{2}+zy^{2}+zx^{2}+xz^{2}+xy^{2}+yx^{2}+x^{3}+y^{3}+z^{3}+3xyz\right) \geq \left( x+y+z\right) ^{2}\frac{\left( y+z...$ .

So the proof must be modified:

$2\left( yz^{2}+zy^{2}+zx^{2}+xz^{2}+xy^{2}+yx^{2}+x^{3}+y^{3}+z^{3}+3xyz\right) -\left( x+y+z\right) ^{2}\frac{\left( y+z\rig...$ ,

But $-y^{3}z^{2}-y^{3}x^{2}-y^{2}z^{3}-y^{2}x^{3}-z^{3}x^{2}-z^{2}x^{3}+z^{4}x+z^{4}y+x^{4}z+x^{4}y+y^{4}z+y^{4}x \geq 0$ from Muirhead, hence we are done.

Thanks again, Peter.

Darij

**Posted:** Fri Jul 02, 2004 3:16 am

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arthur

a transformation like that $a=x+y$ , etc isn't called Ravi's transformation ?

**Posted:** Fri Jul 02, 2004 3:30 am

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

maybe, but who's interested in the name of that? it's not a theorem, thus you do not have to cite it, thus it is senseless to know the name Mr. Green

btw it makes no sense to give it a name since that transformation has been used for long times

Peter

**Posted:** Fri Jul 02, 2004 3:49 am

darij grinberg

Yes, a = y+z, b = z+x and c = x+y is called the Ravi transformation, but this kind of naming is historical nonsense. BTW, the substitution I used ( $a^2=y+z$ , $b^2=z+x$ , $c^2=x+y$ ) yields a nice parametrization of all acute triangles. But I hope you won't call it the Darij substitution!

Darij

**Posted:** Fri Jul 02, 2004 4:33 am

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Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

i wouldn't have done if you didn't mention...but thus if it this year of use at the IMO i will call it the darij substitution Mr. Green

(just gronau will be wondering but anyway it won't influence my result).

Peter

**Posted:** Fri Jul 02, 2004 4:36 am

treegoner

Thank you Darij for your beautiful substitution.
Here is my proof:
Initally, I want to introduce you my lemma:
Let ABC be an acute triangle, m,n,p >0. Then
$(mcosA + ncosB + pcosC)^2 \leq \sum (n^2+p^2-m^2)sin^2C$
Proof:
It is just Bulhiacopski inequality with the application of $\sum \frac { cosA}{sinBsinC} = 2$ and $2cosAsinBsinC = sin^2B +sin^2C - sin^2A.$
$LHS=(\sum (m\sqrt{cosAsinBsinC}) \sqrt \frac {cosA}{sinBsinC})^2 \leq RHS$
Now by replacing $m = bc, n = ca, p=ab$ , we have our inequality.

**Posted:** Fri Jul 02, 2004 9:22 am

_________________
Life doesn't always follow what I have planned or expected. No matter what way my life leads me, I am still thankful.

darij grinberg

**Posted:** Fri Jul 02, 2004 12:25 pm

_________________
Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...

treegoner

In my country, it is called Bunjacovsky.
But I wonder why that inequality is called Cauchy Schwarz ? What about Bunjacovsky? And do you know anything else about him? Thank you. Razz

**Posted:** Fri Jul 02, 2004 3:52 pm

_________________
Life doesn't always follow what I have planned or expected. No matter what way my life leads me, I am still thankful.

darij grinberg

Well actually, whether it's "Bunjacovsky", or "Bunjakovsky", or "Buniakovski", or "Bunjakovski", does not really matter, but you wrote "Bulhiacopski", which sounds a bit strange Mr. Green

Actually, if I remember correctly, the inequality itself, in its elementary form, is due to Cauchy. Later, Schwarz and Bunjakovski (independently) found the integral version of the inequality, but they didn't contribute anything to the elementary variant, so the elementary inequality can be just called the "Cauchy inequality" (and I think some people actually call it so).

Darij

**Posted:** Sat Jul 03, 2004 12:21 am

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Now the die is cast, the first step taken, a glimmer of hope lights up our lives
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manlio

Treegoner solution is elegant and neat , but also Darij solution is very interesting even if there are many computations to check ; last thing to tell is that to prove your last expression, Darij, you must not invoke Muirhead but simply AM-GM as

$(x^3+y^3) \geq xy(x+y)$

**Posted:** Sun Jul 04, 2004 3:09 am

darij grinberg

Oh thanks, I really involved some too heavy arsenal...

Darij

**Posted:** Sun Jul 04, 2004 6:34 am

_________________
Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

the argument "you do not have to use muirhead, you can use AM-GM" is strictly speaking nonsense since muirhead is an application of AM-GM...
(BTW $x^3+y^3\geq x^2y+xy^2$ was an olympiad problem this year in germany(in a lower round)...darij first wanted to use muirhead to prove it but than decided to write $(x+y)(x-y)^2\geq 0$ Smile

)

Peter

**Posted:** Sun Jul 04, 2004 6:58 am

darij grinberg

**Posted:** Sun Jul 04, 2004 7:52 am

_________________
Now the die is cast, the first step taken, a glimmer of hope lights up our lives
Visions of the past, dreams forsaken forming right under our eyes
We are alive...