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Source: Iran 2005

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1275 posts

#1 h Aug 27, 2005, 1:03 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Prove that in acute-angled traingle ABC if $r$ is inradius and $R$ is radius of circumcircle then: $a^2+b^2+c^2\geq 4(R+r)^2$

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MysticTerminator

3697 posts

MysticTerminator

#2 h Aug 27, 2005, 3:44 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

It is known $R + r = R(\cos A + \cos B + \cos C)$ so what we want to prove is $\sin^2 A + \sin^2 B + \sin^2 C \ge (\cos A + \cos B + \cos C)^2$ or, if $x^2 + y^2 + z^2 + 2xyz = 1, x^2 + y^2 + z^2 + xy + xz + yz \le \frac{3}{2}$ . This is 19d in Old and New Inequalities.

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3498 posts

#3 h Aug 27, 2005, 4:24 PM • 1 Y

Y by Adventure10

If $\cos^2{A}+\cos^2{B}+\cos^2{C} \leq \frac34$ is true, then the problem is true.

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6555 posts

#4 h Aug 27, 2005, 5:05 PM • 2 Y

Y by Adventure10, Mango247

shobber wrote:

If $\cos^2{A}+\cos^2{B}+\cos^2{C} \leq \frac34$ is true, then the problem is true.

But $\cos^2 A+\cos^2 B+\cos^2 C\leq\frac34$ is not true, exactly the opposite is true: $\cos^2 A+\cos^2 B+\cos^2 C\geq\frac34$ .

Here is another proof of the inequality $a^2+b^2+c^2\geq 4\left(R+r\right)^2$ : As MysticTerminator noted, this inequality is equivalent to $\sin^2 A+\sin^2 B+\sin^2 C\geq\left(\cos A+\cos B+\cos C\right)^2$ . Since $\sin^2 x=1-\cos^2 x$ for any angle x, this becomes

$\left(1-\cos^2 A\right)+\left(1-\cos^2 B\right)+\left(1-\cos^2 C\right)\geq\left(\cos A+\cos B+\cos C\right)^2$
$\Longleftrightarrow\ \ \ 3-\left(\cos^2 A+\cos^2 B+\cos^2 C\right)\geq\left(\cos A+\cos B+\cos C\right)^2$
$\Longleftrightarrow\ \ \ \left(\cos A+\cos B+\cos C\right)^2+\left(\cos^2 A+\cos^2 B+\cos^2 C\right)\leq 3$
$\Longleftrightarrow\ \ \ \left(\cos B+\cos C\right)^2+\left(\cos C+\cos A\right)^2+\left(\cos A+\cos B\right)^2\leq 3$ .

But this inequality was proved on http://www.mathlinks.ro/Forum/viewtopic.php?t=14567 . (Another of my memorial inequalities: When it was posted in July 2004, I wasn't able to solve it; all I could do was, after Fuzzylogic gave a nice solution, to rewrite it in a different way. Later, in February 2005, the same inequality was given to me by Prof. Gronau at the IMO training camp. At that time, I had absolutely forgotten about the solutions given on ML and started solving it again. After 4 hours, I came up with a solution - it was exactly the same as the one posted by Fuzzylogic...)

darij

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nttu

#5 h Aug 28, 2005, 5:08 AM • 1 Y

Y by Adventure10

As I knowed , this ineq can be written in the form : $p^2 \geq 2R^2 +8Rr+3r^2$ , which is found by A.W.Walker
Because $R \geq 2r$ and $2R^2+10Rr-r^2+2(R-2r)\sqrt{R(R-2r)} \geq p^2$ , so we just need to prove one in two ineqs :
$(2R+r)^2 \geq 2R^2+8Rr+3r^2$ or $2R^2+10Rr-r^2+2(R-2r)\sqrt{R(R-2r)} \geq 2R^2+8Rr+3r^2$ $\iff R^2-2Rr-r^2 \geq 0$ or $R^2-2Rr-r^2 \leq 0$ , which is true
The equality holds for $R=2r$ or $R=(\sqrt2 +1)r$
This ineq has an equivalent form : $(HA+HB+HC)^2 \leq AB^2+AC^2+BC^2$ , where $H$ is the orthocenter of triangle $ABC$

PS : This is not my solution , it's get from my document .

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486 posts

nttu

#6 h Aug 28, 2005, 10:53 AM • 1 Y

Y by Adventure10

Here are my solutions for this problem :
First solution :
I'll prove that : $(HA+HB+HC)^2 \leq a^2+b^2+c^2$
We have $AH.AA' = bc.cosA = \frac{b^2+c^2-a^2}{2}$ , where $A'$ is the feet of the altitude from $A$ to $BC$
So $AH.AA' +BH.BB' +CH.CC' = \frac{a^2+b^2+c^2}{2}$
Apply Cauchy's ineq , we have :
$(HA+HB+HC)^2 = (\sum \sqrt{AH.AA'}.\sqrt{\frac{AH}{AA'}})^2 \leq \frac{a^2+b^2+c^2}{2}. \frac{AH}{AA'} = a^2+b^2+c^2$
Second solution :
We have : $2cosA.cosB = \sqrt{sin2A.cotA.sin2B.cotB} \leq \frac{1}{2}(sin2AcotB+sin2BcotA)$
We just need to show : $\sum{cotA(sin2B+sin2C)= -2(cos2A+cos2B+cos2C)} = 3-2(cos^2A+cos^2B+cos^2C)$ , which is true because $cotA(sin2B+sin2C)= -2cos(C+B).cos(C-B) = -(cos2B+cos2C)$

Now , we have over 8 solutions for this problem

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398 posts

#7 h Sep 3, 2005, 7:45 AM • 2 Y

Y by Adventure10, Mango247

MysticTerminator wrote:

$\sin^2 A + \sin^2 B + \sin^2 C \ge (\cos A + \cos B + \cos C)^2$ or, if $x^2 + y^2 + z^2 + 2xyz = 1, x^2 + y^2 + z^2 + xy + xz + yz \le \frac{3}{2}$ . This is 19d in Old and New Inequalities.

How to prove it? Can you show?

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3952 posts

#8 h Sep 3, 2005, 1:10 PM • 1 Y

Y by Adventure10

I can only prove that $xy+yz+zx\leq\frac{3}{4}$

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7054 posts

#9 h Sep 11, 2005, 7:34 PM • 1 Y

Y by Adventure10

Remark.

$4(R+r)^2\le a^2+b^2+c^2\Longleftrightarrow 2R^2+8Rr+3r^2\le s^2\ \ (1)$

$16Rr-5r^2\le 2R^2+8Rr+3r^2\Longleftrightarrow 2R^2-8Rr+8r^2\ge 0\Longleftrightarrow 2(R-2r)^2\ge 0.$

Therefore, if the triangle $ABC$ is acute, then the Blundon's inequality $16Rr-5r^2\le s^2$ is more weak than the inequality (1).

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1906 posts

#10 h Sep 12, 2005, 5:47 PM • 2 Y

Y by Adventure10, Mango247

It's not Blundon's inequality. In fact this inequality is due to Gerrstern.

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7054 posts

#11 h Sep 12, 2005, 11:01 PM • 2 Y

Y by Adventure10, Mango247

O.K. What is the Blundon's inequality then ?

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3952 posts

#12 h Sep 13, 2005, 1:39 PM • 2 Y

Y by Adventure10, Mango247

I think $s^2\geq 16Rr-5r^2$ is Columbien -Doucet ineq(1872)

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3660 posts

Arne

#13 h Sep 13, 2005, 1:53 PM • 2 Y

Y by Adventure10, Mango247

Guys, isn't it clear that some inequalities just get lots of different names? You are all talking about the same inequality, please don't worry about the name

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26 posts

borz

#14 h Sep 13, 2005, 4:25 PM • 2 Y

Y by Adventure10, Mango247

What is the maximum value of $\cos{\alpha}+\cos{\beta}+\cos{\gamma}$ when $\alpha + \beta + \gamma =360 ^o$
I see that it is 3 ( $\alpha=0, \beta = 0 \gamma = 360$ )
But what if this are the angles between circumradiuses and sides of a triangle?

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904 posts

#15 h Dec 28, 2011, 10:20 PM • 3 Y

Y by Polynom_Efendi, Adventure10, Mango247

We have $\triangle ABC$ acute. Therefore, $a^2, b^2, c^2$ are sides of some triangle. Therefore, we can let $a^2 = y+z, \ b^2 = z+x, \ c^2 = x+y$ for some positive reals $x,y,z$ . Also, let $x$ be the largest among $x,y,z$ so that $a$ is the smallest side, and therefore, $h_a$ the largest altitude.
We have, $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{x}{\sqrt{(x+y)(x+z)}}$
and therefore, $\sin A = \sqrt{\frac{xy+yz+zx}{(x+y)(x+z)}}$
Thus, we have, $R = \frac{a}{2 \sin A} = \sqrt{\frac{(x+y)(y+z)(z+x)}{2(xy+yz+zx)}}$
Also, $1+ \frac rR = \cos A + \cos B + \cos C = \frac{\sum x\sqrt{(y+z)}}{\sqrt{(x+y)(y+z)(z+x)}}$
Now, let us state the problem in terms of $x,y,z$ . We have to prove,
$2(x+y+z) \ge 4(R+r)^2 = (2R)^2 (1+ \frac rR)^2 = \frac{(x+y)(y+z)(z+x)}{(xy+yz+zx)} \cdot \frac{(\sum x\sqrt{(y+z)})^2}{(x+y)(y+z)(z+x)}$
$\iff 2(x+y+z)(xy+yz+zx) \ge (\sum x\sqrt{(y+z)})^2$
Now, we have by Cauchy Schwarz Inequality,
$(\sum x\sqrt{(y+z)})^2 \le (\sum x(y+z)) (\sum x) = 2(xy+yz+zx)(x+y+z)$
We are through.

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