Ukraine 2007

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30189 posts

arqady

#1 h Jun 26, 2007, 4:32 PM • 2 Y

Y by Adventure10, Mango247

$\{a,b,c\}\subset\left(\frac{1}{\sqrt6},+\infty\right)$ such that $a^{2}+b^{2}+c^{2}=1.$ Prove that
$\frac{1+a^{2}}{\sqrt{2a^{2}+3ab-c^{2}}}+\frac{1+b^{2}}{\sqrt{2b^{2}+3bc-a^{2}}}+\frac{1+c^{2}}{\sqrt{2c^{2}+3ca-b^{2}}}\ge2(a+b+c).$

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211 posts

#2 h Jun 26, 2007, 5:34 PM • 2 Y

Y by Adventure10, Mango247

Hm... Does this still hold?

$7\sum a^{4}+29\sum a^{2}b^{2}\geq 18\sum a^{3}b+9\sum ab^{3}+9\sum a^{2}bc$

:

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Yulia

#3 h Sep 16, 2007, 1:03 PM • 2 Y

Y by Adventure10, Mango247

$\frac{1+a^{2}}{\sqrt{2a^{2}+3ab-c^{2}}}+\frac{1+b^{2}}{\sqrt{2b^{2}+3bc-a^{2}}}+\frac{1+c^{2}}{\sqrt{2c^{2}+3ac-b^{2}}}=$
$=\frac{a^{2}+b^{2}+c^{2}+a^{2}}{\sqrt{2a^{2}+3ab-c^{2}}}+\frac{a^{2}+b^{2}+c^{2}+b^{2}}{\sqrt{2b^{2}+3bc-a^{2}}}+\frac{a^{2}+b^{2}+c^{2}+c^{2}}{\sqrt{2c^{2}+3ac-b^{2}}}\geq\frac{(4a+4b+4c)^{2}}{4(\sqrt{2a^{2}+3ab-c^{2}}+\sqrt{2b^{2}+3bc-a^{2}}+\sqrt{2c^{2}+3ac-b^{2}})}$
by Cauchy-Schwarz

$\sqrt{2a^{2}+3ab-c^{2}}+\sqrt{2b^{2}+3bc-a^{2}}+\sqrt{2c^{2}+3ac-b^{2}}\leq 3\sqrt{\frac{2a^{2}+3ab-c^{2}+2b^{2}+3bc-a^{2}+2c^{2}+3ac-b^{2}}{3}}=3\sqrt{\frac{1+3(ab+bc+ac)}{3}}$
by AM-QM
So
$\frac{(4a+4b+4c)^{2}}{4(\sqrt{2a^{2}+3ab-c^{2}}+\sqrt{2b^{2}+3bc-a^{2}}+\sqrt{2c^{2}+3ac-b^{2}})}\geq \frac{4(a+b+c)^{2}}{3\sqrt{\frac{1+3(ab+bc+ac)}{3}}}$
Now we should prove that
$2(a+b+c)\geq 3\sqrt{\frac{1+3(ab+bc+ac)}{3}}$
$4(a^{2}+b^{2}+c^{2}+2(ab+bc+ac))\geq 3(1+3(ab+bc+ac))$
$1\geq ab+bc+ac$ , which obviously true, because $a^{2}+b^{2}+c^{2}=1$
So
$\frac{4(a+b+c)^{2}}{3\sqrt{\frac{1+3(ab+bc+ac)}{3}}}\geq 2(a+b+c)$

Hope that's right. And what is your proof, Michael?

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30189 posts

arqady

#4 h Sep 16, 2007, 4:38 PM • 2 Y

Y by Adventure10, Mango247

Yulia wrote:

And what is your proof, Michael?

Very ugly!

$\sum_{cyc}\frac{1+a^{2}}{\sqrt{2a^{2}+3ab-c^{2}}}\geq2(a+b+c)\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\left(\frac{2a^{2}+b^{2}+c^{2}}{4}\cdot\frac{1}{\sqrt{2a^{2}+3ab-c^{2}}}\right)\geq\frac{a+b+c}{2}.$
But $\sum_{cyc}\left(\frac{2a^{2}+b^{2}+c^{2}}{4}\cdot\frac{1}{\sqrt{2a^{2}+3ab-c^{2}}}\right)\geq\frac{1}{\sqrt{\sum\left(\frac{2a^{2}+b^{2}+c^{2}}{4}\cdot(2a^{2}+3ab-c^{2})\right)}}.$
Thus, it remains to prove that
$16(a^{2}+b^{2}+c^{2})^{3}\geq\sum_{cyc}(a+b+c)^{2}(2a^{2}+b^{2}+c^{2})(2a^{2}+3ab-c^{2})\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(13a^{6}-12a^{5}b-9a^{5}c+32a^{4}b^{2}+38a^{4}c^{2}-27a^{4}bc-11a^{3}b^{3}-$
$-26a^{3}b^{2}c-23a^{3}c^{2}b+25a^{2}b^{2}c^{2})\geq0\Leftrightarrow$
$\Leftrightarrow2\sum_{cyc}(a-b)^{6}+9\sum_{cyc}(a^{6}+a^{2}b^{2}c^{2}-2a^{4}bc)+$
$+3\sum_{cyc}(a^{5}c+a^{3}b^{2}c-2a^{4}bc)+1.5\sum_{cyc}(a^{4}b^{2}+a^{4}c^{2}-2a^{4}bc)+$
$+29\sum_{cyc}(a^{3}b^{3}-a^{3}b^{2}c-a^{3}c^{2}b+a^{2}b^{2}c^{2})+$
$+\sum_{cyc}(0.5a^{4}b^{2}+6.5a^{4}c^{2}+6a^{3}c^{2}b-13a^{2}b^{2}c^{2})\geq0.$

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99 posts

#5 h Sep 16, 2007, 5:29 PM • 4 Y

Y by bobaboby1, hsiangshen, Adventure10, and 1 other user

It is problem from Ukranian TST 2007 number 1.
Here is my solution. It is also author's solution.
We use inequality $\frac{a^{2}}{b}\ge 2a-b$
$\sum_{cyc}\sqrt{\frac{(1+a^{2})^{2}}{2a^{2}+3ab-c^{2}}}\ge\sum_{cyc}\sqrt{2+c^{2}-3ab}\ge\sum_{cyc}\sqrt{\frac{a^{2}+b^{2}+6c^{2}}{2}}\ge\sum_{cyc}\frac{a+b+6c}{4}= 2(a+b+c)$

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329 posts

cancer

#6 h Sep 17, 2007, 6:35 AM • 2 Y

Y by Adventure10, Mango247

Shishkin wrote:

It is problem from Ukranian TST 2007 number 1.

Can you post all the problems from this TST?

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1145 posts

#7 h Sep 18, 2007, 8:42 PM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

$\{a,b,c\}\subset\left(\frac{1}{\sqrt6},+\infty\right)$ such that $a^{2}+b^{2}+c^{2}= 1.$ Prove that
$\frac{1+a^{2}}{\sqrt{2a^{2}+3ab-c^{2}}}+\frac{1+b^{2}}{\sqrt{2b^{2}+3bc-a^{2}}}+\frac{1+c^{2}}{\sqrt{2c^{2}+3ca-b^{2}}}\ge2(a+b+c).$

Another possible solution is with Holder, i.e.

$\left(\frac{1+a^{2}}{\sqrt{2a^{2}+3ab-c^{2}}}+\frac{1+b^{2}}{\sqrt{2b^{2}+3bc-a^{2}}}+\frac{1+c^{2}}{\sqrt{2c^{2}+3ca-b^{2}}}\right)^{2}\geq\frac{4^{3}}{\sum{(1+a^{2})(2a^{2}+3ab-c^{2})}}.$

Hence it remains to prove that $16\geq (\sum{a})^{2}\sum{(1+a^{2})(2a^{2}+3ab-c^{2})}$ , which is true.

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2948 posts

#8 h Apr 24, 2008, 9:59 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

arqady wrote:

$\{a,b,c\}\subset\left(\frac {1}{\sqrt6}, + \infty\right)$ such that $a^{2} + b^{2} + c^{2} = 1.$ Prove that
$\frac {1 + a^{2}}{\sqrt {2a^{2} + 3ab - c^{2}}} + \frac {1 + b^{2}}{\sqrt {2b^{2} + 3bc - a^{2}}} + \frac {1 + c^{2}}{\sqrt {2c^{2} + 3ca - b^{2}}}\ge2(a + b + c).$

By the Cauchy Schwarz Inequality, we get
$\sum \frac{a^2}{\sqrt{2a^2+3ab-c^2}} \ge \frac{(a+b+c)^2}{\sum \sqrt{2a^2+3ab-c^2}} \ge \frac{(a+b+c)^2}{\sqrt{3(\sum a^2+3\sum ab)}} \ge \frac{(a+b+c)^2}{\sqrt{3 \cdot \frac{4}{3}(a+b+c)^2}} =\frac{1}{2}(a+b+c)$
and
$\sum \frac{1}{\sqrt{2a^2+3ab-c^2}} \ge \frac{9}{\sum \sqrt{2a^2+3ab-c^2}} \ge \frac{9}{\sqrt{3(\sum a^2+3\sum ab)}} \ge \frac{9}{\sqrt{3\cdot \frac{4}{3}(a+b+c)^2}} =\frac{9}{2(a+b+c)} =\frac{9(a+b+c)}{2(a+b+c)^2} \ge \frac{9(a+b+c)}{2\cdot 3(a^2+b^2+c^2)} =\frac{3}{2}(a+b+c)$
Adding up these inequalities, we can get the result.

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107 posts

micf

#9 h Aug 25, 2008, 12:29 PM • 2 Y

Y by Adventure10, Mango247

Shishkin wrote:

$\sum_{cyc}\sqrt {\frac {a^{2} + b^{2} + 6c^{2}}{2}}\ge\sum_{cyc}\frac {a + b + 6c}{4}$

Can anybody explain in to me?

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316 posts

#10 h Aug 25, 2008, 1:09 PM • 3 Y

Y by hsiangshen, Adventure10, Mango247

micf wrote:

Shishkin wrote:

$\sum_{cyc}\sqrt {\frac {a^{2} + b^{2} + 6c^{2}}{2}}\ge\sum_{cyc}\frac {a + b + 6c}{4}$

Can anybody explain in to me?

$(a^{2} + b^{2} + 6c^{2})(1+1+6)\ge\ (a+b+6c)^2$

$\rightarrow \sum_{cyc}\sqrt {\frac {a^{2} + b^{2} + 6c^{2}}{2}}\ge\sum_{cyc}\frac {a + b + 6c}{4}$

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107 posts

micf

#11 h Aug 25, 2008, 1:24 PM • 2 Y

Y by Adventure10, Mango247

Thank you

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404 posts

#12 h Oct 27, 2011, 2:15 PM • 2 Y

Y by Adventure10, Mango247

I solve it by using Carlson inequality first,and then
by a series of complicated calculation,and using:
3 $\sum-{cyc}a^{3}b\le\ (a^{2}+b^{2}+c^{2})^2$
the problem can be solved

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