(a^2+b^2)/(2a^5b^5) + (81a^2b^2)/4 + 9ab \geq 81

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(a^2+b^2)/(2a^5b^5) + (81a^2b^2)/4 + 9ab \geq 81

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Source: Bulgaria JBMO TST 2018, Day 1, Problem 2

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#1 h Jun 25, 2018, 7:45 AM • 3 Y

Y by BabaAuRhum, Adventure10, Mango247

For all positive reals $a$ and $b$ , show that
$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab > 18.$

This post has been edited 3 times. Last edited by Amir Hossein, Jun 25, 2018, 8:20 AM

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#2 h Jun 25, 2018, 7:56 AM • 2 Y

Y by Adventure10, Mango247

Amir Hossein wrote:

For all positive reals $a$ and $b$ , show that
$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab \geq 81.$

It's wrong $a=b=1.$

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#3 h Jun 25, 2018, 8:02 AM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

Taken from Pregătire Matematică Olimpiade Juniori. Translated by yours truly.

I have attached the competition here. Maybe the problem is wrong?

Attachments:

baraj-2-2018.pdf (110kb)

baraj-1-2018.pdf (93kb)

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fastlikearabbit

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fastlikearabbit

#5 h Jun 25, 2018, 8:12 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Amir Hossein wrote:

For all positive reals $a$ and $b$ , show that
$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab > 18.$

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#6 h Jun 25, 2018, 8:20 AM • 2 Y

Y by Adventure10, Mango247

fastlikearabbit wrote:

Amir Hossein wrote:

For all positive reals $a$ and $b$ , show that
$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab > 18.$

That makes more sense. Thanks. Fixed.

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Darghy

#7 h Jun 25, 2018, 8:32 AM • 6 Y

Y by Amir Hossein, BabaAuRhum, A-Thought-Of-God, Adventure10, Mango247, AylyGayypow009

By repeated use of AM-GM,
$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab \geq \frac{2ab}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab = \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + 9ab = \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + \frac{9ab}{2} + \frac{9ab}{2} \geq 4 \sqrt[4]{\frac{9^4}{2^4}} = 18$
For equality to hold, we must have $a=b$ , $\frac{9ab}{2} = \frac{81a^2b^2}{4}$ and $\frac{9ab}{2} = \frac{1}{a^4b^4}$ , which is not possible.

This post has been edited 1 time. Last edited by Darghy, Jun 25, 2018, 8:32 AM

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