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 $a=b=1.$

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)

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

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.

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.