I can't come up with the intended solution to $\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18$ I was going trough some easy algebra problems when I encountered
$$
\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18.
$$
As you can see the problem is easily solvable with AM > GM
I fairly quickly came up with this solution:
$$
\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab
 = \frac{a^2}{2a^5b^5} + \frac{b^2}{2a^5b^5} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{9ab}{2} + \frac{9ab}{2}
$$
and using AM-GM
$$
\frac{a^2}{2a^5b^5} + \frac{b^2}{2a^5b^5} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{9ab}{2} + \frac{9ab}{2} \ge 7\sqrt[7]{\frac{a^{10}b^{10}9^8}{a^{10}b^{10}2^{10}3^3}} \approx 20
$$
(Yes I was able to do that by hand and later check with my calculator)
I am not sure that this is the intended solution though

Here is the intended solution (this is from a Junior Balkan Mathematical Olympiad TST of Bulgaria).
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.
I did not know before that there is a simple AM-GM solution which gives a better integer constant than $18$, so nice work!
