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
 A: 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!
A: Let functions $A$ and $G$ denote the arithmetic and geometric means of their arguments.  Then:
$$x := \frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab$$
$$= A(\frac{a^2}{a^5b^5}, \frac{b^2}{a^5b^5}) + \frac{81a^2b^2}{4} + 9ab$$
$$= A(\frac{1}{a^3b^5}, \frac{1}{a^5b^3}) + \frac{81a^2b^2}{4} + 9ab$$
$$\ge G(\frac{1}{a^3b^5}, \frac{1}{a^5b^3}) + \frac{81a^2b^2}{4} + 9ab$$
$$= \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + 9ab$$
Note that $a$ and $b$ now only occur in same-exponent pairs.  So, for convenience, let $p = ab$.
$$x \ge \frac{1}{p^4} + \frac{81p^2}{4} + 9p$$
$$=A(\frac{2}{p^4}, \frac{81p^2}{2}) + 9p$$
$$\ge G(\frac{2}{p^4}, \frac{81p^2}{2}) + 9p$$
$$= \frac{9}{p} + 9p$$
If $p \ge 2$, then $x > 9p \ge 18$.
If $p \ge \frac{1}{2}$, then $x > \frac{9}{p} \ge 18$.
Now, let's consider $\frac{1}{2} < p < 2$.  Then $\frac{1}{2} < \frac{1}{p} < 2$ as well.  So,
$$\frac{9}{2} < \frac{9}{p} < 18$$
$$\frac{9}{2} < 9p < 18$$
Adding these gives:
$$18 < \frac{9}{p} + 9p < 36$$
So $x \ge \frac{9}{p} + 9p > 18$.
Finally, just in case anyone nitpicks strict versus non-strict inequality at the boundaries between cases:
$$p = \frac{1}{2} \implies \frac{9}{p} + 9p = 22.5 > 18$$
$$p = 2 \implies \frac{9}{p} + 9p = 22.5 > 18$$
Combining the cases of $0 < p < \frac{1}{2}$, $p = \frac{1}{2}$, $\frac{1}{2} < p < 2$, $p = 2$, and $p > 2$; we can conclude that $\forall p > 0$ (and so $\forall a,b > 0$), $x > 18$, Q.E.D.
