# 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 $$\frac{a^2}{2a^5b^5} + \frac{b^2}{2a^5b^5} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{81a^2b^2}{12} + \frac{9ab}{2} + \frac{9ab}{2} \ge 7\sqrt{\frac{a^{10}b^{10}9^8}{a^{10}b^{10}2^{10}3^3}} \approx 20$$ (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

• Is it explicitly stated that $a$ and $b$ are positive?
– Dan
Aug 26, 2022 at 18:22
• Yeah sorry for not stating that in the original question Aug 26, 2022 at 18:58

Here is the intended solution (this is from a Junior Balkan Mathematical Olympiad TST of Bulgaria).

By repeated use of AM-GM, $$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab \geq \frac{2ab}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab = \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + 9ab = \frac{1}{a^4b^4} + \frac{81a^2b^2}{4} + \frac{9ab}{2} + \frac{9ab}{2} \geq 4 \sqrt{\frac{9^4}{2^4}} = 18$$ 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!

• Yeah I am Bulgarian too Aug 26, 2022 at 18:58
• The lowest value I've been able to obtain experimentally is 20.006494. So 20 is indeed the largest integer constant that works.
– Dan
Aug 30, 2022 at 1:07

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,

$$\frac{9}{2} < \frac{9}{p} < 18$$ $$\frac{9}{2} < 9p < 18$$

$$18 < \frac{9}{p} + 9p < 36$$
So $$x \ge \frac{9}{p} + 9p > 18$$.
$$p = \frac{1}{2} \implies \frac{9}{p} + 9p = 22.5 > 18$$ $$p = 2 \implies \frac{9}{p} + 9p = 22.5 > 18$$
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.