I can't come up with the intended solution to $\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab>18$

Question

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

Answer 1

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!

Answer 2

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.