<s>[INST] Maximum value of $a+b$ given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{20}$ What is the maximum value of $a+b$ given that $\frac{1}{a} + \frac{1}{b} = \frac{1}{20}$ here $a,b \in \mathbb{Z^+}$?

What I have gotten so far:
From the above, $\frac{a+b}{ab} = \frac{1}{20} \implies 20(a+b) = ab $
I noticed that $(a-1)(b-1) = ab - a-b+1 = k > 0$ when $a,b > 1$, because when $a, b = 1$ is clearly not a solution. $$\therefore 20(a+b) = ab = a+b - 1 + k > 0 \implies 19(a+b) = (a-1)(b-1) - 1 $$
Using AM-GM I got that $\frac{a+b}{2} \geq \sqrt{ab} \implies ab = 2(a+b) \geq 40\sqrt{ab} \implies \sqrt{ab} \geq 40 \implies ab \geq 1600$.
When $ab = 1600$, one obvious answer is $a=b=40$.

Here, I am stuck... How do I find the maximum value of $a+b$ or perhaps prove that maximum occurs when $a=b=40$?
 [/INST] From $20(a+b) = ab$, we have $b = \frac{20a}{a-20}$. This is an integer, so $a - 20 \mid 20a$.
Let $g = \gcd(a, 20)$. Then $a = ga'$ and $a' - \frac{20}{g} \mid 20a'$. But $\gcd(a'-\frac{20}{g}, 20a') = 1$ (since if not we have a contradiction with $g$ being the greatest divisor of $a$ and $20$), so we must have $a'-\frac{20}{g} = 1$.
Rearranging and substituting back in, $a = 20+g$ and $b=\frac{20(20+g)}{g} = \frac{400}{g} + 20$.
Therefore the problem is to select which of the divisors of $20$, that is $\{1,2,4,5,10,20\}$, maximises $40 + g + \frac{400}{g}$.
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