<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] You can get a much more convenient condition by using $(a-r)(b-r)=ab-ar-br+r^2$ for a value of $r$ other than $1$.

Notice that from $20(a+b)=ab$ you can derive $ab-20a-20b=0$ and if you choose $r=20$ you should see $(a-20)(b-20)=400$. You should now find it easy to finish.

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