<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] $a=b=40$ is clearly a solution. If one number is larger than $40$, the other must be less than $40$. But also neither number can be smaller than $21$. So one approach would be to let $a$ range from $21$ to $39$ and see what $b$ arise for each $a$ and if they are integers. And then take note of any values of $a+b$ and see if they beat $40+40$.
Actually right away you find that $a=21$ gives $b=420$. And if we drop the integer condition and apply calculus to the problem, we see that $a+b\to\infty$ as $a\to0$ from $a=40$. So this establishes that $21+420$ is optimal.
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