# What is the maximum value of $a + b + c$, given $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5}$

What is the maximum value of $a + b + c$, where $a, b, c\in \mathbb{Z}$, and $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5}$$

Note: I could solve the question if the question asks "minimum" instead of "maximum". The answer would be calculated as 45 with arithmetic mean - harmonic mean inequality, where all $a, b, c$ are equal to 15, and that would be the minimum value of $a+b+c$. However, the question asks for the maximum value. I could find some other valid solutions, such as $a=6$, $b=31$, $c=930$, giving the sum equal to 967. I cannot prove whether any larger integer solutions exist or not.

• are the variables $a,b,c$ assumed to be positive? Commented Mar 25, 2018 at 18:33
• Is it clear that there is a maximum? If you drop the integrality, then there clearly is not.
– lulu
Commented Mar 25, 2018 at 18:38
• @lulu Exactly this makes the problem interesting. :)
– user
Commented Mar 25, 2018 at 18:41
• Assuming all three are positive, it's a finite problem. Order so that $a≤b≤c$. Remark that $6≤a≤15$. For fixed $a$ look at $\frac 15-\frac 1a$. That is meant to be $\frac 1b+\frac 1c$ and again there are only finitely many things $b$ might be.
– lulu
Commented Mar 25, 2018 at 18:44
• Note: that argument works even if the variables might be negative. Now order them as $|a|≤|b|≤|c|$. You lose the inequality $6≤|a|$ but you still have $|a|≤15$ which is the important one.
– lulu
Commented Mar 25, 2018 at 18:49

Lacking any insight, what follows is a purely mechanical approach. We'll show that there are only finitely many possibilities for $a,b,c$. We will not assume that they are all positive.
Taking any solution, sort it so that $|a|≤|b|≤|c|$. We remark that $$\frac 15=\big \vert \frac 1a+\frac 1b+\frac 1c\big \vert≤ \frac 1{|a|}+\frac 1{|b|}+\frac 1{|c|}≤\frac 3{|a|}\implies |a|≤15$$
Thus there are only finitely many possible values for $a$.
Fix a choice of $a$. Now we have $\frac 1b+\frac 1c=\frac 15-\frac 1a$ and a similar argument shows that there are only finitely many choices for $b$. As $a,b$ determine $c$ we are done.
Note: I did the search via computer and it appears that the OP has the optimal solution in $(6,31,930)$. However I strongly advise checking this more carefully than I have done.