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. 

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.
