If a=b=c then a+b+c=3a and 3/a=1/2, so 6=a=b=c and a+b+c=18.
Given a and b a value of c can always be found; 1/c=1/2-1/a-1/b; 1/c=(ab-2b-2a)/2ab, c=2ab/(ab-2b-2a). So, unless there are restrictions on a, b and c, there may be no definable max or min.
1/b+1/c=1/2-1/a; b+c=bc(a-2)/2a; a+b+c=a+bc(a-2)/2a=a+b^2(a-2)/(ab-2(a+b)). Note that the denominator is the same as the one for defining c.
If we assume we have positive integers only, then the denominator ab>2(a+b) and ab-2(a+b) must be a factor of 2ab. ab>2a+2b implies ab-2a>2b, a(b-2)>2b, a>2b/(b-2) or b>2a/(a-2) and a, b>2. If b=3, a>6 (and vice versa). So if b=3 and a=7, c=42 and a+b+c=52. Check: 1/7+1/3+1/42=(6+14+1)/42=21/42=1/2. If b=4, a>4, so put a=5, then c=20. Check: 1/5+1/4+1/20=(4+5+1)/20=10/20=1/2 and a+b+c=29. Put a=6 and b=4, c=2*6*4/4=12. Check: 1/6+1/4+1/12=(2+3+1)/12=6/12=1/2, a+b+c=22. We already know that if a=b=c=6, the sum is 18, the minimum so far. The maximum so far is 52.
We can start to organise the positive integer solutions into sets where a≤b≤c.
So far we have {3 7 42}, {4 5 20}, {4 6 12}, {6 6 6} and with these we can associate sums: 52, 29, 22, 18. We have identified a sequence of integers 3, 4, 5, 6, 7. We can continue the sequence, looking for more. If we introduce a=8, we get b>8/3, so b≥3. This is the same as a≥3 and b=8. Put a=3, then c=24 giving us {3 8 24} (1/3+1/8+1)/24=12/24=1/2), and the sum 35. And a=3, b=9 gives c=2*3*9/3=18: {3 9 18}, sum=30. The sequence of sums becomes 18, 22, 29, 30, 35, 52.
We can continue with a=3, b=10, c=15; sum=28. As a and b get further apart, the denominator ab-2(a+b) gets larger making c smaller, so the sum is also reduced. Also, b is catching up with c. It's beginning to look like 52 is the maximum sum, with 18 as the minimum. The minimum value of the denominator is 1 to maximise c:
ab-2(a+b)=1; ab-2a-2b-1=0; b(a-2)=2a+1; b=(2a+1)/(a-2); (a,b)=(3,7) or (7,3) are the only positive integer solutions, so c=42 and 7+3+42=52 is the maximum value of a+b+c.