If a^2 + b^2 + C^2 = 9 then what is the max value of 2(a + b + c) - abc where a,b,c are real numbers?

Question

If
abcde=1 (where a,b,c,d and e are all positive real numbers)
then
what is the minimum value of a+b+c+d+e?

Answer 1

a^2 + b^2 + C^2 = 9 implies a = +/-1, b = +/- 2 = c.
so to maximize 2(a + b + c) - abc we need one variable among a,b,c to be negative. But only if 1 is negative we can get the maximum because the term 2(a + b + c) will be biggest only if 1 is negative. So if a = -1 & b = c = 2.
2(-1+2+2) - (-1)(2)(2) = 6 + 4 = 10. ie., 10 is the largest value of 2(a + b + c) - abc with a^2 + b^2 + C^2 = 9 as the constraint.