if 2a+3b+4c=1, what is minimum value of (1/a)+(1/b)+(1/c)

Answer 1

You don't say if there are any restrictions to a, b, and c (positive values only, integers only, etc). I will assume there are no restrictions.

Since we have 2a + 3b + 4c = 1, we can say that c = (1 - 2a - 3b)/4

This makes the function 1/a + 1/b + 1/c actually the same as:

1/a + 1/b + 1/[(1 - 2a - 3b)/4] =

1/a + 1/b + 4/(1 - 2a - 3b) which is a function of just "a" and "b"

As "a" and "b" get larger and larger, the value gets closer and closer to zero. For instance for a = 100 and b = 100 the value is 1/100 + 1/100 + 4/(-499) = 0.01198 For a = 1000 and b = 1000, the value is 0.0011998.

If we keep "a" or "b" at a value of 1, and make the other value increasingly large, the value gets closer and closer to 1.

If we use fractions less than 1, the value starts getting large. For instance for a = 1/10 and b = 1/10, the value becomes 28. For a = 1/100 and b = 1/100, the value is 204. So we see that is not the way to a minimum value.

If we make "a" and "b" negative numbers, and we start getting more and more negative, we see the value approaches zero from the other side. For instance a = -10 and b = -10 gives a value of -0.122 and for a = -100 and b = -100, the value becomes -0.012

But when we make either "a" or "b" a number between -1 and zero, we see interesting results. For a = -0.1 and b = -0.1, the value is -17.3. For a = -0.001 b = -0.001, the value is -1996 and for a = -0.00001 and b = -0.00001, the value is -199,996.

Therefore, we can conclude that there is no minimum value. We can get closer and closer to negative infinity as we let "a" and "b" approach zero from the negative side.