We are given the equation:

$a + b + c = 1$

We are tasked with finding the lowest possible value of:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

Step 1: Using the Arithmetic Mean-Harmonic Mean (AM-HM) inequality

The AM-HM inequality states that for non-negative real numbers $a$, $b$, and $c$:

$\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

In this case, we know that $a + b + c = 1$, so applying this inequality gives:

$\frac{1}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

Rearranging, we find:

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9$

Thus, the minimum possible value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is 9.

Step 2: Achieving the equality

Equality in the AM-HM inequality occurs when $a = b = c$. Since $a + b + c = 1$, we must have:

$a = b = c = \frac{1}{3}$

Substituting these values into $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$, we get:

$\frac{1}{\frac{1}{3}} + \frac{1}{\frac{1}{3}} + \frac{1}{\frac{1}{3}} = 3 + 3 + 3 = 9$

Thus, the lowest value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is indeed 9, and it is achieved when $a = b = c = \frac{1}{3}$.

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Do you want more details or have any questions?

Here are 5 related questions for deeper exploration:

1. How does the AM-HM inequality generalize to more variables?
2. Can this problem be solved using another inequality, like the Cauchy-Schwarz inequality?
3. How would the result change if $a + b + c = 2$ instead of 1?
4. What is the geometric interpretation of the minimum value found in this problem?
5. How can you use optimization techniques, like Lagrange multipliers, to solve this type of problem?

Tip: Always consider applying inequalities like AM-HM or Cauchy-Schwarz in problems involving sums and reciprocals. These can often provide quick insights into extremal values.