Let $a,$ $b,$ and $c$ be positive real numbers.  Find the minimum value of
\[\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.\]
By AM-GM,
\[\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge 3 \sqrt[3]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3.\]Equality occurs when $a = b = c,$ so the minimum value is $\boxed{3}.$
3
Level 2
Intermediate Algebra
Problem: Let $a,$ $b,$ and $c$ be positive real numbers.  Find the minimum value of
\[\frac{a}{b} + \frac{b}{c} + \frac{c}{a}.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum value of the expression \(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. The AM-GM inequality states that for any non-negative real numbers \(x_1, x_2, \ldots, x_n\), the following holds:

\[
\frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n}
\]

with equality when all \(x_i\) are equal.

In our case, we set \(x_1 = \frac{a}{b}\), \(x_2 = \frac{b}{c}\), and \(x_3 = \frac{c}{a}\). Thus, we can apply AM-GM to these three terms:

\[
\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq \sqrt[3]{\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}}
\]

Calculating the product on the right side, we have:

\[
\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} = 1
\]

Therefore, we can simplify the inequality:

\[
\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq \sqrt[3]{1} = 1
\]

Multiplying both sides by 3 gives:

\[
\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3
\]

This shows that the minimum value of \(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\) is at least 3. 

To determine if this minimum can be achieved, we check the condition for equality in the AM-GM inequality, which occurs when all terms are equal:

\[
\frac{a}{b} = \frac{b}{c} = \frac{c}{a}
\]

Let \(k = \frac{a}{b} = \frac{b}{c} = \frac{c}{a}\). From these equalities, we can express \(a\), \(b\), and \(c\) in terms of a single variable. For instance, if we let \(b = 1\), then \(a = k\) and \(c = \frac{1}{k}\). 

Substituting these into the expression gives:

\[
\frac{k}{1} + \frac{1}{\frac{1}{k}} + \frac{\frac{1}{k}}{k} = k + k + \frac{1}{k} = 2k + \frac{1}{k}
\]

To minimize \(2k + \frac{1}{k}\), we can take the derivative and set it to zero. The derivative is:

\[
2 - \frac{1}{k^2}
\]

Setting this equal to zero gives \(2 = \frac{1}{k^2}\), or \(k^2 = \frac{1}{2}\), leading to \(k = \frac{1}{\sqrt{2}}\). 

Substituting \(k = 1\) (the case where \(a = b = c\)) directly into the original expression yields:

\[
\frac{1}{1} + \frac{1}{1} + \frac{1}{1} = 3
\]

Thus, the minimum value of \(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\) is achieved when \(a = b = c\), confirming that the final answer is \(3\).<|END_OF_LATENT|> The final answer is $3$.