Let
\[S = \frac{c}{a} + \frac{a}{b + c} + \frac{b}{c}.\]Then
\[S + 1 = \frac{c}{a} + \frac{a}{b + c} + \frac{b}{c} + 1 = \frac{c}{a} + \frac{a}{b + c} + \frac{b + c}{c}.\]By AM-GM,
\begin{align*}
S + 1 &= \frac{c}{a} + \frac{a}{b + c} + \frac{b + c}{c} \\
&\ge 3 \sqrt[3]{\frac{c}{a} \cdot \frac{a}{b + c} \cdot \frac{b + c}{c}} \\
&= 3.
\end{align*}Note that equality occurs if and only if
\[\frac{c}{a} = \frac{a}{b + c} = \frac{b + c}{c} = 1.\]Since $b$ and $c$ are positive,
\[\frac{b + c}{c} > 1,\]which tells us that equality cannot occur.  Therefore, $S + 1 > 3,$ which means $S > 2.$

We claim that $S$ can take on all real numbers that are greater than 2.  Let $c = a,$ so
\[S = 1 + \frac{a}{b + a} + \frac{b}{a}.\]As $b$ approaches 0, this expression approaches 2.  This tells us that we can make this expression arbitrarily close to 2 as we want.

On the other hand, as $b$ becomes very large, the expression also becomes very large.  This tells us that can we can make this expression arbitrarily large.  Hence, by a continuity argument, $S$ can take on all values in $\boxed{(2,\infty)}.$
Question:
Let $a,$ $b,$ and $c$ be positive real numbers.  Find the set of all possible values of
\[\frac{c}{a} + \frac{a}{b + c} + \frac{b}{c}.\]
Answer:
Let's think step by step.

user
(2,\infty)
Let $a,$ $b,$ and $c$ be positive real numbers.  Find the set of all possible values of
\[\frac{c}{a} + \frac{a}{b + c} + \frac{b}{c}.\]
test