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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, $$\begin{array}{rl}S+1& =\frac{c}{a}+\frac{a}{b+c}+\frac{b+c}{c}\\ & \ge 3\sqrt[3]{\frac{c}{a}\cdot \frac{a}{b+c}\cdot \frac{b+c}{c}}\\ & =3.\end{array}$$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 $\boxed{(2,\infty )}.$