6. let $a, b, c$ be positive real numbers with $a b c=1$. Determine all possible values that the expression

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

can accept.

## Solution
Let $A$ be the given expression. The following applies because $a b c=1$

$$
\begin{aligned}
A & =\frac{1+a}{1+a+a b}+\frac{a(1+b)}{a(1+b+b c)}+\frac{a b(1+c)}{a b(1+c+c a)} \\
& =\frac{1+a}{1+a+a b}+\frac{a+a b}{1+a+a b}+\frac{a b+1}{1+a+a b} \\
& =\frac{2(1+a+a b)}{1+a+a b}=2
\end{aligned}
$$

The expression is therefore always equal to 2, regardless of $a, b, c$.