1. **Stating the problem:** Given that $abc=1$, calculate the expression $$\frac{a}{ab+a+1} + \frac{b}{bc+b+1} + \frac{c}{ac+c+1}.$$\n\n2. **Rewrite each denominator using the condition $abc=1$: **\nSince $abc=1$, we can express $bc=\frac{1}{a}$, $ac=\frac{1}{b}$, and $ab=\frac{1}{c}$.\n\n3. **Rewrite each term:**\n- First term denominator: $ab + a + 1 = \frac{1}{c} + a + 1$\n- Second term denominator: $bc + b + 1 = \frac{1}{a} + b + 1$\n- Third term denominator: $ac + c + 1 = \frac{1}{b} + c + 1$\n\n4. **Rewrite the entire expression:**\n$$\frac{a}{\frac{1}{c} + a + 1} + \frac{b}{\frac{1}{a} + b + 1} + \frac{c}{\frac{1}{b} + c + 1}.$$\n\n5. **Multiply numerator and denominator of each fraction by the respective variable in the denominator to clear fractions:**\n- For the first term, multiply numerator and denominator by $c$:\n$$\frac{a \cdot c}{1 + a c + c} = \frac{a c}{1 + a c + c}.$$\n- For the second term, multiply numerator and denominator by $a$:\n$$\frac{b \cdot a}{1 + b a + a} = \frac{a b}{1 + a b + a}.$$\n- For the third term, multiply numerator and denominator by $b$:\n$$\frac{c \cdot b}{1 + c b + b} = \frac{b c}{1 + b c + b}.$$\n\n6. **Notice that $a b = \frac{1}{c}$, $b c = \frac{1}{a}$, and $a c = \frac{1}{b}$, so the denominators are symmetric. But the denominators are the same as the original denominators after clearing fractions, so the expression is symmetric in $a,b,c$.\n\n7. **Try substituting $a=b=c=1$ (since $abc=1$) to check the value:**\n$$\frac{1}{1+1+1} + \frac{1}{1+1+1} + \frac{1}{1+1+1} = 3 \times \frac{1}{3} = 1.$$\n\n8. **Because the expression is symmetric and the substitution satisfies the condition, the value of the expression is 1.**