Make common denominators and combine the fractions: $\displaystyle \frac{a(1+b)(1+c)+b(1+a)(1+c)-c(1+a)(1+b)}{(1+a)(1+b)(1+c)}$.

You only have to show that the numerator is positive, as the denominator is clearly positive (because $\displaystyle a,b,c>0$).

Multiply out the numerator: $\displaystyle a(1+b)(1+c)+b(1+a)(1+c)-c(1+a)(1+b)$ $\displaystyle =a+ab+ac+abc+b+ab+bc+abc-c-ac-bc-abc$ $\displaystyle =a+b-c+2ab+abc$

Now we use the fact that $\displaystyle a+b-c\geq0$ to say that $\displaystyle a+b-c+2ab+abc\geq 2ab+abc$.

Clearly $\displaystyle 2ab+abc\geq0$, because $\displaystyle a$, $\displaystyle b$, and $\displaystyle c$ are all positive. So we're done.