If $\frac{1}{a+3}+\frac{1}{b+4}+\frac{1}{c+5}=\frac{7}{12}$ for positive integer $a$, $b$, $c$, then find $\frac{a}{a+3}+\frac{b}{b+4}+\frac{c}{c+5}$ 
Given $a, b, c$ are positive integers, and that
  $$\frac{1}{a + 3} + \frac{1}{b+4} + \frac{1}{c+5} = \frac{7}{12},$$
  compute:
  $$\frac{a}{a+3} + \frac{b}{b+4} + \frac{c}{c+5}$$

Source (Romanian Math Magazine, Gazeta Matematica S:E19.333, this problem marked as targetting 6th graders)
I tried writing the terms as $1 - \frac{1}{a+3}$ and similarly for the others, but I get $3 - \frac{3}{a+3} - \frac{4}{b + 4} - \frac{5}{c + 5}$ which I don't find very helpful.
Many thanks in advance! 

It is given that $a,b,c \geq 1$. If $a \geq 2$. Then:
$$\frac{1}{a+3}+\frac{1}{b+4}+\frac{1}{c+5} \leq \frac{1}{5}+\frac{1}{5}+\frac{1}{6}=\frac{17}{30} < \frac{7}{12}$$
So $a=1$ and 
$$\frac{1}{b+4}+\frac{1}{c+5}=\frac{1}{3}$$
If $b \geq 3$, we get:
$$\frac{1}{b+4}+\frac{1}{c+5} \leq \frac{1}{7}+\frac{1}{6} = \frac{13}{42} < \frac{1}{3}$$
So $b\in \{1,2\}$. Checking $b=1$, we don't get an integer $c$, so $b=2$ and $c=1$. The final answer is:
$$\frac{a}{a+3}+\frac{b}{b+4}+\frac{c}{c+5} = \frac{3}{4}$$
