Calculate the maximum value of $\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$ where $3a + 4b + 5c = 12$ 
$a$, $b$ and $c$ are positives such that $3a + 4b + 5c = 12$. Calculate the maximum value of $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}$$

I want to know if there are any other solutions that are more practical. This came from my homework my teacher gave today. I have given my solution down below if you want to check out.

Also, we can use C-S and AM-GM:
$$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c}\leq$$
$$\leq\frac{ab}{(1+2)^2}\left(\frac{1^2}{ab}+\frac{2^2}{a+b}\right)+\frac{2ca}{9}\left(\frac{1}{ca}+\frac{4}{c+a}\right)+\frac{3bc}{9}\left(\frac{1}{bc}+\frac{4}{b+c}\right)=$$
$$=\frac{2}{3}+\frac{4}{9}\cdot\frac{ab}{a+b}+\frac{8}{9}\cdot\frac{ca}{c+a}+\frac{4}{3}\cdot\frac{bc}{b+c}\leq$$
$$\leq\frac{2}{3}+\frac{4}{9}\cdot\frac{a+b}{4}+\frac{8}{9}\cdot\frac{c+a}{4}+\frac{4}{3}\cdot\frac{b+c}{4}=\frac{2}{3}+\frac{3a+4b+5c}{9}=2.$$
