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.

We have that $$3a + 4b + 5c = 12$$
$$\implies (a + b) + 2(c + a) + 3(b + c) = 12 \implies \sqrt{ab} + 2\sqrt{ca} + 3\sqrt{bc} \le 6$$
In addition to that, $$\frac{ab}{ab + a + b} + \frac{2ca}{ca + c + a} + \frac{3bc}{bc + b + c} \le \frac{ab}{ab + 2\sqrt{ab}} + \frac{2ca}{ca + 2\sqrt{ca}} + \frac{3bc}{bc + 2\sqrt{bc}}$$
$$ = \frac{\sqrt{ab}}{\sqrt{ab} + 2} + \frac{2\sqrt{ca}}{\sqrt{ca} + 2} + \dfrac{3\sqrt{bc}}{\sqrt{bc} + 2} = 6 - 2\left(\frac{1}{\sqrt{ab} + 2} + \frac{2}{\sqrt{ca} + 2} + \frac{3}{\sqrt{bc} + 2}\right)$$
$$\le 6 - \frac{72}{\sqrt{ab} + 2\sqrt{ca} + 3\sqrt{bc} + 12} \le 6 - \frac{72}{6 + 12} = 2$$
The equality sign occurs when $a = b = c = 1$.
