<s>[INST] Calculate the maximum value of $\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$ where $x + 2y + 3z = 2$ 
$x$, $y$ and $z$ are positives such that $x + 2y + 3z = 2$. Calculate the maximum value of $$ \sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$$

This problem is brought to you by a recent competition. There should be different answers that are more creative than the one I have provided. Oh well...
 [/INST] Let $x=\frac{2}{3}a$, $y=\frac{1}{3}b$ and $z=\frac{2}{9}c$.
Thus, $a+b+c=3$ and by AM-GM we obtain:
$$\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}=\sum_{cyc}\sqrt{\frac{bc}{bc+3a}}=$$
$$=\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{c}{a+c}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{b}{a+b}+\frac{a}{b+a}\right)=\frac{3}{2}.$$
The equality occurs for $a=b=c=1$, which says that we got a maximal value.
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