$x$, $y$ 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...
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We have that
$$\sqrt{\frac{3yz}{3yz + x}} + \sqrt{\frac{3zx}{3zx + 4y}} + \sqrt{\frac{xy}{xy + 3z}}$$
$$ = \sqrt{\frac{3yz}{3yz + (2 - 2y - 3z)}} + \sqrt{\frac{3zx}{3zx + 2(2 - 3z - x)}} + \sqrt{\frac{xy}{xy + (2 - x - 2y)}}$$
$$ = \sqrt{\frac{3yz}{(1 - y)(2 - 3z)}} + \sqrt{\frac{3zx}{(2 - 3z)(2 - x)}} + \sqrt{\frac{xy}{(2 - x)(1 - y)}}$$
$$\le \frac{1}{2}\left(\frac{2y}{2 - 3z} + \frac{3z}{2 - 2y}\right) + \frac{1}{2}\left(\frac{3z}{2 - x} + \frac{x}{2 - 3z}\right) + \frac{1}{2}\left(\frac{x}{2 - 2y} + \frac{2y}{2 - x}\right)$$
$$ = \frac{1}{2}\left(\frac{2y + 3z}{2 - x} + \frac{3z + x}{2 - 2y} + \frac{x + 2y}{2 - 3z}\right)$$
$$ = \frac{1}{2}\left[(x + 2y + 3z - 2)\left(\frac{1}{2 - x} + \frac{1}{2 - 2y} + \frac{1}{2 - 3z}\right) - 3\right] = \frac{3}{2}$$
The equality sign occurs when $\left\{ \begin{align} x(2 - x) = 2y(2 - 2y) = 3z(2 - 3z)\\ x + 2y + 3z = 2\end{align} \right.$
$\iff \left\{ \begin{align} (x - 1)^2 = (2y - 1)^2 = (3z - 1)^2\\ x + 2y + 3z = 2 \end{align} \right.$$\implies x = 2y = 3z = \dfrac{x + 2y + 3z}{1 + 1 + 1} = \dfrac{2}{3}$
$\iff (x, y, z) = \left(\dfrac{2}{3}, \dfrac{1}{3}, \dfrac{2}{9}\right)$.
Let $x=\frac{2}{3}a$, $y=\frac{1}{3}b$ and $z=\frac{2}{9}c$.
Thus, $a+b+c=3$ and by AM-GM we obtain: $$\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.