<s>[INST] Calculate the maximum value of $\sum_{cyc}\frac{1}{\sqrt{a^2 + b^2}}$ where $a, b, c > 0$ and $abc = a + b + c + 2$. 
$a$, $b$ and $c$ are positives such that $abc = a + b + c + 2$. Caculate the maximum value of $$\large \frac{1}{\sqrt{a^2 + b^2}} + \frac{1}{\sqrt{b^2 + c^2}} + \frac{1}{\sqrt{c^2 + a^2}}$$

This problem is adapted from a recent competition. Here's what I got, and I am pretty proud of it. (Actually not.)
$$abc = a + b + c + 2$$
$$\iff abc + (ab + bc + ca) + (a + b + c) + 1 = ab + bc + ca + 2(a + b + c) + 3$$
$$\iff (a + 1)(b + 1)(c + 1) = (a + 1)(b + 1) + (b + 1)(c + 1) + (c + 1)(a + 1)$$
$$\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} = 1$$
And I'm done.
 [/INST] Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x+y}{z}$ and by AM-GM twice we obtain:
$$\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}=\sum_{cyc}\frac{xy}{\sqrt{y^2(y+z)^2+x^2(x+z)^2}}\leq$$
$$\leq\sum_{cyc}\frac{xy}{\sqrt{2xy(x+z)(y+z)}}=\frac{1}{\sqrt2}\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq$$
$$\leq\frac{1}{2\sqrt2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{3}{2\sqrt2}.$$
The equality occurs for $x=y=z=1,$ which says that we got a maximal value.
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