<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] So, by Cauchy-Schwarz, we have $a^2+b^2 \geq (a+b)^2/2$, so we just have to prove that
$\displaystyle \sum_{cyc} \frac{1}{a+b} \leq 3/4$ (It's obvious that this maximum occurs when $a=b=c=2$, I will prove it.)
Letting, $a=2k,b=2l,c=2m$ we have to prove that $\displaystyle \sum_{cyc} \frac{1}{k+l} \leq 3/2$ subject to the condition $4klm=k+l+m+1$.
By expanding and letting $k+l+m=p, kl+lm+mk, klm=r$ we have to prove that $3pq-3r \geq 2p^2+2q$.
Substituting $r=(p+1)/4$, we have to prove that $q(3p-2) \geq \frac{8p^2+3p+3}{4}$. 
Now, it is known that $q^2 \geq 3pr=\frac{3p^2-3p}{4}$ so it suffices to show that $\sqrt{3p^2+3p}\cdot (3p-2) \geq \frac{8p^2+3p+3}{2}$ which is equivalent to :
$(p-3)(44p^3+48p^2-9p+3)$.
By AM-GM, $p+1=4r \leq 4p^3/27$ so easily $p \geq 3$.
So, $p-3 \geq 0$ and $44p^3+48p^2-9p+3 >48p^2-9p=p(48p-9)>0$ and the problem is solved.
</s>