MOP 2011 inequality If $a,b,c$ are positive integers prove that 

$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$

My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that 
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} \le a+b+c$
After trying out few values I realized it is wrong.So I decided to tackle it all at your once. I tried few well known inequalities and played around but nothing came useful out of it. Any help??

Another way.
By AM-GM $$\sum_{cyc}\sqrt{a^2-ab+b^2}\leq\sum_{cyc}\left(\frac{a^2-ab+b^2}{a+b}+\frac{a+b}{4}\right)=$$
$$=\sum_{cyc}\left(\frac{a^2+2ab+b^2-3ab}{a+b}+\frac{a+b}{4}\right)=\frac{5}{2}(a+b+c)-3\sum_{cyc}\frac{ab}{a+b}=$$
$$=4(a+b+c)-3\sum_{cyc}\left(\frac{ab}{a+b}+\frac{a+b}{4}\right)\leq$$
$$\leq 4(a+b+c)-3\sum_{cyc}\sqrt{ab}\leq4(a+b+c)-9\sqrt[3]{abc}.$$
