Using Am-Gm, we have: $\sum \frac{a^2+3b^2}{ab^2(4-ab)} \ge 2 \sum \frac{a+b}{ab(4-ab)} \ge 4\sum \frac{1}{\sqrt{ab}(4-ab)}$ , so it remains to prove: $\sum \frac{1}{\sqrt{ab}(4-ab)} \ge 1$ Using cauchy schwarz, we have: $ \sum \frac{1}{\sqrt{ab}(4-ab)} \ge \frac{9}{4\sum \sqrt{ab}- \sum (ab)^{\frac{3}{2}}}$ So we need to prove: $9+\sum (ab)^{\frac{3}{2}} \ge 4\sum \sqrt{ab}$ But using Am-Gm, we have: $\sum \left( (ab)^{\frac{3}{2}}+1+1 \right) \ge 3\sum \sqrt{ab}$ and from well known inequalities: $3(xy+xz+yz) \le (x+y+z)^2 \le 3(x^2+y^2+z^2)$ it follows: $\sum \sqrt{ab} \le \sqrt{3\sum ab} \le \sum a=3$ Hence, inequality is proven. Equality holds for $a=b=c=1$. $\blacksquare$

Posted also there: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=282119&hilit=Turkey+NMO+2007

Also posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=440599

\[\sum{\frac{a^2+3b^2}{ab^2(4-ab)}}\geq \sum{\frac{2ab+2b^2}{ab^2(4-ab)}}=\sum{\frac{2(a+b)}{ab(4-ab)}}\overset{?}{\geq} 4\]\[\sum{\frac{a+b}{ab(4-ab)}}\overset{?}{\geq} 2\iff \sum{\frac{a^2c^2}{4ac-a^2bc}}+\sum{\frac{b^2c^2}{4bc-ab^2c}}=\sum{\frac{ac+bc}{4-ab}}\overset{?}{\geq} 2abc\]\[\sum{\frac{a^2c^2}{4ac-a^2bc}}+\sum{\frac{b^2c^2}{4bc-ab^2c}}\geq \frac{(\sum{ab})^2}{4\sum{ab}-3abc}+\frac{(\sum{ab})^2}{4\sum{ab}-3abc}=\frac{2(\sum{ab})^2}{4\sum{ab}-3abc}\overset{?}{\geq} 2abc\]Set $\sum{ab}=3v^2$ and $abc=w^3$ where $1\geq v\geq w$. \[9v^4\overset{?}{\geq} 12v^2w^3-3w^3\iff 3v^4\overset{?}{\geq} w^3(4v^2-1)\]Which is true since $w^3(4v^2-1)\leq v^3(4v^2-1)=4v^5-v^3=4v^5+(v^5-v^3)\leq 4v^5\leq 4v^2$ as desired.$\blacksquare$