65th Czech and Slovak Mathematical Olympiad

We use the well known inequality $a^2+b^2\ge 2ab$ to adjust the given one: $$4a=(a+c)(b^2+ac)=a(b^2+c^2)+c(a^2+b^2)\ge a(b^2+c^2)+2abc=a(b+c{)}^2.$$ We can see, that $b+c\le 2$, and also that the equality holds if and only if $0<a=b<2$ and $c=2-b>0$. That's it.