Uzbekistan wrote: Let a,b,c Postive real numbers such that $a+b+c\geq 6$. Find the minimum value $A=\sum_{cyc}{a^2}$+$\sum_{cyc}{\frac{a}{b^2+c+1}}$ Here is my solution, we easy have:$a^2+b^2+c^2\geq 2(a+b+c) \geq 12 (1)$ then,applying AM-GM inequality $\frac{a^2}{14}+\frac{2(b^2+c+1)}{49}+\frac{a}{b^2+c+1}\geq \frac{3a}{7}$ and using (1) we are done.The minimum is $\frac{6}{7}$.

trenkialabautroj wrote: Uzbekistan wrote: Let a,b,c Postive real numbers such that $a+b+c\geq 6$. Find the minimum value $A=\sum_{cyc}{a^2}$+$\sum_{cyc}{\frac{a}{b^2+c+1}}$ Here is my solution, we easy have:$a^2+b^2+c^2\geq 2(a+b+c) \geq 12 (1)$ then,applying AM-GM inequality $\frac{a^2}{14}+\frac{2(b^2+c+1)}{49}+\frac{a}{b^2+c+1}\geq \frac{3a}{7}$ and using (1) we are done.The minimum is $\frac{6}{7}$. Considering about the $a^2+b^2+c^2$ term, It's $12+\frac{6}{7}$ small mistake a=b=c=2

my water solution. by AG, $A=\sum\frac{c}{a^2+b+1}+\frac{2}{49}\sum(a^2+b+1)+\frac{47}{49}\sum a^2-\frac{2}{49}\sum b-\frac{6}{49} \ge \frac{2}{7}\sqrt 2\sum\sqrt c+\frac{47}{49}\sum a^2-\frac{18}{49}$ let $a=2t,b=2u,c=2v$then$t+u+v=3$ by weighted AG, $\frac{47}{49}a^2+\frac{2}{7}\sqrt{2a}=\frac{188}{49}t^2+\frac{4}{7}\sqrt t\ge\frac{216}{49}t^{\frac{390}{216}}$ then by power mean ineq $A\ge \frac{216}{49}\sum t^{\frac{390}{216}}-\frac{18}{49}\ge\frac{216}{49}*3-\frac{18}{49}=\frac{90}{7}$.