Prove the inequality $\sum_{cyc}\frac{a}{1+\left(b+c\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$ 
Let $a>0$, $b>0$ and $c>0$ such that $a+b+c=3$. Prove the inequality 
  $$\frac{a}{1+\left(b+c\right)^2}+\frac{b}{1+\left(c+a\right)^2}+\frac{c}{1+\left(a+b\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}.$$


$$LHS=\frac{a}{1+\left(3-a\right)^2}+\frac{b}{1+\left(3-b\right)^2}+\frac{c}{1+\left(3-c\right)^2}$$
We have inequality: $$\frac{a}{1+\left(3-a\right)^2}\le \frac{9}{25}a-\frac{4}{25}\Leftrightarrow -\frac{\left(a-1\right)^2\left(9a-40\right)}{25\left(\left(a-3\right)^2+1\right)}\le 0\forall 0<a\le 1$$
$$\Rightarrow LHS\le \frac{9}{25}\left(a+b+c\right)-\frac{4}{25}\cdot 3=\frac{3}{5}$$
Need prove: $$\frac{3}{5}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$$
$$\Leftrightarrow a^2+b^2+c^2+12abc\le 5\left(a^2+b^2+c^2\right)$$
$$\Leftrightarrow 12abc\le 4\left(a^2+b^2+c^2\right)\Leftrightarrow 3abc\le a^2+b^2+c^2$$
By AM-GM: $$3abc\le \frac{\left(a+b+c\right)^3}{9}=3=\frac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2$$
Right or Wrong. I think it's wrong. Help me
 A: We need to prove that
$$\sum_{cyc}\left(\frac{a}{1+(b+c)^2}-a\right)\leq\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}-3$$ or
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}\geq\frac{36abc}{a^2+b^2+c^2+12abc}.$$
Now, by C-S
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}=\sum_{cyc}\frac{a}{1+\frac{1}{(b+c)^2}}=$$
$$=\sum_{cyc}\frac{a^2}{a+\frac{a}{(b+c)^2}}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}\left(a+\frac{a}{(b+c)^2}\right)}=\frac{9}{3+\sum\limits_{cyc}\frac{a}{(b+c)^2}}.$$
Thus, it remains to prove that
$$a^2+b^2+c^2+12abc\geq4abc\left(3+\sum\limits_{cyc}\frac{a}{(b+c)^2}\right)$$ or
$$a^2+b^2+c^2\geq4abc\sum\limits_{cyc}\frac{a}{(b+c)^2}$$ or
$$\sum_{cyc}\left(\frac{a}{bc}-\frac{4a}{(b+c)^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a(b-c)^2}{bc(b+c)^2}\geq0.$$
Done!
A: We need to prove that
$$\sum_{cyc}\left(\frac{a}{1+(b+c)^2}-a\right)\leq\frac{3(a^2+b^2+c^2)}{a^2+b^2+c^2+12abc}-3$$ or
$$\sum_{cyc}\frac{a(b+c)^2}{1+(b+c)^2}\geq\frac{36abc}{a^2+b^2+c^2+12abc},$$
which seems better, but I did not find something nice here.
By the way, our inequality is obviously true after homogenization and full expanding.
Indeed, we need to prove that
$$\sum_{cyc}\frac{a}{(a+b+c)^2+9(b+c)^2}\leq\frac{a^2+b^2+c^2}{(a^2+b^2+c^2)(a+b+c)+36abc}$$ or
$$\sum_{sym}(10a^7b+130a^6b^2+250a^5b^3+130a^4b^4-68a^6bc-159a^5b^2c+255a^4b^3c-225a^4b^2c^2-323a^3b^3c^2)\geq0,$$
which is true by Muirhead:
$$\sum_{sym}10a^7b\geq\sum_{sym}10a^6b^2;$$
$$\sum_{sym}68a^6b^2\geq\sum_{sym}68a^6bc;$$
$$\sum_{sym}72a^6b^2\geq\sum_{sym}72a^5b^3;$$
$$\sum_{sym}159a^5b^3\geq\sum_{sym}159a^5b^2c;$$
$$\sum_{sym}163a^5b^3\geq\sum_{sym}163a^4b^2c^2;$$
$$\sum_{sym}62a^4b^4\geq\sum_{sym}62a^4b^2c^2;$$
$$\sum_{sym}68a^4b^4\geq\sum_{sym}68a^3b^3c^2$$ and
$$\sum_{sym}255a^4b^3c\geq\sum_{sym}255a^3b^3c^2.$$
After summing of these inequalities we'll get the needed inequality.
Done!
