An inequality with substitution If $a,b,c$ positive real numbers, then I have to prove $ \frac {1}{18} \sum\limits_{cycl}^{} \frac{a^2}{b^2} + \sum\limits_{cycl}^{} \frac {a}{2a+b+c} \ge \frac {11}{12}$
We have that $\frac {1}{18} \sum\limits_{cycl}^{} \frac{a^2}{b^2} \ge \frac {3}{18} = \frac {1}{6}$
If we assume that $x=2a+b+c,y=a+2b+c,z=a+b+2c,$ then $a=\frac{3x-y-z}{4},b=\frac{3y-x-z}{4},c=\frac{3z-x-y}{4},$ and
$\sum\limits_{cycl}^{} \frac {a}{2a+b+c} = \frac{1}{4} [9 - \sum\limits_{cycl}^{} (\frac {x}{y}+\frac {y}{x})]$
Since
$\sum\limits_{cycl}^{} (\frac {x}{y}+\frac {y}{x}) \ge 6$
so we have
$ \frac{3}{4} \ge \sum\limits_{cycl}^{} \frac {a}{2a+b+c} = \frac{1}{4} [9 - \sum\limits_{cycl}^{} (\frac {x}{y}+\frac {y}{x})]$
and I stuck there. Thank you
 A: Let $c=\max\{a,b,c\}$.
Since by C-S $$\sum_{cyc}\frac{a}{2a+b+c}=\sum_{cyc}\frac{a^2}{2a^2+ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2+ab+ac)}=\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+ab)},$$
it's enough to prove that
$$\frac{1}{18}\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)+\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+ab)}\geq\frac{11}{12}$$ or
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-3\geq\frac{27}{2}-\frac{9(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}$$ or
$$\frac{a^2}{b^2}+\frac{b^2}{a^2}-2+\frac{b^2}{c^2}+\frac{c^2}{a^2}-\frac{b^2}{a^2}-1\geq\frac{9\sum\limits_{cyc}(a^2-ab)}{2\sum\limits_{cyc}(a^2+ab)}$$ or
$$\frac{(a^2-b^2)^2}{a^2b^2}+\frac{(c^2-a^2)(c^2-b^2)}{a^2c^2}\geq\frac{9((a-b)^2+(c-a)(c-b))}{2\sum\limits_{cyc}(a^2+ab)}.$$
Id est, it's enough to prove that
$$\frac{(a+b)^2}{a^2b^2}\geq\frac{9}{2\sum\limits_{cyc}(a^2+ab)}$$ and
$$\frac{(a+c)(b+c)}{a^2c^2}\geq\frac{9}{2\sum\limits_{cyc}(a^2+ab)}.$$ 
Both these inequalities we can prove by AM-GM.
Indeed, $$2(a+b)^2\sum_{cyc}(a^2+ab)\geq2(a+b)^2(a^2+ab+b^2)\geq2\cdot4\cdot3a^2b^2>9a^2b^2 $$ and
$$2(a+c)(b+c)\sum_{cyc}(a^2+ab)\geq2(a+c)c(a^2+ac+c^2)\geq2c(a+a)\cdot3ac=12a^2c^2>9a^2c^2.$$
Done!
