Prove that $\sum\limits_{cyc}\frac{a}{b(3+a-b)}\ge 1$ Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 3$. Prove that
$$\sum_{cyc}\frac{a}{b(3+a-b)}\ge1$$
I tried applying the Cauchy-Schwarz inequality by doing:
$$\sum_{cyc}\frac{a}{b(3+a-b)}=\sum_{cyc}\frac{a^2}{ab(3+a-b)}\ge \frac{(a+b+c)^2}{\sum_{cyc}ab(3+a-b)}=\frac{9}{\sum_{cyc}ab(3+a-b)}.$$ 
We need therefore to prove that
$$\sum_{cyc}ab(3+a-b)\le 9.$$ 
Thanks for any help.

By the AM-GM inequality we have
$$
3=3\root{3}\of{\frac{a}{b}\frac{b}{c}\frac{c}{a}}\leq \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\tag{1}
$$
By the Cauchy-Schwarz inequality we have
$$
\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\leq
\left(\sum_{cyc}\frac{a}{b(2a+c)}\right)\cdot\left(\sum_{cyc}(2a+c)\right)\tag{2}
$$
But $$\sum_{cyc}(2a+c)=9\tag{3}$$ So combining $(1)$, $(2)$ and $(3)$ we get
$$1\leq \sum_{cyc}\frac{a}{b(2a+c)}.$$
which is the same as the desired inequality since $2a+c=3+a-b~$.
