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.

Your way works!
By C-S
$$\sum_{cyc}\frac{a}{b(3+a-b)}=\sum_{cyc}\frac{a}{b(2a+c)}=\sum_{cyc}\frac{a^2}{ab(2a+c)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}ab(2a+c)}.$$
Thus, it remains to prove that
$$(a+b+c)^3\geq3\sum\limits_{cyc}ab(2a+c)$$ or
$$\sum_{cyc}(a^3-3a^2b+3a^2c-abc)\geq0,$$ 
which is obvious.
Done!
