An elementary inequality for 3 positive real numbers Let  $a, b, c > 0$ and $n \ge 0$ . Show that 
$$\frac{a^2}{b^2 + nab} + \frac{b^2}{c^2 + nbc} + \frac{c^2}{a^2 + nca} \ge \frac{(a+b+c)^2}{(n+1)(ab+bc+ca)} $$
My first attempt was direct computations in the hope that this will lead to some simplifications or a known inequality , but this was not the case.
Another idea is to use induction over n .
For $n = 0$ the inequality become simpler :
$$\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \ge \frac{(a+b+c)^2}{ab+bc+ca} $$
But still direct computation or manipulating simpler inequalities (like those between arithmetic , geometric and harmonic means ) don't seems to work . 
Any suggestions are welcome. Thank you !

By Holder
$$\sum_{cyc}\frac{a^2}{b^2+nab}=\frac{\sum\limits_{cyc}\frac{a^2}{b(b+na)}\sum\limits_{cyc}ab\sum\limits_{cyc}(b+na)}{\sum\limits_{cyc}ab\sum\limits_{cyc}(b+na)}\geq$$
$$\geq\frac{(a+b+c)^3}{\sum\limits_{cyc}ab\sum\limits_{cyc}(b+na)}=\frac{(a+b+c)^2}{(n+1)(ab+ac+bc)}.$$
