Inequality with $ab + bc + ca = 1$ 
As in the title, let $a>0$, $b>0$ and $c > 0$ be such that $ab + bc + ca = 1.$ Define $S = \frac{a+b+c}{\sqrt{bc} + \sqrt{ca} + \sqrt{ab}}.$ Prove that
  $$S^2 \ge \sum_{\text{cyc}} \frac{1}{(2a+b)(2a+c)} \ge 1.$$

Edit: this is from an old mock usamo, which no longer exists on the internet, and to which I don't have solutions. Inequalities are not my strong point, so I'm hoping this will be entertaining to someone here :)
 A: The right inequality.
$$\sum_{cyc}\frac{1}{(2a+b)(2a+c)}-1=\sum_{cyc}\frac{1}{(2a+b)(2a+c)}-\frac{1}{ab+ac+bc}=$$
$$=\sum_{cyc}\left(\frac{1}{(2a+b)(2a+c)}-\frac{1}{3(ab+ac+bc)}\right)=$$
$$=\frac{1}{3}\sum_{cyc}\frac{ab+ac+2bc-4a^2}{(2a+b)(2a+c)}=$$
$$=\frac{1}{3}\sum_{cyc}\frac{(c-a)(2a+b)-(a-b)(2a+c)}{(2a+b)(2a+c)}=$$
$$=\frac{1}{3}\sum_{cyc}(a-b)\left(\frac{1}{2b+a}-\frac{1}{2a+b}\right)=$$
$$=\sum_{cyc}\frac{(a-b)^2}{3(2a+b)(2b+a)}\geq0.$$
The left inequality.
We'll use the identity
$$\sum_{cyc}\frac{1}{(2a+b)(2a+c)}-1=\frac{1}{3}\sum_{cyc}\frac{(a-b)^2}{(2a+b)(2b+a)},$$
which we got during the proof of the right inequality.
Indeed, let $a\geq b\geq c$.
Thus, by C-S $$b^2\left(\frac{(a+b+c)^2}{(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})^2}-\sum_{cyc}\frac{1}{(2a+b)(2a+c)}\right)\geq$$
$$\geq b^2\left(\frac{(a+b+c)^2}{3(ab+ac+bc)}-\sum_{cyc}\frac{1}{(2a+b)(2a+c)}\right)=$$
$$= b^2\left(\frac{(a+b+c)^2}{3(ab+ac+bc)}-1-\left(\sum_{cyc}\frac{1}{(2a+b)(2a+c)}-1\right)\right)=$$
$$= b^2\left(\frac{\sum\limits_{cyc}(a^2-ab)}{3(ab+ac+bc)}-\frac{1}{3}\sum_{cyc}\frac{(a-b)^2}{(2a+b)(2b+a)}\right)=$$
$$= b^2\left(\frac{\sum\limits_{cyc}(a-b)^2}{6(ab+ac+bc)}-\frac{1}{3}\sum_{cyc}\frac{(a-b)^2}{(2a+b)(2b+a)}\right)=$$
$$=\frac{b^2}{6}\sum_{cyc}\frac{(a-b)^2((2a+b)(2b+a)-2(ab+ac+bc))}{(2a+b)(2b+a)}=$$
$$=\frac{b^2}{6}\sum_{cyc}\frac{(a-b)^2(2a^2+2b^2+3ab-2ac-2bc)}{(2a+b)(2b+a)}\geq$$
$$\geq\frac{b^2}{3}\sum_{cyc}\frac{(a-b)^2(a^2+b^2+ab-ac-bc)}{(2a+b)(2b+a)}\geq$$
$$\geq\frac{b^2(a-c)^2(a^2+c^2+ac-ab-bc)}{3(2a+c)(2c+a)}+\frac{b^2(b-c)^2(b^2+c^2+bc-ab-ac)}{3(2b+c)(2c+b)}\geq$$
$$\geq\frac{a^2(b-c)^2(a^2+ac-ab-bc)}{3(2a+c)(2c+a)}+\frac{b^2(b-c)^2(b^2+bc-ab-ac)}{3(2b+c)(2c+b)}=$$
$$=\frac{(b-c)^2(a-b)}{3}\left(\frac{a^2(a+c)}{(2a+c)(2c+a)}-\frac{b^2(b+c)}{(2b+c)(2c+b)}\right)\geq$$
$$\geq\frac{(b-c)^2(a-b)}{3}\left(\frac{a^2(b+c)}{(2a+c)(2c+a)}-\frac{b^2(b+c)}{(2b+c)(2c+b)}\right)=$$
$$=\frac{c(b+c)(a-b)^2(b-c)^2(5ab+2ac+2bc)}{3(2a+c)(2c+a)(2b+c)(2c+b)}\geq0.$$
Done!
