An algebraic inequality involving $\sum_{cyc} \frac1{(a+2b+3c)^2}$ I was reading through the proof of an inequality posted on a different website and the following was mentioned as being easily proven by AM-GM:
Let $a,\ b,\ c>0$, then $$\frac{1}{(a+2b+3c)^2} + \frac{1}{(b+2c+3a)^2} + \frac{1}{(c+2a+3b)^2} \le \frac{1}{4(ab + bc+ca)}$$
I checked that the inequality holds numerically, but I can't find a solution (definitely nothing obvious using AM-GM). I played around with a simpler version of the inequality:
$$\frac{1}{(a+2b)^2} + \frac{1}{(b+2a)^2} \le \frac{2}{9}\frac{1}{ab}$$
which I can prove by brute force algebra and a little AM-GM. However, brute force seems impractical for the three variable inequality. Any suggestions?

Hint: Using AM-GM, 
$$(a+c+2(b+c))^2=(a+c)^2+4(a+c)(b+c)+4(b+c)^2\geqslant 6(a+c)(b+c)+3(b+c)^2=3(b+c)(2a+b+3c)$$
Hence it is enough to show:
$$\sum_{cyc} \frac{ab+bc+ca}{(b+c)(2a+b+3c)} \leqslant \frac34$$
$$\iff \sum_{cyc}\frac{(b+c)^2+2c^2}{(b+c)(2a+b+3c)}\geqslant\frac32$$
which follows almost directly from CS inequality, as its:
$$LHS \geqslant \frac{\left(\sum_{cyc} (b+c) \right)^2+2(\sum_{cyc} c)^2 }{\sum_{cyc} (b+c)(2a+b+3c)} = \frac{4(a+b+c)^2+2(a+b+c)^2 }{4(a+b+c)^2}=\frac32$$
