An interesting inequality with condition If $a,b,c$ positive reals and $\frac{a}{b+c} \ge 2$ I have to prove that
$(ab+bc+ca)\left(\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \right)\geq \frac{49}{18}$
We may assume that $a\geq b \geq c.$ Firstly, let's show that
$\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2}\geq \frac{1}{4ab}+\frac{2}{(a+c)(b+c)}.$ This can be rewritten as
$\left(\frac{1}{a+c}-\frac{1}{b+c}\right)^2\geq \frac{(a-b)^2}{4ab(a+b)^2},$ or equivalently $4ab(a+b)^2\geq (a+c)^2(b+c)^2.$ This is obvious, since $4ab\geq (b+c)^2$ and $(a+b)^2\geq (a+c)^2.$
Thus, it remains to prove that
$(ab+bc+ca)\left(\frac{1}{4ab}+\frac{2}{(a+c)(b+c)} \right)\geq \frac{49}{18}.$ Using the identities
$\frac{ab+bc+ca}{4ab}=\frac{1}{4}+\frac{c(a+b)}{4ab}, \quad \frac{2(ab+bc+ca)}{(a+c)(b+c)}=2 -\frac{2c^2}{(a+c)(b+c)},$ this becomes
$\frac{c(a+b)}{4ab}\geq \frac{2c^2}{(a+c)(b+c)}+\frac{17}{36}.$
Then I stuck. Any idea please?
 A: This can be solved in a brute force way:
$$\frac{a}{b+c}\ge2\implies a=2b+2c+x$$
..where $x$ is some non-negative value. The inequality:
$$(ab+bc+ca)\left(\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \right)-\frac{49}{18}\ge0$$
...becomes:
$$((2b+2c+x)b+bc+c(2b+2c+x))\left(\frac{1}{(b+c)^2}+\frac{1}{(2b+3c+x)^2}+\frac{1}{(3b+2c+x)^2} \right)-\frac{49}{18}\ge0$$
This can be writen as: 
$$\frac AB\ge 0\tag{1}$$
where
$$A=654 b^5 c+462 b^5 x+2783 b^4 c^2+3620 b^4 c x+851 b^4 x^2+4276 b^3 c^3+8748 b^3 c^2 x+4260 b^3 c x^2+572 b^3 x^3+2783 b^2 c^4+8748 b^2 c^3 x+6854 b^2 c^2 x^2+1932 b^2 c x^3+167 b^2 x^4+654 b c^5+3620 b c^4 x+4260 b c^3 x^2+1932 b c^2 x^3+352 b c x^4+18 b x^5+462 c^5 x+851 c^4 x^2+572 c^3 x^3+167 c^2 x^4+18 c x^5$$
$$B=18 (b+c)^2 (3 b+2 c+x)^2 (2 b+3 c+x)^2$$
$A,B$ are positive so (1) is obviously true.
There is a similar problem which I find to be much more interesting:

For all positive $a,b,c$:$$(ab+bc+ca)\left(\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \right)\ge\frac94$$

