$(ab + bc + ca) \left(\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)}\right)\ge \frac {9}{(p + 1)(q + 1)}$ One of my friends showed me this inequality.
$$(ab + bc + ca)  \left(\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)}\right)\ge \frac {9}{(p + 1)(q + 1)}$$ for every positive numbers $a, b, c, p$ and $q.$
My first idea was to use Bergstrom's Inequality, so I would have $$\frac {1}{(a + pb)(a + qb)} + \frac {1}{(b + pc)(b + qc)} + \frac{1}{(c + pa)(c + qa)} \ge \frac {9}{(a ^ 2 + b ^ 2 + c ^ 2) (pq + 1) + pq (ab + bc + ca)} \ge \frac {9}{(a ^ 2 + b ^ 2 + c ^ 2)(pq + p + q + 1)} = \frac {9}{(a ^ 2 + b ^ 2 + c ^ 2)(p + 1)(q + 1)},$$ but in the end we get to $$\frac {ab + bc + ca}{a ^ 2 + b ^ 2 + c ^ 2} \ge 1,$$ which is obviously false.
I am eager to hear your advice.
 A: The AM-GM:
$\dfrac{(q+1)(a+pb)+(p+1)(a+qb)}2 \geqslant \sqrt{(p+1)(q+1)(a+pb)(a+qb)}$
$$\implies \frac1{(a+pb)(a+qb)}\geqslant \frac{4(p+1)(q+1)}{[(q+1)(a+pb)+(p+1)(a+qb)]^2}=\frac{4(p+1)(q+1)}{[(p+q+2)a+(2pq+p+q)b]^2}$$
Defining $k$ using $2pq+p+q=k(p+q+2)$, the above can be rewritten as
$$\frac1{(a+pb)(a+qb)}\geqslant \frac{4(p+1)(q+1)}{(p+q+2)^2(a+kb)^2}$$
Using this, it is enough to show that (with $\sum$ denoting cyclic sums):
$$\left(\sum ab\right)\cdot \sum \frac1{\left(a+kb \right)^2} \geqslant \frac{9(p+q+2)^2}{4(p+1)^2(q+1)^2} = \frac9{(k+1)^2} \tag{1}$$
We can make this symmetric using the substitution $x = a+kb, y = b+kc, z = c + ka$, to instead prove:
$$\sum \frac1{x^2}\geqslant \frac{9(k^2-k+1)}{(k^2+1)\sum xy - k\sum x^2}$$
For this, it may be expedient to use the so called "uvw" method - let $3u = \sum x, 3v^2=\sum xy, w^3=xyz$, so that $u\geqslant v\geqslant w$.    Note as the RHS is positive, we also have $(k^2+1)\sum xy > k\sum x^2 \implies (k+1)^2v^2>3ku^2$.  Now the inequality is
$$\frac{(3v^2)^2-6uw^3}{w^6}\geqslant \frac{3(k^2-k+1)}{(k+1)^2v^2-3ku^2}$$
$$(k^2-k+1)w^6 +2u[(k+1)^2v^2-3ku^2]w^3-3v^4[(k+1)^2v^2-3ku^2]\leqslant 0$$
As the LHS is a quadratic in $w^3$ with both leading coefficients positive, it is maximised when $w$ attains its maximum, viz. $w=v$, which implies $w=v=u \implies x=y=z \implies a=b=c$, hence it is enough to check inequality $(1)$ for this case, which is obviously true.
