Proving $\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} + \frac2{a^2+b^2+c^2}\ge\frac{11}{ab+bc+ca}$ with $pqr$ method 
Let $a, b, c > 0$. Prove that
$$\sum\limits_{\mathrm{cyc}}\frac1{ab}+\frac2{\sum\limits_{\mathrm{cyc}} a^2}\ge\frac{11}{\sum\limits_{\mathrm{cyc}} ab}.$$

These problems give us the sense that some part of it is inverted(otherwise it’s a simple $\text{AM-GM}$ problem. In addition I suppose we need $\text{Schure}$ inequality to solve them. Let $\begin{cases}p=a+b+c\\q=ab+bc+ca\\r=abc\end{cases}$. Then $p^2\ge3q$, and $p^3+9r\ge4pq$ (Schur).
Our problem becomes $24qr+p^3q\ge11rp^2+2pq^2$. However this doesn’t seem correct because in the two inequalities above $r$ has never appeared on the right side of $\ge$ but in the problem, it did. So perhaps we need more relationships with $p,q,r$.
 A: Let $p=a+b+c, q = ab + bc + ca, r = abc$.
The desired inequality is written as
$$\frac{p}{r} + \frac{2}{p^2 - 2q} \ge \frac{11}{q}.$$
Using $q^2 \ge 3pr$, it suffices to prove that
$$\frac{3p^2}{q^2} + \frac{2}{p^2 - 2q} \ge \frac{11}{q}$$
or
$$3p^4 + 24q^2 - 17p^2 q \ge 0$$
or
$$(p^2 - 3q)(3p^2 - 8q) \ge 0$$
which is true using $p^2 \ge 3q$.
We are done.
A: Short solution for the reader in hurry: We have to show $(1)$ below, it follows by adding $(2)$ and $(3)$ below.

I will write $e_1$, $e_2$, $e_3$ for the elementary symmetric polynomials in the variables $a,b,c$ (instead of $p,q,r$). (This is because $p$ is in my world always the product $e_3$, and i use $s$ for the sum $e_1$. I tried to adapt to the given notation, no chance in this train.) The inequality to be shown is symmetric w.r.t. the action of the permutation group of the variables $a,b,c$. Usually, each solution will try to use only symmetric expressions to proceed. The solution below will do this, too. Now there is a question of the choice of the weapons, how they should be presented. Shall we work in the world of $a,b,c$ - or in the world of $p,q,r$? If the expressions (to start with) encountered below in the proof and/or the "visualization" of the inequality are/is easily or naturally done in the $p,q,r$ world, then i would take $p,q,r$. However, i could not do so.
We have to show:
$$
\begin{aligned}
0&\le e_1^3e_2 - 11e_1^2 e_3 - 2e_1e_2^2 + 24e_2e_3\ ,\qquad\text{ or equivalently}
\\[2mm]
0&\le \sum a^4b + \sum a^3b^2 -8\sum a^3bc + 4\sum a^2b^2c\ .
\end{aligned}
$$
$$
\tag{$1$}
$$
Here, the following convention is used. The sum $\sum a^4b$ has six terms, we use (all) permutations to generate further monomials.
The sum $\sum a^3b^2$ has also six similar terms.
In the sum $\sum a^3bc$ we do not repeat monomials, so this is explicitly $abc(a^2+b^2+c^2)$, three terms (after expanding).
The final sum $\sum a^3bc$ follows the same convention, no repeating of monomials,
it is $abc(ab+bc+ca)$, three terms. (Convert to $p,q,r$, if this way to think has to be translated.)
We can draw...
Let us "visualize" this inequality in the plane of triples $(i,j,k)$ of the involved weights $i,j,k$, we have total weight five, $i+j+k=5$:
                  WEIGHTS                                    COEFFICIENTS 

                    500                                            .
                410     401                                    1       1
            320     311     302                            1      -8       1
        230     221     212     203                    1       4       4       1
    140     131     122     113     104            1      -8       4      -8       1
050     041     032     023     014     005    .       1       1       1       1       .

The geometry is slightly broken above, if the weights were drawn inside an equilateral triangle, the $-8$ coefficients would be seen inside a hexagon of bounding coefficients $1,1,1,1,4,4$.
Such a visualization is not always leading to a proof, but it shows the complication in the problem, and may be a guide for the progress done in the next steps.
So far we did almost nothing in the direction of a solution... Let's start.

It is natural to build expressions "using squares" that lead to weights $(4,1,0)$, and its permuted cousins. Here is what we need:
$$
\tag{$2$}
0\le \sum (a-b)^2(a-c)^2(b+c) = \sum a^4b -8\sum a^3bc +6\sum a^2b^2c\ .
$$
(If the above would be easier in the $p,q,r$ world, the answer will differ.)
(Drawing the weights and coefficients again, we see the finish immediately:)
It remains to see
$$
\tag{$3$}
0\le \sum a^3b^2 - 2\sum a^2b^2c
$$
to conclude. How to see this? The $2$-coefficients, drawn in the $(i,j,k)$ plane, are incarcerated in the cage of the $1$ coefficients. To see an individual domination for $a^2b^2c$ it is a good idea to search for a line through $(2,2,1)$ with useful coefficients on both sides. We take of course $(4,0,1)$ and $(0,4,1)$, i.e. $a^4c$ and $b^4c$. Although we could also go with $(3,2,0)$ and $(0,2,3)$.
(For this last step, breaking the symmetry makes some sense...)
$\square$
