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 $p,q,r$.
Half year edit! We show that $abc\sum ab\sum a^2\left(\text{left}-\text{right}\right)\ge0$, or
\[\frac1{81}\sum\left(59a^2b-96abc\right)(b-c)^2+\sum a\left(b^2+\frac{bc}9+\frac{20ac}9-c^2-\frac{7ab}3\right)^2\ge0.\]
Best Answer
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.