A while ago I conjectured this inequality and its (little) sister on AOPS. Here is another related inequality in the opposite direction which I strongly suspect is true, although I don't have a proof:
Consider three positive real numbers $a$, $b$ and $c$. Prove (or disprove) that the following inequality holds:
$$5+\frac{3(a^2+2b^2+c^2)}{(a+b)(b+c)} \geq \frac{9(a+b)(b+c)(c+a)}{(a+b+c)(ab+bc+ca)}$$
The idea of reducing to a symmetric inequality does not work, that is:
$$\frac{3(a^2+b^2+c^2-ab-bc-ca)}{(a+b)(b+c)} \geq \frac{9(a+b)(b+c)(c+a)-8(a+b+c)(ab+bc+ca)}{(a+b+c)(ab+bc+ca)}$$
and trying to prove
$$\frac{3(a^2+b^2+c^2-ab-bc-ca)}{a^2+b^2+c^2+ab+bc+ca} \geq \frac{9(a+b)(b+c)(c+a)-8(a+b+c)(ab+bc+ca)}{(a+b+c)(ab+bc+ca)}$$
does not work because this inequality is false at $c = 0$, but the proposed inequality is true at $c = 0$.
How can we prove this inequality?
The inequality $$\frac{3(a^2+b^2+c^2-ab-bc-ca)}{a^2+b^2+c^2+ab+bc+ca} \geq \frac{9(a+b)(b+c)(c+a)-8(a+b+c)(ab+bc+ca)}{(a+b+c)(ab+bc+ca)}$$ is true!
It's just $$\sum_{cyc}(2a^4b+2a^4c-2a^3b^2-2a^3c^2+7a^3bc-7a^2b^2c)\geq0,$$ which is true by Muirhead.