Let $a=b=c=\frac{1}{3}$. Hence, $M\geq2$. Thus, it remains to show that $\left(\frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}\right)(a-bc)(b-ac)(c-ab)\le2abc$, which is true, but my proof is still ugly: it's obviously true after homogenization and full expanding (Muirhead only).

The problem is from Silk road mathematical competition 2012 (hold in Kazakhstan).Not a big competition . Only 2,3 countries take part.

i don't understand! This problem was posted 5 or 6 times(with the solutions and ideas). Here, my solution but last time: Let $a=\tan \frac{B}{2} \tan \frac{C}{2} $, ..., where $A,B,C$ are angles of a triangle $ABC$. Now, we need to prove \[ \Bigl( \sum_{cyc} \frac{1}{\sin^2 \frac{C}{2} \cos^2 \frac{C}{2}} \Bigr) \cos A \cos B \cos C \le 2. \] It's clear! We can use by $ \cos A \cos B \le \sin^2 \frac{C}{2} $ ... and now \[ LHS\le \sum_{cyc} \frac{\cos C}{\cos^2 \frac{C}{2}}\le 2 . \]

We claim that the answer is $M=2.$ By letting $a = b = c= \frac13,$ we obtain that $M \geq 2.$ We will now show that $M = 2$ actually works, that is: $$2abc \geq (\sum_{cyc} \frac{1}{(b+c)^2})(a-bc)(b-ca)(c-ab),$$ whenever $a, b, c \in (0, 1)$ with $a+b+c = 1.$ Observe that $\frac{1}{(b+c)^2} \leq \frac{1}{4bc},$ and so $\sum_{cyc} \frac{1}{(b+c)^2} \leq \frac{1}{4abc},$ where we used $a+b+c = 1.$ Therefore, we now want that: $$8a^2b^2c^2 \geq (a-bc)(b-ca)(c-ab),$$which upon rearranging and dividing $abc$ is equivalent to: $$9abc + \frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b} \leq 1 + a^2 + b^2 + c^2.$$ Rearranging and factoring, the above is equivalent to: $$\prod_{cyc} (\frac{a}{bc} - 1) \leq 8 \Leftrightarrow \prod_{cyc} (a^2+ab+ac-bc) \leq 8a^2b^2c^2.$$ Observe that $a^2 + ab + ac - bc = 2a^2 - (a-b)(a-c).$ Let's assume WLOG that $a \geq b \geq c.$ Then, we know that $c^2 + ca + cb - ab \leq 2c^2,$ and so it suffices to show that $(a^2+ab+ac-bc)(b^2+ba+bc-ac) \leq 4a^2b^2.$ Notice that every term is positive, with the exception of possibly $c^2 + ca + cb -ab.$ Therefore, we can assume WLOG that $c^2 + ca + cb - ab \geq 0 \Leftrightarrow (a-c)(b-c) \leq 2c^2.$ Let $x = a-c, y = b-c,$ so that $xy \leq 2c^2.$ Then, the claim is equivalent to: $$((c+x)^2 + (c+x)(c+y) + c(c+x) - c(c+y)) ((c+y)^2 + (c+x)(c+y) + c(c+y) - c(c+x)) \geq 0 \Leftrightarrow (2c^2 - xy)(x-y)^2 \geq 0,$$after expansion, and so we're done. $\square$

Pathological wrote: $$\prod_{cyc} (a^2+ab+ac-bc) \leq 8a^2b^2c^2.$$ https://artofproblemsolving.com/community/c6h573307p3374249

For all $a,b,c\in R^+ $ such that $a+b+c=1$ and $$ ( \frac{1}{(a+b)^2}+\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2} )(a-bc)(b-ac)(c-ab)\le M \cdot abc.$$Find min $M$.