<s>[INST] Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$ From Mr. Michael Rozenberg solution:
For $a,b,c>0$$,$ prove that$:$
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$
I found two SOS proof:
1) $$\text{LHS-RHS}={\frac { \left( a-b \right) ^{2}\cdot \text{M}+ab \left( {a}^{2
}-2\,ab+ca+{b}^{2}+bc-2\,{c}^{2} \right) ^{2}}{a+b}}$$
Where $$\text{M}=\left( 2\,ab-ca-bc+{c}^{2} \right) ^
{2}+c \left( -c+a+b \right) ^{2} \left( a+b \right)$$
2) $$\text{LHS-RHS}=c \left( a-b \right) ^{2} \left( a+b-c \right) ^{2}+a \left( b-c \right) ^{2} \left( b+c-a \right) ^{2}+b \left( c-a \right) ^{2} \left( c+a-b \right) ^{2}\geqq 0$$
 [/INST] I think, the shortest way here it's $uvw$. See here: https://artofproblemsolving.com/community/c6h278791
Indeed, this inequality is a linear inequality of $w^3$ and for the proof it's enough to consider two cases:


*

*$c=0$;

*$b=c$, which very easy to make. 
Due to the Nguyenhuyen_AG's post your inequality we can rewrite in the following form.

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0.$ Prove that:
  $$\frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{8abc}{(a+b)(b+c)(c+a)} \geqslant 2.$$

There is the following stronger inequality.

Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0.$ Prove that:
  $$\frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{8abc}{(a+b)(b+c)(c+a)} \geqslant 2+\frac{16(a-b)^2(a-c)^2(b-c)^2}{(ab+ac+bc)(a+b+c)^4}.$$

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