Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$

Question

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) $$\text{LHS-RHS}=c \left( a-b \right) ^{2} \left( a+b-c \right) ^{2}+a \left( b-c \right) ^{2} \left( b+c-a \right) ^{2}+b \left( c-a \right) ^{2} \left( c+a-b \right) ^{2}\geqq 0$$

Answer 1

We write the inequality as $$\frac{(a+b+c)^2}{ab+bc+ca} \geqslant \frac{\displaystyle 4 \sum (a^2b+ab^2)}{(a+b)(b+c)(c+a)},$$ equivalent to $$\frac{a^2+b^2+c^2}{ab+bc+ca}+2 \geqslant \frac{4[(a+b)(b+c)(c+a)-2abc]}{(a+b)(b+c)(c+a)},$$ or $$\frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{8abc}{(a+b)(b+c)(c+a)} \geqslant 2.$$ Which is know.

Answer 2

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:

1. $c=0$;

2. $b=c$, 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: $$\frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{8abc}{(a+b)(b+c)(c+a)} \geqslant 2+\frac{16(a-b)^2(a-c)^2(b-c)^2}{(ab+ac+bc)(a+b+c)^4}.$$