Fun Inequality to work on \[ \left( \sum_{cyc} (a-b)^2 \right)^2 \le \left( \sum_{cyc} (a+b)^2 - 12abc \right)^2 \]\[ \left( \sum_{cyc} (a-b)^2 \right)^2 \le \left( \sum_{cyc} (a+b)^2 \right)^2 + 144a^2 b^2 c^2 - 24abc \left( \sum_{cyc} (a+b)^2 \right) \]\[ 3abc \left( \sum_{cyc} (a+b)^2 - 6abc \right) \le \left( \sum_{cyc} (a^2 + b^2) \right) \left( \sum_{cyc} ab \right) \]Homogenize, \[ 3abc \left( 2 \sum_{cyc} a \right) \le \left( \sum_{cyc} (a^2 + b^2) \right) \left( \sum_{cyc} ab \right) \]\[ 3abc(a + b + c) \le (a^2 +b^2 + c^2)(ab + ac + bc) \] Obviously true by Muirhead Inequality.

We have that $\sum_{cyc}a^2+\sum_{cyc}ab=\sum_{cyc}a+3abc$. We split in $2$ cases if $a+b+c\geq 3abc$,we need to prove $\sum_{cyc}a^2-\sum_{cyc}ab\leq\sum_{cyc}a-3abc$ and using the given relation we have to prove $ab+bc+ca\geq 3abc$ wich is true since $(ab+bc+ca)^2\geq 3abc(a+b+c)\geq 9(abc)^2$ if $a+b+c\leq 3abc$,we need to prove $\sum_{cyc}a^2-\sum_{cyc}ab\leq -\sum_{cyc}a+3abc$ again using the given realtion we have to prove $ab+bc+ca\geq a+b+c$ wich is true since $(ab+bc+ca)^2\geq 3abc(a+b+c)\geq (a+b+c)^2$ $\blacksquare$

I guess the easiest solution: The given condition is $$\sum a^2+\sum ab=\sum a+3abc$$We claim that $$\sum a^2\leq max(\sum a,3abc)$$. Assume to the contrary: Thus $$\sum ab<\sum a$$and $$\sum ab<3abc$$Therefore $$\sum ab<3abc\leq \frac{(\sum ab)^2}{\sum a}$$Therefore $\sum ab>\sum a$,a contradiction to our assumption. Therefore $$\sum (a+b)^2+\sum (a-b)^2=4\sum a^2\leq 4 max(\sum a,3abc)$$,thus implying the problem.

The given condition is $a^2+b^2+c^2+ab+bc+ca=a+b+c+3abc$. We will use the fact that $(ab+bc+ca)^2 \ge 3abc(a+b+c)$ (which follows from AM-GM). If $a+b+c\ge 3abc$, then the desired result is equivalent to $ab+bc+ca \ge 3abc$, which is true by the above fact. The other case is similar.

A pretty useful trick to tackle on absolute value is to square it. Notice that we are given with: $\sum_{\text{cyc}}{a^2}+\sum_{\text{cyc}}{ab}=\frac{1}{2}(\sum_{\text{cyc}}{(a+b)^2})=\sum_{\text{cyc}}{a}+3abc$ $\Longleftrightarrow (\sum_{\text{cyc}}{a^2})^2+2(\sum_{\text{cyc}}{a^2})(\sum_{\text{cyc}}{ab})+(\sum_{\text{cyc}}{ab})^2=(\sum_{\text{cyc}}{a})^2+2(\sum_{\text{cyc}}{a})(3abc)+(3abc)^2$...................(1) and we want to prove that: $\sum_{\text{cyc}}{a^2}-\sum_{\text{cyc}}{ab}=\frac{1}{2}(\sum_{\text{cyc}}{(a-b)^2})=\sum_{\text{cyc}}{a}+3abc$ $\Longleftrightarrow (\sum_{\text{cyc}}{a^2})^2-2(\sum_{\text{cyc}}{a^2})(\sum_{\text{cyc}}{ab})+(\sum_{\text{cyc}}{ab})^2=(\sum_{\text{cyc}}{a})^2-2(\sum_{\text{cyc}}{a})(3abc)+(3abc)^2$..................(2) Comparing (1) and (2), we only need to show $(\sum_{\text{cyc}}{a^2})(\sum_{\text{cyc}}{ab}) \ge (\sum_{\text{cyc}}{a})(3abc)$ after expansion, the rest is trivial by AM-GM or Murihead

Just use this $(ab+bc+ca)^2 \ge 3abc(a+b+c)$ $a^2+b^2+c^2 \ge ab+bc+ca$

Dadgarnia wrote: $a,b$ and $c$ are positive real numbers so that $\sum_{\text{cyc}} (a+b)^2=2\sum_{\text{cyc}} a +6abc$. Prove that $$\sum_{\text{cyc}} (a-b)^2\leq\left|2\sum_{\text{cyc}} a -6abc\right|.$$ Proposed by Shayan Talaei