Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ . 
Given three non-negatve numbers $a, b, c$. Prove that:
$$1+ a^{2}+ b^{2}+ c^{2}+ 4abc\geqq a+ b+ c+ ab+ bc+ ca$$

Let $t= a+ b+ c$, we have to prove
$$\left(\!\frac{1}{t^{3}}- \frac{1}{t^{2}}+ \frac{1}{t}\!\right)\sum a^{3}+ \left(\!\frac{3}{t^{3}}- \frac{3}{t^{2}}\!\right)\left (\!\sum a^{2}b+ \sum a^{2}c\!\right )+ \left(\!\frac{6}{t^{3}}- \frac{6}{t^{2}}- \frac{3}{t}+ 4\!\right)abc\geqq 0$$
If $0< t< 1$ so
$${\rm LHS}\geqq \left(\frac{3}{t}+ 1\right)\left(\frac{3}{t}- 2\right)^{2}abc\geqq 0$$
If $1< t$ so
$${\rm LHS}= \left(\!\frac{3}{t^{2}}- \frac{3}{t^{3}}\!\right)(\!{\rm Schur.3}\!)+ \frac{1}{t}\left(\!\frac{2}{t}- 1\!\right)^{2}(\!{\rm a.m.}- {\rm g.m.}\!)+ \left(\!\frac{3}{t}+ 1\!\right)+ \left(\!\frac{3}{t}- 2\!\right)^{2}abc\geqq 0$$
(Can you find the way without deviding two cases as above?)
 A: Another way.
Since $$\prod_{cyc}(2a-1)^2=\prod_{cyc}((2a-1)(2b-1))\geq0$$ and our inequality is symmetric, we can assume that $$(2a-1)(2b-1)\geq0,$$ which gives
$$c(2a-1)(2b-1)\geq0$$ or
$$4abc\geq 2ac+2bc-c.$$
Thus, $$1+a^2+b^2+c^2+4abc-a-b-c-ab-ac-bc\ge$$
$$\geq1+c^2+a^2+b^2-ab+ac+bc-a-b-2c\geq$$
$$\geq1+c^2+\frac{1}{4}(a+b)^2+(a+b-2)c-a-b=$$
$$=c^2+(a+b-2)c+\left(\frac{a+b}{2}-1\right)^2=\left(c+\frac{a+b}{2}-1\right)^2\geq0.$$
A: It can be characterized as a sum of non-negative polynomials. This decompostion writes it as follow
$$\!1\!+\!a^{2}\!+\!b^{2}\!+\!c^{2}\!+\!4 abc\!-\!a\!-\!b\!-\!c\!-\!ab\!-\!bc\!-\!ca\!=\\=\!abc\left\{\!\left(\!1\!+\!\dfrac{3}{a\!+\!b\!+\!c}\!\right)\left(\!2\!-\!\dfrac{3}{a\!+\!b\!+\!c}\!\right)^{2}\!+\!\dfrac{3\left(\!\sum a^{2}\!-\!\sum ab\!\right)}{(\!a+ b+ c\!)^{3}}\!\right\}\!$$
$$+\sum\limits_{cyc}\left\{\!\left(\!\frac{3}{2}+ 3\times \frac{a}{a+ b+ c}\!\right)\left(\!\frac{1}{2}- \frac{a}{a+ b+ c}\!\right)^{2}
\left(\!b- c\!\right)^{2}\right\}$$
$$+\sum\limits_{cyc}\left\{\left(\!\frac{1}{2}+ \frac{3a}{a+ b+ c}\!\right)\left(\!\frac{1}{2}- \frac{1}{a+ b+ c}\!\right)^{2}\!(\!b- c\!)^{2}\right\}\geqq 0$$
q.e.d / You can also see here $\lceil$ https://h-a-i-d-a-n-g-e-l.hatenablog.com/entry/2019/05/12/145452 $\rfloor$
A: Partial answer:
Let $a = \frac{1}{2} + a'$, $b = \frac{1}{2} + b'$ and $c = \frac{1}{2} + c'$. Then, from Empy2's hint we have:
$$8a'b'c' + (a'+b')^2 + (b'+c')^2 + (c' + a')^2 ≥ 0$$
which is true if $a', b', c' ≥ 0$. This leaves only the cases when any and only one or all three of $a, b, c$ satisfy $0 < a,b,c < \frac{1}{2}$.
A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that
$$1+9u^2-6v^2+4w^3\geq3u+3v^2,$$ which is a linear inequality of $w^3$.
Id est, it's enough to prove our inequality for the extreme value of $w^3$, which happens in the following cases.


*

*$w^3=0$.


Let $c=0$.
Thus, we need to prove that
$$1+a^2+b^2\geq a+b+ab$$ or
$$(a-1)^2+(b-1)^2+(a-b)^2\geq0,$$ which is obvious;


*Two variables are equal.


Let $b=a$.
Thus, we need to prove that:
$$(1+4c)a^2-2(1+c)a+c^2-c+1\geq0$$ for which it's enough to prove that
$$(1+c)^2-(1+4c)(c^2-c+1)\leq0$$ or
$$c(2c-1)^2\geq0$$ and we are done!
