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?)

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 $\rfloor$
