<s>[INST] 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?)
 [/INST] 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, $$1+a^2+b^2+c^2+4abc-a-b-c-ab-ac-bc\ge$$
$$\geq1+c^2+a^2+b^2-ab+ac+bc-a-b-2c\geq$$
$$\geq1+c^2+\frac{1}{4}(a+b)^2+(a+b-2)c-a-b=$$
$$=c^2+(a+b-2)c+\left(\frac{a+b}{2}-1\right)^2=\left(c+\frac{a+b}{2}-1\right)^2\geq0.$$
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