<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] 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!
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