$a,b,c \geq 0$,prove that $a^2+b^2+c^2+abc+5 \geq3(a+b+c) $ 
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
  $$a^2+b^2+c^2+abc+5 \geq3(a+b+c).$$

I'm certain that this problem could be solved by using dirchlet's theory.but I do not know how to apply it exactly.

Since $(a_1)^2(b-1)^2(c-1)^2\geq0$, we can assume that $(a-1)(b-1)\geq0$ or
$c(a-1)(b-1)\geq0$ or $abc\geq ac+bc-c$.
Thus, it remains to prove that
$$a^2+b^2+c^2+ac+bc-c+5\geq3(a+b+c)$$ or
$$c^2+(a+b-4)c+a^2+b^2-3a-3b+5\geq0,$$
for which it's enough to prove that
$$(a+b-4)^2-4(a^2+b^2-3a-3b+5)\leq0$$ or
$$(a-b)^2+2(a-1)^2+2(b-1)^2\geq0.$$
Done!
