$a,b,c \geq 0$,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.
 A: Since $a^2+b^2+c^2+abc+5 \geq3(a+b+c) \iff (a^2-3a)+(b^2-3b)+(c^2-3c)+abc\ge-5 \iff$
$\;\;\;\;\;(a-\frac{3}{2})^2+(b-\frac{3}{2})^2+(c-\frac{3}{2})^2+abc\ge\frac{7}{4},$
it suffices to show that $f(x,y,z)=(x-\frac{3}{2})^2+(y-\frac{3}{2})^2+(z-\frac{3}{2})^2+xyz$ has 
a minimum value of $\frac{7}{4}$
 for $x\ge0, y\ge0, z\ge0.$
$\textbf{1)}$ If we consider the values of $f$ on the cube defined by $0\le x\le3, 0\le y\le3, 0\le z\le3$, 
$f(x,y,z)\ge\frac{9}{4}$ on the boundary of the cube since $x=0$ or $x=3\implies f(x,y,z)\ge\frac{9}{4}$, 
and similarly for the other 4 faces of the cube.
$\textbf{2)}$ At any critical point $(x,y,z)$ in the interior of the cube,
$f_x=2x-3+yz=0$, $f_y=2y-3+xz=0$, and $f_z=2z-3+xy=0$, so
$\;\;\;\;2x+yz=3, \;\;\;2y+xz=3, \;\;\;2z+xy=3$.
Subtracting the 2nd equation from the first gives $(x-y)(2-z)=0$, so either $x=y$ or $z=2$.  However, $z=2\implies4+xy=3\implies xy=-1$, which is not possible since $x>0, y>0$; so $x=y$.
Similarly, subtracting the 3rd equation from the 2nd gives $(y-z)(2-x)=0$, so either $y=z$ or $x=2$.  As above, $x=2\implies4+yz=3\implies yz=-1$, which is not possible for $y>0,z>0$, so $y=z$.
Substituting into the 1st equation gives $2x+x^2=3$, so $x^2+2x-3=0\implies(x+3)(x-1)=0\implies x=1$
since $x>0$; 
so $(1,1,1)$ is the only critical point in the interior of the cube.
Since $f(1,1,1)=\frac{7}{4}$, f has a minimum value of $\frac{7}{4}$ in the cube.
$\textbf{3)}$ Since $f(x,y,z)>\frac{9}{4}$ if $x>3$ or $y>3$ or $z>3$,
it follows that f has a minimum value of $\frac{7}{4}$ for $x\ge0,y\ge0,z\ge0$.
A: I am not sure if you can apply  "dirchlet's theory". But another way is:
for $f(x)=px^2+qx+r, \Delta=q^2-4pr$, if  $p>0 \cap\Delta \le 0 \implies f(x) \ge 0$
WLOG,let $ a \ge b \ge c$
$a^2+b^2+c^2+abc+5 \geq3(a+b+c) \iff a^2+(bc-3)a+b^2+c^2+5-3(b+c) \ge 0 \iff \Delta_a =(bc-3)^2-4(b^2+c^2+5-3(b+c)) \le 0 \iff (4-c^2)b^2+(6c-12)b+4c^2-12c+11 \ge 0 \iff 4-c^2>0 \cap \Delta_b=(6c-12)^2-4(4-c^2)(4c^2-12c+11) \le 0 \iff 4-c^2>0 \cap (c-1)^2(c-2)(c+1) \le 0 $
which is true. the "=" will hold $(c=1, \Delta_b=0) \cap (b=1,\Delta_a=0) \cap (a-1)^2=0  \implies a=b=c=1$
it remains $c^2 \ge 4$ or $c\ge 2$ case:
$abc\ge c^2a\ge4a ,abc\ge c^2b \ge 4b ,abc\ge 4c \implies abc\ge \dfrac{4(a+b+c)}{3} \implies a^2+b^2+c^2+abc+5 \geq3(a+b+c) \iff a^2+b^2+c^2+5 \ge \dfrac{5(a+b+c)}{3} \iff (a^2+ \dfrac{25}{36} \ge \dfrac{5a}{3} )\cap( b^2+ \dfrac{25}{36}\ge \dfrac{5b}{3}) \cap  (c^2+ \dfrac{25}{36}\ge \dfrac{5c}{3}) \cap (5-3\times\dfrac{25}{36}=\dfrac{35}{12}>0)$
which means the inequality is always true and only ">" is  hold.
A: 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!
