<s>[INST] $a,b,c \geq 0$,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.
 [/INST] 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 $x\ge0,y\ge0,z\ge0$.
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