<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] 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.
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