<s>[INST] $a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$ $a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$  
My work:
I tried using Cauchy-Schwarz inequality to find that,
$(a^2+b^2+c^2)(1^2+1^2+1^2)\ge (a+b+c)^2$
$(a^2+b^2+c^2)\ge \dfrac13(a+b+c)^2$
$\sqrt{(a^2+b^2+c^2)}\ge \dfrac{1}{\sqrt{3}}(a+b+c)\ge \dfrac{1}{\sqrt{3}}abc$
which is not what I need and neither I can use it to prove the required inequality. Please help.
 [/INST] here is another way 
$$(a^2+b^2+c^2)^2=\sum\limits_{cyc} a^4 + \sum\limits_{cyc} 2a^2b^2 \ge abc(a+b+c) + 2abc(a+b+c) =3abc(a+b+c)\ge3a^2b^2c^2$$
inequality $\sum\limits_{cyc} a^4 \ge \sum\limits_{cyc} a^2b^2 \ge abc(a+b+c)$
can be proved using AM-GM or Cauchy-Schwarz. 
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