<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] We have $a^2 + b^2 + c^2 \geq \frac{(a+b+c)^2}{3} \geq \frac{(abc)^2}{3}$ and $a^2+b^2+c^2 \geq 3 \sqrt[3]{a^2b^2c^2}$ by AM-GM. Take the $1/4$-th power of the first inequality and the $3/4$-th power of the second inequality, and multiply (this is allowed since everything is positive). The result is $a^2+b^2+c^2 \geq \sqrt{3}{abc}$.
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