Prove $a^2b+b^2c+c^2a \ge\sqrt{3(a^2+b^2+c^2)}$ if $abc=1$ 
if $a,b,c$ are positive real numbers that $abc=1 $,Prove:$$a^2b+b^2c+c^2a \ge\sqrt{3(a^2+b^2+c^2)}$$
  Additional info: We should only use AM-GM and Cauchy inequalities.

Things I have done so far: for $a^2b+b^2c+c^2a $ minimum I can say: $$a^4b^2+b^2c^3a + b^2c^3a \ge 3 \sqrt{a^6b^6c^6}=3$$
So like this I can conclude that: $$(a^2b+b^2c+c^2a)^2=\sum \limits_{cyc}a^4b^2+\sum \limits_{cyc}b^2c^3a + \sum \limits_{cyc}b^2c^3a \ge 27$$ 
So $a^2b+b^2c+c^2a \ge 3\sqrt 3$. So my idea is to show that $3 \ge \sqrt{a^2 + b^2 + c^2}$. And I stuck here.

Expand $(a^2b+b^2c+c^2a)^2$ and use the fact that $abc=1$ to reduce it and then factorise to get,
$$ a^2(a^2b^2+2c) + b^2(b^2c^2+2a) + c^2(c^2a^2+2b) $$
Using the fact $abc=1$ again,
$$ a^2\left(\frac{1}{c^2}+2c\right) + b^2\left(\frac{1}{a^2}+2a\right) + c^2\left(\frac{1}{b^2}+2b\right) $$
and then it's easy to show that $\frac{1}{x^2}+2x\geq 3$ for all $x>0$.
