Let $a,b,c$ be positive real numbers such that $abc =1$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that $$a^2+b^2+c^2\geq a+b+c$$.
Also, state the condition for equality.
My Attempt, 
$a,b,c$ are real and positive numbers,  then 
$$(a-1)^2+(b-1)^2+(c-1)^2\ge 0$$
$$a^2-2a+1+ b^2-2b+1+c^2-2c+1\ge 0$$
$$a^2+b^2+ c^2-2(a+b+c)+3\ge 0$$.
I have made a start in this way, but I am not sure if this works. Please help me, with any simple and beautiful method. 

By AM-GM $$6(a^2+b^2+c^2)=\sum_{cyc}(4a^2+b^2+c^2)\geq6\sum_{cyc}\sqrt[6]{a^8b^2c^2}=6(a+b+c)$$
and we are done!
