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. 

Note that from $$(a-1)^2+(b-1)^2+(c-1)^2 \ge 0$$we have $$a^2+b^2+ c^2 \ge 2(a+b+c)-3$$
By AM-GM $$a+b+c \ge 3 \sqrt[3]{abc}=3 \implies a+b+c-3 \ge 0$$
So $$a^2+b^2+ c^2 \ge 2(a+b+c)-3=a+b+c+(a+b+c-3) \ge a+b+c$$
EDIT
Here is a simple proof of AM-GM when $n=3$. We will prove $$\frac{a+b+c}{3} \ge \sqrt[3]{abc}$$
Let $x=\sqrt[3]{a}, y=\sqrt[3]{b}, z=\sqrt[3]{c}$. The problem is equivalent to proving $$x^3+y^3+z^3-3xyz \ge 0$$
From here, we know $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
This can simplify to $$x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)((x-y)^2+(y-z)^2+(z-x)^2) \ge 0$$
So we are done.
