TITLE: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$.
QUESTION [8 upvotes]: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that
  $a^2 + b^2 + c^2 \ge a + b + c$.

I'm supposed to prove this use AM-GM, but can't figure it out. Any hints?

REPLY [2 votes]: Using C-S and then AM-GM gives something shorter:
$$3(a^2+b^2+c^2) \ge (a+b+c)^2 \Rightarrow a^2+b^2+c^2 \ge (a+b+c)\frac{(a+b+c)}{3} $$
$$\ge (a+b+c)\sqrt[3]{abc}=a+b+c$$
And we're done.