Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $a^2 + b^2 + c^2 \ge a + b + c$. 
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?

Solution from Thomas Mildorf notes
first we homogenize inequality with multipliying $(abc)^\frac{1}{3}$ and we get the following inequality $$a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{4}{3}c^\frac{1}{3}+a^\frac{1}{3}b^\frac{1}{3}c^\frac{4}{3}\leq a^2+b^2+c^2$$ then we can apply $A.O\geq H.O$ $$\frac{a^2+a^2+a^2+a^2+b^2+c^2}{6}\geq a^\frac{4}{3}b^\frac{1}{3}c^\frac{1}{3}$$ and we can get in the same manner other term and after adding them we are done
