AM-GM Inequality. Let a, b, c be positive real numbers. Prove that $ \frac {a+b+c}{3} \cdot (a^2+b^2+c^2) \ge a^2b + b^2c + c^2a$.

Question

Prove that $$ \frac {a+b+c}{3} \cdot \frac{a^2+b^2+c^2}{3} \ge \frac{a^2b + b^2c + c^2a}{3}$$.
given that a,b,c are positive real numbers. Solve only using AM-GM.
So far I have tried expanding LHS, and simplifying to:$$a^3+b^3+c^3+ab^2+bc^2+ca^2 \ge 2(a^2b + b^2c + c^2a)$$ However this has not lead me anywhere. Source: Basics Of Inequality

Answer 1

HINT

(You made an excellent start.)

Apply the AM-GM inequality to $\frac{c^3+ca^2}{2}$