Inequality from Riasat book

Question

Let $a,b,c\geq0$. Prove $$(a+b+c)(a^2+b^2+c^2) \geq 3(a^2b+b^2c+c^2a)$$ I tried to write LHS as $((a+b)^3+(b+c)^3+(c+a)^3)/3)$ but I got nothing interesting.
I'm pretty sure I have to use the fact that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$ but i wouldn't know how to exploit because in the LHS i have symmetric terms but not like RHS and not how to get rid of them

Answer 1

We have $$(a+b+c)(a^2+b^2+c^2) \geq 3(a^2b+b^2c+c^2a) \tag{1}$$ if and only if $$a^3 + b^3 + c^3 + ab^2 + bc^2 + ca^2 \ge 2(a^2 b + b^2 c + c^2 a) \tag{2}$$ Written concisely, the above inequality is $$\sum_{\text{cyc}} a^3 + \sum_{\text{cyc}} ab^2 \ge 2 \sum_{\text{cyc}} a^2b \tag{3}$$ Using the AM-GM inequality, $$a^3 + ab^2 = a(a^2 + b^2) \ge 2a^2b \tag{4}$$ giving $$\sum_{\text{cyc}} a^3 + \sum_{\text{cyc}} ab^2 \ge 2 \sum_{\text{cyc}} a^2b \tag{5}$$ as required.