Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$ So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$.
Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$
For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I don't know what to do after that...

Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$a^2b+b^2c+c^2a+abc=a\cdot ab+b\cdot bc+c\cdot ca+abc\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$
$=y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4$. 
